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Circle equations on the SAT test

Studying circle equations

On the SAT test circle equations topic is the last topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Circle equations topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

Circle equations- summary

A standard form of a circle contains 2 squares of binomials that are equal to a squared radius, so that a circle with a center (h,k) and a radius r has a standard form equation of (x-h)2+(y-k)2=r2.

To find the location of the circle in the xy plane, we need to draw it according to its standard equation.

If we open brackets of the standard form equation, we get the expended form: x2-2hx+y2-2ky+(h2+k2-r2)=0, in which there are 3 constants h2, k2 and -r2.

Rewriting the expended equation to a standard form: We can identify the first 4 components of the equation since each one of them has a different type of variable (x2,x,y2 and y) but the 3 constants will be given by a single number. We can find the 3 constants by completing 2 quadratic equations of x and y (finding k and h) with square of sum and square of difference formulas.

A standard form equation of a circle in the xy plane

A standard form of a circle contains 2 squares of binomials that are equal to a squared radius, so that a circle with a center (h,k) and a radius r has a standard form equation of (x-h)2+(y-k)2=r2.

For example: If the center of the circle is located at the point (2,5) and the radius of the circle is equal to 3 then h=a and k=5 the standard equation will be (x-2)2+(y-5)2=32.

The figure below represents a circle in the xy plane and its standard equation.

The center of the circle O has the coordinates of (h,k) and its standard equation is (x-h)2+(y-k)2=r2.

Drawing a circle in the xy plane using its standard form equation

We can be asked about the location of the circle, like in which quadrant the circle is located. To find the location of the circle in the xy plane, we need to draw it according to its standard equation.

Finding the location of the circle in the xy plane steps:
Step 1: Draw the axes and write the numbers of the quadrants in the xy plane.
Remember that the axes divide the coordinate plane into 4 quadrants, the first quadrant is the top right quadrant (for positive x values and positive y values) and moving counterclockwise. The quadrants are names using the Roman numbers I (+,+), II (-,+) , III (-,-) and IX(+,-).

Step 2: Draw the center of the circle (the center coordinates are h and k values from the equation).

Step 3: Draw the circle by adding the value of the radius around the center (the radius value is written in the equation).

Step 4: See in which quadrant the circle is located.

Consider the following example:

In which quadrant is located the circle that has an equation (x+5)2+(y+5)2=22?

The center of the circle coordinates are (-5,-5) and its radius is equal to 2.

The figure below presents the circle in the xy plane.

The circle is in the third quadrant.

An expanded equation of a circle

If we open brackets of the standard form equation, we get the expended form:
(x-h)2+(y-k)2=r2
x2+h2-2hx+y2+k2-2ky-r2=0

Note that there are 3 constants in the equations which are h2, k2 and -r2, therefore, we get the equation
x2-2hx+y2-2ky+(h2+k2-r2)=0

Finding a standard form equation of a circle from its expanded form

To find the center and the radius of the circle, we need to rewrite the expended equation to a standard form.

We can identify the first 4 components of the equation since each one of them has a different type of variable (x2,x,y2 and y) but the 3 constants will be given by a single number. We can find the 3 constants by completing 2 quadratic equations of x and y (finding k and h) with square of sum and square of difference formulas.

We know that the square of sum formula is (a+b)2=a2+b2+2ab.

We know that the square of difference formula is (a-b)2=a2+b2-2ab.

Consider the following example:

We are given the expression x2-6x.

What constant c should be added to this expression, so that we get a quadratic binomial in a form of (x-c)2?

We know that the square of difference formula is (a-b)2=a2+b2-2ab and we are given that a2=x2 and -2ab=-6x

a2=x2

x=a, therefore the first number is x

-2ab=-6x

We found that x=a therefore -2xb=-6x.

We can divide by -2x getting b=-6x/-2x.

b=3, therefore the second number is 3 and the quadratic binomial is (x-3)2

b2=9, therefore the constant that is added to the expression x2-6x to complete a quadratic binomial is 9.

(x-3)2=x2-6x+9

Consider the following example:

We are given the expression 2x2+20x,

What constant c should be added to this expression, so that we get a quadratic binomial in a form of (x-c)2?

First, we should factor out the number 2 to make the coefficient of x2 equal 1, so that the expression will be 2(x2+10x).

We know that the square of sum formula is (a+b)2=a2+b2+2ab and we are given that a2=x2 and 2ab=10x

a2=x2

a=x, therefore the first number is x

2ab=10x

We found that x=a, therefore 2bx=10x

b=5, therefore the second number is 5 and the quadratic binomial is 2(x+5)2

b2=25, therefore the constant that is added to the expression 2x2+20x to complete a quadratic binomial is 25*2=50.

2(x+5)2=2(x2+10x+25)

2(x+5)2=2x2+20x+50

Consider the following example:

What is the center and the radius of the circle equation x2-2x+y2-6y+1=0?

The circle standard equation is (x-h)2+(y-k)2=r2, where the center of the circle is (h,k) and the radius is r.

The given constant of 1 combines 3 constants so that c12+c22-c32=1
c1 for completing the expression x2-2x to quadratic binomial
c2 for completing the expression y2-6y to quadratic binomial
-c3 that is equal to the radius

Note that the sign of c3 is negative since the radius is transferred from the second side of the equation.

We need to complete x2-2x to x2-2x+c12=(x-c1)2 getting x2-2x+12=(x-1)2, therefore c1=1 and c12=1.

We need to complete y2-6y to y2-6y+c22=(x-c2)2 getting y2-6y+32=(y-3)2, therefore c2=3 and c22=9.

We know that c12+c22-c32=1, therefore 1+9-c32=1 and c32=9, c3=3.

x2-2x+1  +  y2-6y+9    +1-10  =0

(x-1)2          (y-3)2          -9     =0

(x-1)2          (y-3)2                  =32

The standard equation is (x-1)2+(y-3)2=32, therefore the center of the circle is (1,3) and the radius of the circle is 3.

(x-1)2+(y-3)2=32

x2+1-2x+y2+9-6y-9=0

x2-2x+y2-6y+1=0

Consider the following example:

What is the center and the radius of the circle equation 4x2-4x+9y2-12y+1=0?

4x2-4x+1  +  9y2-12y+9  +1-10  =0

(2x-1)2           (3y-3)2        -9     =0

(2x-1)2           (3y-3)2                =32

The standard equation is (2x-1)2+(3y-3)2=32, therefore the center of the circle is (1,3) and the radius of the circle is 3.

You just finished studying circle equations topic, the last topic of additional topics in math!

Angles, arc lengths and trig functions on the SAT test

Studying angles, arc lengths and trig functions

On the SAT test Angles, arc lengths and trig functions topic is the sixth topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Before learning arc lengths and trig functions topic learn the topics circle theorems and right triangle trigonometry (from additional topics in math).

Arc lengths and trig functions topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

Angles, arc lengths and trig functions- summary

Angles, arc length and trig functions topic includes 3 parts:
Calculation of arc lengths and sector areas in radians.
Calculation of sine, cosine and tangent in radians.

A radian is defined as the angle subtended from the center of a circle which intercepts an arc equal in length to the radius of the circle. To find the number of radians in a circle, we need to divide the length of the circumference by the radius: 2πr/r=2π, meaning that the number of radians of arc in a circle is 2π.

The relationship between radian and degree measures:
2π radians is equal to 360 degrees, meaning that one radian is equal to 360/2π=180/π=180/3.14≈57 degrees.

_________________        _________________
π                                 180°

The relationship between central angle in radians, arc length and sector area:
central angle    =         arc length              =     sector area
_____________       ____________________         ____________
2π                   circle circumference           circle area

Special right triangles in circles:
In these questions we are given a circle which center is located at the axis intersection point (0,0).

Special right triangles are right triangles whose sides are in a particular ratio.
Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.

The radian measures of angles of special right triangles are:

We can draw a right triangle from any point on the circle so that the hypotenuse is equal to the radius of the circle and the sides of the triangle are equal to x and y coordinates of the point.

We can check the ratios between the sides of the triangle to see if the triangles are special right triangles. If the triangles are special right triangles, we know the values of their angles.

Calculating trigonometric functions (sin, cos and tan) with radian angle measures in right triangles in circles:
We calculate trigonometric functions under the assumption of unit circle, meaning that the radius is equal to 1. Since the hypotenuse is equal to the radius, we know that the hypotenuse is equal to 1.

The trigonometric functions values are:
sin(A)=opposite/hypotenuse= opposite/1=opposite.

The relationship between radian and degree measures

Degrees in a circle: The number of degrees of arc in a circle is 360.

Radians in a circle: To find the number of radians in a circle, we need to divide the length of the circumference by the radius: 2πr/r=2π, meaning that the number of radians of arc in a circle is 2π.

Note that the underlined sentences above are provided at the beginning of each SAT math section.

Therefore, the connection is that 2π radians is equal to 360 degrees, meaning that one radian is equal to 360/2π=180/π=180/3.14≈57 degrees.

The figure below presents the connection between radian and degree measures.

A radian is defined as the angle subtended from the center of a circle (marked in red) which intercepts an arc equal in length to the radius of the circle (the radii and the equal arc are marked in blue).

Since 2π radians is equal to 360 degrees, we can calculate radian measure given degree measure or calculate degree measure given radian measure using the following ratios:

________________        _________________
2π                               360°

We can simplify the proportion getting:

_________________        _________________
π                                 180°

Consider the following example:

________________         ________________
π                                 180°

Represent radian measure by the variable x.

x    =    100
__         ____
π          180

180x=100π
x=100/180π
100 degrees are equal to 0.55π radians.

Consider the following example:

________________         ________________
π                                 180°

Represent degree measure by the variable x.

3.5   =    x
___        ___
π          180

πx=180*3.5
x=180*3.5/π
x=200°
3.5 radians are equal to 200 degrees.

We can also calculate without using the proportion using the fact that 1 radian is equal to 57 degrees:
3.5*57=200

Calculating angles, arc length and sector areas with radians

We can measure arc length and sector areas with radians instead of degrees.

We know that the relationship between central angle in degrees, arc length and sector area is given by the following ratios:

central angle    =            arc length              =     sector area
_____________        _____________________          ____________
360°                  circle circumference            circle area

We also know that the number of radians of arc in a circle is 2π, therefore we can substitute 360 degrees by 2π:

The relationship between central angle in radians, arc length and sector area is given by the following ratios:

central angle    =         arc length              =     sector area
_____________       ____________________         ____________
2π                circle circumference           circle area

Consider the following example:

The central angle of a circle is equal to 0.5π, the circumference of the circle is equal to 10 centimeters.

What is the measure of the arc formed by this angle?

What is the measure of the sector area formed by this angle?

Calculating the arc length:

central angle    =         arc length
_____________        ______________________
2π                    circle circumference

Represent arc length by the variable x and plug the given data into the ratios equation:

0.5 π  =   x
____     ___
2π        10

x=0.5*10/2
x=2.5
The arc length is 2.5 centimeters.

0.5 π  = 2.5
____     ___
2π        10

0.5/2=2.5/10
1/4=1/4

Calculating the sector area:
The circle area formula is A=πr2.
A=π*102
A=100π
A=100*3.14=314

central angle    =    sector area
_____________        __________
2π                      circle area

Represent sector area by the variable x and plug the given data into the ratios equation:

0.5π  =     x
____      _____
2π         100π

x=0.5*100π/2
x=25π
The sector area is 25π.

0.5π  =   25π
____      _____
2π         100π

0.5/2=25/100
1/4=1/4

Special right triangles in circles

In these questions we are given a circle which center is located at the axis intersection point (0,0)

We can draw a right triangle from any point on the circle so that the hypotenuse is equal to the radius of the circle and the sides of the triangle are equal to x and y coordinates of the point.

We can check the ratios between the sides of the triangle to see if the triangles are special right triangles. If the triangles are special right triangles, we know the values of their angles.

Special right triangles measures

Special right triangles are right triangles whose sides are in a particular ratio.
Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.
Special right triangles with their side sizes length are given at the beginning of each SAT section.

30°, 60°, 90° triangle:
In a 30°, 60°, 90° right triangle the side opposite the 30° angle is half the length of the hypotenuse and the side opposite to 60° angle is equal to the length of a side opposite to 30° angle multiplied by √3.
In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.

45°, 45°, 90° triangle:
In a 45°, 45°, 90° right triangle the sides opposite the 45° angles are equal and the hypotenuse is equal to the side opposite to 45° angle multiplied by √2.
In 45°, 45°, 90° triangle the sides are s, s and s√2.

The following graphs present the special right triangles with the side sizes length.

Radian measures of angles of special right triangles

Since 2π radians is equal to 360 degrees we get:

________________         ________________
π                                180°

x   =  30
__     ___
π      180

x=30π/180
x=π/6

Since 2π radians is equal to 360 degrees we get:

________________         ________________
π                                 180°

Representing radian measure by the variable x and plugging the degree measure into the proportion:

x  = 45
__   ___
π    180

x=45π/180
x=π/4

Since 30*1.5=45 we can also multiply the radian measure of 30° angle by 1.5 getting:
1.5*π/6= π/4.

Since 2π radians is equal to 360 degrees we get:

________________         ________________
π                                 180°

Representing radian measure by the variable x and plugging the degree measure into the proportion:

x  = 60
__   ___
π    180

x=60π/180
x=π/3

Since 30*2=60 we can also multiply the radian measure of 30° angle by 2 getting:
2*π/6= π/3.

Since 2π radians is equal to 360 degrees we get:

________________         ________________
π                               180°

Representing radian measure by the variable x and plugging the degree measure into the proportion:

x  = 90
__   ___
π    180

x=90π/180
x=π/2

Since 30*3=90 we can also multiply the radian measure of 30° angle by 3 getting:
3*π/6= π/2.

The following figures present special right triangles with their side lengths and angles in degrees and radians (the radian measures are marked in red).

Calculating side lengths and radian angle measures in special right triangles in circles

We are given a circle which center is located at the axis intersection point (0,0)

We can draw a right triangle from any point on the circle so that the hypotenuse is equal to the radius of the circle and the sides of the triangle are equal to x and y coordinates of the point.

We can check the ratios between the sides of the triangle to see if the triangles are special right triangles. If the triangles are special right triangles, we know the values of their angles.

Consider the following example:

The points coordinates are A(3,3) and b(-4,4/√3). Both points are located on a circle and the center of the circle is located at the axis intersection point (0,0).

What is the size of the angle BOD?

What is the size of the angle AOC?

The figure above presents 2 points A and B that are located on a circle. The center of the circle is located at the axis intersection point (0,0). The coordinates of point A are (3,3) and the coordinates of point B are (-4,4/√3).

Drawing right triangles from points on a circle and calculating side length from the coordinates of the points:

The lines AO and BO are radii of the circle.

The line AC in drawn from point A to create a right triangle ACO, so that the angle ACO is equal to 90°. Since the angle ACO is equal to 90°, the length of the side AC is equal to y coordinate of point A so that AC=3. In addition, the length of the side CO is equal to x coordinate of point A so that CO=3.

The line BD in drawn from point A to create a right triangle BDO, so that the angle BDO is equal to 90°. Since the angle BDO is equal to 90°, the length of the side BD is equal to the y coordinate of point B so that BD=4/√3. In addition, the length of the side DO is equal to the absolute value of the x coordinate of point B so that DO=|-4|=4.

Calculating the angles of special right triangles:

In the triangle ACO the side lengths are AO=CO=3, therefore the triangle ACO is an isosceles triangle. An isosceles triangle is a special right triangle and we know its angles measures are 45°,45° and 90° and the radian measures are π/4, π/4 and π/2.

Note that we can also calculate the angle measures: Since the triangle ACO is an isosceles triangle, the angles CAO and AOC are equal. Since the angle ACO is equal to 90° and the sum of the angles in a triangle is 180°, the angles CAO and AOC are equal to 45°.

In the triangle BDO the side lengths are BD=4/√3 and DO=4, therefore the triangle ACO is a special right triangle and its angles measures are 30°,60° and 90° and the radian measures are π/6, π/3 and π/2. Therefore, the angle BOD=30°= π/6 and the angle DBO=60°=π/3.

Calculating trigonometric functions (sin, cos and tan) with radian angle measures in right triangles in circles

Special right triangles with their side sizes length are given at the beginning of each SAT section.

The following figure presents 2 special right triangles and their angles in degrees, like given in the SAT (note that the radian measures are not given).

We calculate trigonometric functions under the assumption of unit circle, meaning that the radius is equal to 1. Since the hypotenuse is equal to the radius, we know that the hypotenuse is equal to 1.

Remember the trigonometric functions values:
sin(A)=opposite/hypotenuse= opposite/1=opposite.

Calculating the radian angles measures of triangles

At the beginning of each SAT math section, it is given that:
The number of degrees of arc in a circle is 360.
The number of radians of arc in a circle is 2π.

Therefore, we know that 360°=2π and π=180°.
The angle of 30°: 30=360/12=2π/12=π/6.
The angle of 45°: 45=360/8=2π/8=π/4.
The angle of 60°: 60=360/6=2π/6=π/3.
The angle of 90°: 90=360/4=2π/4=π/2.
The angle of 120°: 120=360/3=2π/3.
The angle of 135°: 135=360*3/8=2π*3/8=3π/4.
The angle of 180°: 180=360/2=2π/2=π.

Special right triangle with 30°, 60°, 90° angles

Calculating the side lengths of the triangle:
We know the side lengths from the beginning of each math SAT section: x, 2x and x√3. Since the hypotenuse is equal to 1 (unit circle) we know that 2x=1 and x=1/2. Therefore, the sides are 1/2, √3/2 and 1.

Calculating the sine, cosine and tangent of the angle of π/6 radians:
We need to look at the given 30°, 60°, 90° special right triangle, we saw that its sides are 1/2, √3/2 and 1.
sin (π/6)=opposite/1=(1/2)/1=1/2.
tan (π/6)=opposite/adjacent=(1/2)/(√3/2)=1/√3 multiply by √3/√3 getting (1*√3)/(√3*√3)= √3/√9=√3/3.

Calculating the sine, cosine and tangent of the angle of π/3 radians:
We need to look at the given 30°, 60°, 90° special right triangle, we saw that its sides are 1/2, √3/2 and 1.
sin(π/3)=opposite/1=√3/2.

Special right triangle with 45°, 45°, 90° angles

Calculating the side lengths of the triangle:

We know the side lengths from the beginning of each math SAT section: s, s and s√2. Since the hypotenuse is equal to 1 (unit circle) we know that s√2=1 and s=1/√2. Therefore, the sides are 1/2, 1/2 and 1.

Calculating the sine, cosine and tangent of the angle of π/4 radians:
We need to look at the given 45°, 45°, 90° special right triangle.
sin(π/4)=opposite/1=1/√2 multiplying by √2/√2 getting (1*√2)/( √2*√2)=√2/2.
cos(π/4)=adjacent/1=1/√2 multiplying by √2/√2 getting (1*√2)/( √2*√2)=√2/2.

Calculating the sine, cosine and tangent of the angle of 0 radians:
sin(0)=opposite/1: If the angle is close to 0, the side that is opposite to the angle is also close to 0, therefore sin(0)=0/1=0.
cos(0)=adjacent/1= If the angle is close to 0, the side that is adjacent to the angle is almost equal to the hypotenuse which is equal to 1, therefore cos(0)=1/1=1.
tan(0)=opposite/adjacent= If the angle is close to 0, the side that is opposite to the angle is also close to 0 and the side that is adjacent to the angle is equal to the hypotenuse which is equal to 1 therefore tan(0)=0/1=0.

Calculating the sine, cosine and tangent of the angle of π/2 radians:
We know that 2π=360°, therefore π/2=90°.
sin(π/2)=opposite/1=If the angle is close to 90°, the side that is opposite to the angle is almost equal to the hypotenuse which is equal to 1, therefore sin(π/2)=1/1=1.
cos(π/2)=adjacent/1= If the angle is close to 90°, the side that is adjacent to the angle is close to 0, therefore cos(π/2)=0/1=0.
tan(π/2)=opposite/adjacent= If the angle is close to 90°, the side that is opposite to the angle is 1 and the side that is adjacent to the angle is 0. We can’t divide by 0, therefore tan(π/2) is not defined.

Angle measures bigger than 90°:

We can convert these angles to angles smaller than 90° using 2 formulas:
We also know that tan(α)=sin(α)/cos(α)

Calculating the sine, cosine and tangent of the angle of 2π/3 radians:
We know that 2π=360°, therefore 2π/3=360°/3=120°>90°.
sin(2π/3)= sin(π-2π/3)=sin{(3π-2π)/3}=sin(π/3), we found that sin(π/3)=√3/2.
cos(2π/3)=-cos(π-2π/3)= -cos{(3π-2π)/3}=-cos(π/3), we found that cos(π/3)=1/2, therefore -cos(π/3)=-1/2.
tan(2π/3)= sin(2π/3)/cos(2π/3)=(√3/2)/(-1/2)=-(√3/2)*2=-√3.

Calculating the sine, cosine and tangent of the angle of 3π/4 radians:
We know that 2π=360°, therefore π=180° and 3π/4=180°*3/4=135°>90°.
sin(3π/4)= sin(π-3π/4)=sin{(4π-3π)/4}=sin(π/4), we found that sin(π/4)=√2/2.
cos(3π/4)=-cos(π-3π/4)=-cos{(4π-3π)/4}=-cos(π/4), we found thar cos(π/4)=√2/2, therefore -cos(π/4)=-√2/2.
tan(3π/4)=sin(3π/4)/cos(3π/4)=(√2/2)/(-√2/2)=-1.

Calculating the sine, cosine and tangent of the angle of π radians:
Since 2π=360°, the angle of π radians is equal to 180°.
sin(π)=sin(π-π)=sin(0)=0.
cos(π)=-cos(π-π)=-cos(0)=-1.
tan(π)=sin(π)/cos(π)=0/-1=0.

You just finished studying angles, arc lengths and trig functions topic, the sixth topic of additional topics in math!

Continue studying the next additional topic in math- circle equations.

Circle theorems on the SAT test

Studying circle theorems

On the SAT test circle theorems topic is the fifth topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Circle theorems topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

Circle theorems- summary

Circle theorems topic includes two parts:
Calculating arc length and sector areas using central angles measured in degrees.
Calculating angle measures in degrees inside a circle.

The following formulas are provided at the beginning of each SAT math section:
Circumference of a circle formula is C=2πr.
Area of a circle formula is A=πr2.
Number of degrees of arc in a circle is 360.

Pi (π) is the ratio of the circumference of any circle to the diameter of that circle. Regardless of the circle’s size, this ratio will always equal π (approximately 3.14).

A triangle that one of its angles is a central angle has 2 radii as its sides and it is therefore an isosceles triangle (one of its angles is a central angle and the other 2 angles are equal).

The intersection point of the diameters is at the center of the circle and it divides each diameter to 2 radii. The intersection creates 4 central angles: 2 pairs of equal vertical angles and 4 pairs of supplementary angles (the sum of supplementary angles is 180°).

The relationship between central angle, arc length and sector area is given by the following ratios:

central angle    =             arc length              =     sector area
______________       _____________________          ____________
360°                 circle circumference            circle area

Calculating circumference and area of a circle

The following formulas are provided at the beginning of each SAT math section:
Circumference of a circle formula is C=2πr.
Area of a circle formula is A=πr2.

Units of measurement:
Arc length and circumference measure distance in units such as inches or centimeters.
Area and sector measure area is square units such as square inches or square centimeters.
Angles are measured in degrees.

Consider the following example:

The radius of the circle is equal to 10 centimeters.

What are the circumference and the area of the circle?

Circumference of a circle:
The circumference formula is C=2πr.
C=2*10*π=20π

Area of a circle:
The area formula is A=πr2.
A=πr2=π*102=100π

The relationship between central angle, arc length and sector area

A central angle has a vertex at the center of a circle O and sides located on the circle circumference in two points A and B. The central angle ∠AOB determines a portion of the circumference (an arc) AC and a potion from the area (sector).

A minor arc: Each central angle divides a circle into two arcs. The smaller of the two arcs is called the minor arc and the larger of the two arcs is called the major arc (a minor arc subtends an angle less than or equal to 180°). Note that when dealing with arcs, we always look at the minor arc.

The relationship between central angle, arc length and sector area is given by the following ratios:

central angle    =             arc length              =     sector area
______________       _____________________          ____________
360°                 circle circumference            circle area

This relationship presents 3 equal ratios:

Ratio 1: The angle as a portion of the total degrees of the circle (360°) =

Ratio 2: The arc length as a portion from the length of the circle circumference (C=2πr) =

Ratio 3: The sector area as a portion from the total area of the circle (A=πr2).

Note that we can calculate the denominators of the arc and the sector ratios given the value of the radius r.

In the figure below we have the following ratios:

∠AOC     =        arc AC          =    sector ACO area
_______      _______________        ________________
360°         circumference               circle area

________     ________________        _________________
360°            circumference             circle area

∠BOD     =         arc BD            =     sector BDO area
________      _______________          __________________
360°           circumference                circle area

∠BOC     =          arc BC           =     sector BCO area
________       ______________           _________________
360°          circumference                 circle area

Consider the following example:

A central angle of a circle is equal to 120 degrees and the radius of the circle is equal to 5 centimeters.

What is the arc length of the central angle?

Calculating the circumference of the circle:
C=2πr
C=2*5*π=10π

Calculating the arc length with the ratios:
Ratio 1: The angle as a portion of the total degrees of the circle (360°).
Ratio 2: The arc length as a portion from the length of the circle circumference (C=2πr).

central angle     =           arc length
______________        _____________________
360°                  circle circumference

Representing with the variable x the arc length and plugging the given values into the ratio:

120°   =     x
_____      _____
360°        10π

360x=1200π
x=1200/360π
x=120/36π
x=10/3π=3.33π
x=31/3π=31/3*3.14=10.47
The arc length is 10.47 centimeters.

120/360=1/3=0.33
3.33/10.47=0.32
The answers are different because of rounding differences.

Consider the following example:

The arc length of a circle is 2 centimeters and the radius of the circle is 1.6 centimeters.

What is the area of the sector of the arc?

Calculating the circumference of the circle:
C=2πr
C=2*1.6*π=3.2π=10.

Calculating the area of the circle:
The area formula is A=πr2.
A=π*1.62=2.56π=8.

Calculating the sector area with the ratios:
Ratio 1: The arc length as a portion from the length of the circle circumference (C=2πr).
Ratio 2: The sector area as a portion from the total area of the circle (A=πr2).

Representing with the variable x the sector area and plugging the given values into the ratio:

arc length              =     sector area
____________________        _____________
circle circumference           circle area

1.6   =   x
____      ___
10         8

1.6*8=10x
x=1.6*8/10
x=1.28
The sector area is 1.28 centimeters.

1.6/10=0.16
1.28/8=0.16

Angle relationships in circle

This subject includes 2 angle types:

Angles of isosceles triangles in a circle.

Angles created by intersection of diameters in a circle.

Angles of isosceles triangles in a circle

At the beginning of each SAT math section, it is provided that the number of degrees of arc in a circle is 360. In other words, the sum of central angle measures in a circle is equal to 360°.

A triangle that one of its angles is a central angle has 2 radii as its sides and it is therefore an isosceles triangle. Meaning that one of its angles is a central angle and the other 2 angles are equal.

The sum of angles in a triangle is 180°, if we are given the value of the central angle or the value of one of the equal angles, we can calculate the other angle values.

Consider the following example:

In the figure below, the point O is the center of a circle. The angle ∠AOB is equal to 100°.

What is the value of the angle ∠OAB?

Since the angle ∠AOB is a central angle, the sides OA and OB are radii of the circle and therefore the triangle ABO is an isosceles triangle.

In an isosceles triangle the angles opposite the two equal sides are equal, therefore ∠OAB=∠ABO.

The sum of angles in a triangle is equal to 180°, therefore ∠OAB+∠ABO+∠AOB=180°.

We are given that ∠AOB=100° and we know that ∠OAB=∠ABO, therefore 100°+∠OAB+∠ABO=180°.

Represent by x the equal angles ∠OAB and ∠ABO we get an equation that we can solve:
100+2x =180
2x=80
x=40
The value of the angle ∠OAB is 40 degrees.

40+40+100=180
180=180

Consider the following example:

In the figure below, the point O is the center of a circle.
The angle ∠AOB is twice bigger than the angle ∠COA.
The angle ∠COB is 5 times bigger than the angle ∠COA (the angle ∠COB is bigger than 180°).

What is the value of the angles ∠AOB and ∠COB?

Represent by x the angle ∠COA.

We are given that the angle ∠AOB is twice bigger than the angle ∠COA, therefore the angle ∠AOB=2x.
We are given that the angle ∠COB is 5 times bigger than the angle ∠COA, therefore the angle ∠COB=5x.

The sum of central angle measures in a circle is equal to 360° and we are given that the point O is the center of the circle, therefore we can write an equation and solve it:
x+2x+5x=360
8x=360
x=45

We know that the angle ∠AOB=2x, therefore the angle ∠AOB=90°.
We know that the angle ∠COB=5x, therefore the angle ∠COB=5*45=225 degrees.

Angles created by intersection of diameters in a circle

The intersection point of the diameters is at the center of the circle and it divides each diameter to 2 radii.

The intersection creates 4 central angles: 2 pairs of equal vertical angles and 4 pairs of supplementary angles (the sum of supplementary angles is 180°).

Given the value of one of the 4 central angles, we can calculate the other central angles.

In the figure below AB and CD are diameters of the circle.

What are the central and supplementary angles created by the diameters?

The intersection creates 4 equal radii AO, CO, DO and BO.

The intersection point O is at the center of the circle and it divides each diameter to 2 radii: the diameter AB is divided to radii AO and OB; the diameter CD is divided to radii CO and OD.

The intersection creates 4 central angles: ∠COA, ∠AOD, ∠DOB and ∠BOC.

The intersection creates 2 pairs of equal vertical angles: ∠COA=∠BOD; ∠COB=∠AOD.

The intersection creates 4 pairs of supplementary angles: ∠COA+∠AOD=180°; ∠AOD+∠DOB=180°;  ∠DOB+∠BOC=180°; ∠BOC+∠COA=180°.

Consider the following example:

In the figure below, O the is center of the circle and the chords AB and CD intersect at point O. The angle ∠BOD is equal to 40°.

What is the ratio between the angles ∠ACO and ∠ADO?

Given the value of one of the 4 central angles, we can calculate the other central angles:

We are given that ∠BOD=40°

∠AOC=∠BOD (vertical angles are equal), therefore, ∠AOC=40°.

∠BOD+∠AOD=180° (the sum of supplementary angles is 180°), therefore ∠AOD=180°-40°=140°.

Calculating the angles in triangle ACO:

We are given that O the is center of the circle and the chords AB and CD intersect at point O, therefore AB and CD are diameters and create 4 radii CO=BO=AO=DO.

Since CO=AO the triangle ACO is isosceles and the angles ∠ACO and ∠CAO are equal (represented by x).

The sum of angles in a triangle is equal to 180°, therefore ACO+CAO+AOC=180°.

Plugging the values we get an equation that we can solve:
x+x+40=180
2x=140
x=70
The angle ∠ACO is equal to 70 degrees.

Calculating the angles in triangle ADO:

We are given that O the is center of the circle and the chords AB and CD intersect at point O, therefore AB and CD are diameters and create 4 radii CO=BO=AO=DO.

Since AO=DO the triangle ADO is isosceles and the angles ∠DAO and ∠ADO are equal (represented by x).

The sum of angles in a triangle is equal to 180°, therefore DAO+ADO+AOD=180°.

We found that the angle ∠ACO is equal to 140 degrees.

Plugging the values, we get an equation that we can solve:
x+x+140=180
2x=40
x=20
The angle ∠ADO is equal to 20 degrees.

The ratio between the angles ∠ACO and ∠ADO is 70/20=7/2=31/2

Note that we need to write the value of an angle ∠ACO in the nominator and not the opposite.

You just finished studying circle theorems topic, the fifth topic of additional topics in math!

Continue studying the next additional topic in math- angles, arc lengths and trig functions.

Complex numbers on the SAT test

Studying complex numbers

On the SAT test complex numbers topic is the first topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu).

Complex numbers topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

Complex numbers- summary

A complex number has a real component and an imaginary component, it is written in a form of a + bi, where a and b are real numbers, and i is an imaginary number satisfying i2 = −1.

An imaginary number is a number that, when squared, has a negative result (i2 = −1). Since an imaginary number i is equal to a square root of a negative number -1 it does not have a tangible value (negative numbers don’t have real square roots since a square is either positive or zero).

The unit imaginary number, i, equals the square root of minus 1, so that i=√-1. As said above, when squared imaginary number has a negative result, so that i2=(√-1)2=-1.

For example: 3i is an imaginary number, and its square is (3√-1)2=9*-1=-9.

To add or subtract complex numbers, combine like terms (real terms with real terms and imaginary terms with imaginary terms).

To multiply complex numbers, multiply the numbers with foil formula, replace i2 with -1 and combine like terms.

To divide complex numbers, you need to cancel the denominator by turning the imaginary component in the denominator to a real number. This is done by multiplying the numerator and the denominator by the conjugate of the denominator. The next steps are multiplying the numbers with foil formula, replacing i2 with -1 and combining like terms.

Combine like terms: real terms with real terms and imaginary terms with imaginary terms and write the result as a+bi.

Consider the following example:

If a=6+4i and b=2i-4, that are the values of a-b and a+b?

a=6+4i b=2i-4 a-b=6+4i-(2i-4)= 6+4i-2i+4=10+2i a+b=6+4i+(2i-4)= 6+4i+2i-4=2+6i

Multiplying complex numbers

Remember that since i=√-1, the value of i2 is i2=-1.

If after multiplying we get i2, we can write it as -1 and continue solving.

Steps for multiplying complex numbers:

Step 1: Multiply the numbers with foil formula.
The FOIL formula is y=(x+c)(x+d)= x2+dx+cx+cd= x2+(c+d)x+cd.

Step 2: Replace i2 with -1.

Step 3: Combine like terms (real terms with real terms and imaginary terms with imaginary terms) and write the result as a+bi.

Consider the following example:

If a=6+4i and b=2i-4, that is the value of a*b?

Step 1: Multiplying the numbers with foil formula:

a=6+4i

b=2i-4

a*b=(6+4i)(2i-4)=12i-24+8i2-16i

Step 2: Replacing i2 with -1:

a*b=12i-24+8i2-16i

a*b=12i-24+8*-1-16i

Step 3: Combining like terms:

a*b=-32-4i

Dividing complex numbers

We have a numerator and a denominator as 2 complex numbers in a form of a+bi and we need to simplify the result to a form of one complex number in a form of a+bi (staying without the denominator).

To cancel the denominator, we need to turn the imaginary component in the denominator to a real number, this is done by multiplying the numerator and the denominator by the conjugate of the denominator.

For example:
We learned in the quadratic equations topic that (a+b)(a-b)=a2-b2.
If we multiply the complex number a+bi by a conjugate of a-bi we get (a+bi)(a-bi)=a2-b2i2.
Since we know that i2=-1. The expression a2-b2i2 is equal to a2+b2. This outcome is a real number.

Steps for dividing complex numbers:

Step 1: Multiply the numerator and the denominator by the conjugate of the denominator (conjugate divided by itself is equal to 1 and we can multiply by 1 without changing the original value).

Step 2: Multiply the numbers in the numerator and the denominator with foil formula.

Step 3: Replace i2 in the numerator and the denominator with -1. In the denominator you will be left with real terms without imaginary terms.

Step 4: Combine like terms and write the answer as a complex number in the numerator in a form of a+bi divided by a real number in the denominator.

Consider the following example:

If a=6+4i and b=2i-4, that is the value of a/b?

Step 1: Multiplying the numerator and the denominator by the conjugate of the denominator:

a=6+4i

b=2i-4

a   = 6+4i
__    _____
b      2i-4

The conjugate of the denominator is 2i+4.

a    =  6+4i   =  (6+4i)(2i+4)
__      _____       ____________
b         2i-4        (2i-4) (2i+4)

Step 2: Multiplying the numbers in the numerator and the denominator with foil formula:

a   =  12i+24+8i2+16i
__      _______________
b              4i2-16

Step 3: Replacing i2 in the numerator and the denominator with -1:

a   =  12i+24+8i2+16i  =   12i+24+8*-1+16i
__     ________________       ________________
b               4i2-16                        4*-1-16

Step 4: Combining like terms and writing the answer as a complex number in the numerator divided by a real number in the denominator:

a      =    12i+24+8*-1+16i   =   28i+16   =    28i  +  16  =   -7i  –  4
__           _________________       _______       ____     ____     ___    ___
b                    4*-1-16                     -20            -20      -20        5       5

You just finished studying complex numbers topic, the first topic of additional topics in math!

Continue studying the next additional topic in math- volume word problems.

Congruence and similarity on the SAT test

Studying congruence and similarity

On the SAT test congruence and similarity topic is the third topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Congruence and similarity topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

Congruence and similarity- summary

Congruence and similarity questions include congruent angles and similar triangles topics.

Similar triangles have the same angle measures and their corresponding side lengths are related by a constant ratio.

Sum of angles in a triangle: The sum of the measures in degrees of the angles of a triangle is 180.

Angle measures in triangles: In an isosceles triangle the angles opposite the two equal sides are equal; the angles of an equilateral triangle are equal to 60°.

Congruent angles are angles that have the same angle measure.

Vertical Angles are the angles opposite each other when two lines cross, vertical angles are equal.

Supplementary angles are those angles that measure up to 180 degrees. Angles that lie on the same side of a straight line, are always supplementary.

Alternate angles are angles located in opposite positions relative to a transversal intersecting two parallel lines. Alternate angles are equal.

Corresponding angles are angles located in the same position relative to parallel lines intersected by a transversal. Corresponding angles are equal.

Two intersecting lines create vertical angles and supplementary angles.

Two parallel lines combined with two intersecting lines form 2 similar triangles and equal alternate angles.

A parallel line inside a triangle forms 2 similar triangles and equal corresponding angles.

Angle relationships inside a triangle

Sum of angles in a triangle:
The sum of the measures in degrees of the angles of a triangle is 180. (This statement is provided at the beginning of each SAT math section).
If use the variables x°, y° and z° to represent angles then x°+y°+z°=180°.

Angle measures in an isosceles triangle:
An isosceles triangle is a triangle that has two equal sides, the angles opposite the two equal sides are equal.

Angle measures in an equilateral triangle:
An equilateral triangle is a triangle with all three sides of equal length, the angles of an equilateral triangle are equal to 60°.

The figure below presents a triangle and its angles.

Consider the following example:

In a triangle ABC given that the angle A is equal to 50° and AB=BC.

What is the value of the angles B and C?

The sum of the measures of the angles of a triangle is 180°, therefore the angles A+B+C=180°.
We are given that AB=BC, therefore the angle B is equal to the angle C.
We are given that A=50°.

Therefore, we can represent by x the value of angle B and C and solve an equation with 1 variable:
x+x+50°=180°
2x=180°-50°
2x=130°
x=65°

65°+65°+50°=180°
180°=180°

Consider the following example:

In a triangle ABC given that the angle A is equal to 60° and the angle B is twice bigger than the angle C.

What is the value of the angles B and C?

The sum of the measures of the angles of a triangle is 180°, therefore the angles A+B+C=180°.
We are given that A=60°.
We are given that the angle B is twice bigger than the angle C.

Therefore, we can represent by x the value of angle C and solve an equation with 1 variable:
2x+x+60°=180°
3x=120°
x=40° therefore C=40° and D=40°*2=80°

40°+80°+60°=180°
180°=180°

Angle relationships between intersecting lines- vertical and supplementary angles

Two intersecting lines create vertical angles and supplementary angles.

Vertical Angles are the angles opposite each other when two lines cross (vertical means that they share the same vertex). Vertical angles are equal.

Supplementary angles are those angles that measure up to 180 degrees. Angles that lie on the same side of a straight line, are always supplementary.

The figure below shows the angles created between two intersecting lines.

The angles 1 and 3 (180-α) and the angles 2 and 4 (α) are vertical angles. All pairs of vertical angles are equal.

The angles 1 and 2, the angles 2 and 3, the angles 3 and 4 and the angles 4 and 1 are supplementary angles. The sum of each pair of supplementary angles equals to 180 degrees (180-α+α=180°).

Consider the following example:

In the figure below are presented 4 lines that cross each other: AB, FI, AH and BG. The triangle ABC is an isosceles triangle, so that AS=BC. The angle ∠FEG is equal to 30° and the angle ∠FDH is equal to 120°.

What is the value of the angle BAC?

We are given that the angle ∠FEG is equal to 30°, therefore the angle ∠CED is also equal to 30° (the angles are vertical, therefore they are equal).

We are given that the angle ∠FDH is equal to 120°, therefore the angle ∠CDI is equal to 180°-120°=60° (the angles are supplementary, therefore their sum is 180°).

The sum of angles in a triangle is equal to 180°, therefore the sum of angles in the triangle CDE is 180° and the angle ∠DCE=180°-60°-30°=90°.

We are given that the triangle ABC is an isosceles triangle, therefore the angles ∠ABS and ∠BAC are equal.
The sum of angles in a triangle is equal to 180°, therefore the sum of angles in the triangle ABC is 180°.
We calculated that the angle ∠DCE is equal to 90° the angles ∠ABS and ∠BAC are equal.
From these 3 statements we can conclude that the angle ∠BAC=∠ABC=(180°-90°)/2=90°/2=45°.

Angle relationships between intersecting and parallel lines- alternate angles

Two parallel lines combined with two intersecting lines form 2 similar triangles and equal alternate angles.

Similar triangles have the same shape, but not the same size (they have the same angle measures, but not the same side lengths).
Note that the corresponding side lengths of similar triangles are related by a constant ratio, which is called k. See further details below.

Alternate angles are angles located in opposite positions relative to a transversal intersecting two parallel lines. Alternate angles are equal.

In the figure below two parallel lines AB and DE (marked in blue) combined with two intersecting lines AE and BD form 2 similar triangles ABC and CDE.

The angles x°, y° and z° are equal.

The equal angles x°, y° and z°:

The equal angles z° are vertical angles that are formed by the lines AE and BD.

The equal angles x° are called alternate angles, they are formed by the parallel lines AB and DE and the line AE that crosses them.

The equal angles y° are called alternate angles, they are formed by the parallel lines AB and DE and the line BD that crosses them.

Since the angles x°, y° and z° are equal between the triangle ABC and CDE, the triangles are similar.

Consider the following example:

In the figure below are presented 4 lines that cross each other: HI, DE, EF and BG. The lines HI and DE are parallel. The angle ∠FAH is equal to 40° and the angle ∠EDG is equal to α. In addition, AC=BC.

What is the value of α?

We are given that the angle ∠EDG is equal to α.

Angles that lie on the same side of a straight line, are always supplementary and their sum is equal to 180°. Therefore, the angles ∠EDG and ∠CDE are supplementary and ∠EDG+ ∠CDE=180°.

The angle ∠CDE=180°-α.

Two parallel lines combined with two intersecting lines form 2 similar triangles. We are given that the lines HI and DE are parallel, therefore the triangles ABC and CDE are similar and the angles ∠CDE and ∠ABC are equal and their value is 180°-α (we found that ∠CDE=180°-α).

Note that instead of using similar triangles we can use the statement that two parallel lines combined with two intersecting lines form 2 equal alternate angles. We are given that the lines HI and DE are parallel, therefore the angles ∠CDE and ∠ABC are equal (these angles are alternate) and their value is 180°-α (we found that ∠CDE=180°-α).

We are given that the angle ∠FAH is equal to 40°, therefore the angle ∠BAC is also equal to 40° (the angles are vertical, therefore they are equal).

We are given that AC=BC, therefore the angle ∠BAC=∠ABC.

We found that the angle ∠BAC=∠ABC, the angle ∠BAC=40° and the angle ∠ABC=∠CDE=180°-α. Therefore, the angle ∠ABC=∠CDE=∠BAC=∠FAH=180°-α=40°

We can solve the equation 180°-α=40°, getting α=180°-40°=140°

Angle relationships between parallel lines inside a triangle- corresponding angles

A parallel line inside a triangle (the line is parallel to one of the sides of the triangle) forms 2 similar triangles (a small triangle and a large triangle) and equal corresponding angles.

Corresponding angles are angles located in the same position relative to parallel lines intersected by a transversal. Corresponding angles are equal.

In the figure below a parallel line EB inside a triangle ACD (EB is parallel to CD) forms 2 similar triangles (a small triangle ABE and a large triangle ACD).

The equal angles x° and y°:

The equal angles x° are called corresponding angles, they are formed by the parallel lines BE and CD and the line AD.

The equal angles y° are called corresponding angles, they are formed by the parallel lines BE and CD and the line AC.

Since the angles x°, y° and z° are equal between the triangle ABE and ACD, the triangles are similar.

Consider the following example:

In the figure below are presented 4 lines that cross each other: AD,AC,BF and CE. The lines BF and CE are parallel. The angle ∠BAF is equal to 20° and the angle ADE is equal to α. In addition, AF=AB.

What is the value of α?

Angles that lie on the same side of a straight line, are always supplementary and their sum is equal to 180°. Therefore, the angles ADE and ADC are supplementary and ADE+ ADC=180°.

A parallel line inside a triangle forms 2 similar triangles. We are given that the lines BF and CE are parallel, therefore the triangles ABF and ACD are similar and the angles AFB and ADC are equal and their value is 180°-α (we found that ADC=180°-α).

Note that instead of using similar triangles we can use the statement that a parallel line inside a triangle forms 2 equal corresponding angles. We are given that the lines BF and CE are parallel, therefore the angles AFB and ADC are equal (these angles are corresponding) and their value is 180°-α (we found that ADC=180°-α).

We are given that AB=AF, therefore the angle ABF=AFB. We are also given that the angle BAF=20°.

The sum of angles in a triangle is equal to 180°, therefore the sum of angles in the triangle ABF is 180° and the angle ABF=AFB=(180°-20)/2=160°/2=80°.

We have an equation that we can solve: ABF=AFB=ADC=180°-α=80°.

180°-α=80°

α=180°-80°

α=100°

Note that the angle ACD=ABF=80°.

Calculating side lengths with triangles similarity

Similar triangles have the same shape, but not the same size (they have the same angle measures, but not the same side lengths).

The corresponding side lengths of similar triangles are related by a constant ratio, which is called k.

Note that:
Similar triangles have the same angle measures.
If 2 lines are parallel, the triangles that are formed by them are similar.

The figure below shows two similar triangles: ABC and DEF. The triangle ABC is similar to the triangle DEF.

Similar triangles have the same angle measures, therefore the angle ∠A is equal to the angle ∠D; the angle ∠B is equal to the angle ∠E and the angle ∠C is equal to the angle ∠F.

The corresponding side lengths of similar triangles are related by a constant ratio, which is called k, therefore the following ratios between pairs of the sides exist:
DE=k*AB (marked in orange in the figure above)
DF=k*AC (marked in green in the figure above)
EF=k*BC (marked in black in the figure above)

We can also present all the ratios as being equal to k:

k =  DE   =   DF   =   EF
____     ____      ____
AB        AC        BC

Note that we could write the ratio in an opposite way by switching between the numerator and the denominator (so that the numerator will include the sides of the small triangle and the denominator will include the sides of the big triangle). The parameter k will be smaller than 1 and equal to 1 divided by the k parameter above.
For example: if the ratio k between the sides of the big triangle divided by the sides of the small triangle is 2 (like 4 divided by 2), then the ratio k between the sided of the small triangle divided by the sides of the big triangle will be equal to 1/2 (like 2 divided by 4).

Consider the following example:

In the figure below is given that AB=5 centimeters; CD=27 centimeters; BE is parallel to CD; AC is 3 times bigger than AB.

What is the value of BE?

We are given that AC is 3 times bigger than AB, therefore we can write an equation AC=AB*3.

We are given that AB=5 centimeters, therefore AC=AB*3=5*3=15. Meaning that AC/AB=3.

A parallel line inside a triangle forms 2 similar triangles. We are given that BE is parallel to CD, therefore the triangles ABE and ACD are similar.

The corresponding side lengths of similar triangles are related by a constant ratio, which is called k.

We found AC/AB=3, therefore k=3.

K is also equal to BE/CD, therefore we can write an equation CD/BE=3.

We are given that CD=27 centimeters, therefore we can write an equation 27/BE=3.

27/BE=3

BE=27/3

BE=9 centimeters

Calculating side lengths with triangles similarity when one side is common for both triangles

In these questions one side is common for both triangles, therefore we need the length of two sides instead of three sides to calculate the length of the fourth side using the similar triangles ratio.

Consider the following example:

The figure below presents two similar triangles, ABC and BDC.

The length of AB is 20 centimeters and the length of BC is 15 centimeters.

What is the length of DC?

In this example the side BC is common for both the triangles. Therefore we can you it twice in the similar triangles ratio.

The similar triangles ratio is

k   =   AB     =    BC
____        ____
BC           DC

k   =   20   =  15
____     ____
15        DC

4  =  15
__      __
3      DC

4DC=45

DC=45/4=11 1/4

The length of the side DC is 11 1/4 centimeters.

Note that we could write the ratio in an opposite way by switching between the numerator and the denominator and get the same answer:

k    =   BC     =    DC
____         ____
AB            BC

K   =   15   =    DC
_____      _____
20          15

3   =   DC
___      ___
4         15

4DC=45

DC=45/4=11 1/4

The ratio k in this case will be 15/20=3/4, which is 1 divided by the previous ratio k=4/3.

You just finished studying congruence and similarity topic, the third topic of additional topics in math!

Continue studying the next additional topic in math- right triangle trigonometry and word problems.

Right triangle trigonometry and word problems on the SAT test

Studying right triangle trigonometry and word problems

On the SAT test right triangle trigonometry and word problems topic is the fourth topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Right triangle trigonometry and word problems topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

Right triangle trigonometry and word problems- summary

Right triangle is a triangle with a right angle (equal to 90°). The side opposite the right angle (the longest side of the right triangle) is called a hypotenuse.

Right triangle trigonometry and right triangle word problems require calculating side lengths and angle measures in right triangles.

Pythagorean theorem:

Pythagorean theorem states that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides: a2 + b2 = c2

Pythagorean triples are combinations of side lengths a, b and c that satisfy the Pythagorean theorem. If you remember the triples values, you know the size of the third side without the need to calculate it. The most common Pythagorean triples are: 3, 4 and 5;    5, 12 and 13;    7, 24 and 25.

Trigonometric ratios (sine, cosine and tangent):

Trigonometric ratios (functions) represent connections between angle degrees and side lengths in a right triangle:

• The sine of an angle (sin) in a right triangle is defined as the ratio of the length of the side that is opposite to the angle, to the length of the hypotenuse.
• The cosine of an angle (cos) in a right triangle is defined as the ratio of the length of the side that is adjacent to the angle, to the length of the hypotenuse.
• The tangent of an angle (tan) in a right triangle is defined as the ratio of the length of the side that is opposite to the angle, side that is adjacent to the angle.

Complementary angles are two angles with the sum of 90 degrees. Sine of an angle (α) in a right triangle is equal to cosine of its complementary angle (90-α).

Similar triangles have the same angle measures and their corresponding side lengths are related by a constant ratio therefore they also have similar sine, cosine and tangent.

Special right triangles:

Special right triangles are right triangles whose sides are in a particular ratio.

• In a 30°, 60°, 90° right triangle the side opposite the 30° angle is half the length of the hypotenuse and the side opposite to 60° angle is equal to the length of a side opposite to 30° angle multiplied by √3.
In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.
• In a 45°, 45°, 90° right triangle the sides opposite the 45° angles are equal and the hypotenuse is equal to the side opposite to 45° angle multiplied by √2.
In 45°, 45°, 90° triangle the sides are s, s and s√2.

Pythagorean theorem in right triangles

Pythagorean theorem states that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. (The hypotenuse is the side opposite the right angle).

The formula of the Pythagorean theorem is a2 + b2 = c2. This formula is provided at the beginning of each math section followed by a diagram.

The following figure presents a right triangle and the Pythagorean theorem a2 + b2 = c2.

Calculating side lengths with Pythagorean theorem

If we are given the values of two sides of a right triangle, we can calculate the value of the hypotenuse with the Pythagorean theorem.

If we are given the values of one side and the hypotenuse of a right triangle, we can calculate the value of the second side with the Pythagorean theorem.

Consider the following example:

The length of the sides of a right triangle are 3 centimeters and 4 centimeters.

What is the length of the hypotenuse?

The formula of the Pythagorean theorem is a2+b2=c2.
a=3, b=4, c=?
32 + 42 = c2
c2=9+16
c2=25
c=5
The length of the hypotenuse is 5 centimeters.

32+42=52
9+16=25
25=25

Consider the following example:

The length of a side of a right triangle is 4 centimeters and the length of the hypotenuse is 8 centimeters.

What is the length of the other side?

The formula of the Pythagorean theorem is a2+b2=c2.
a=4, b=?, c=8
42+b2=82
16+b2=64
b2=64-16
b2=48
b=√48
b=√(4*12)
b=√4*√12
b=2√12=6.93

a2+b2=c2
42+(2√12)2=82
16+4*12=64
16+48=64
64=64

Pythagorean triples in a right triangle

Pythagorean triples are combinations of side lengths a, b and c that satisfy the Pythagorean theorem. If you remember the triples values, you know the size of the third side without the need to calculate it.

The most common Pythagorean triple is 3, 4 and 5.
5, 12 and 13
7, 24 and 25

Note that each multiplication of the triple also satisfies the Pythagorean theorem.
For example: The triple 9, 12 and 15 is the triple 3, 4 and 5 multiplies by 3. Therefore, the triple 9, 12 and 15 is a Pythagorean triple.
We can check this statement by plugging into the formula of the Pythagorean theorem:
a2+b2=c2
92+122=152
81+144=225
225=225

Consider the following example:

The length of a side of a right triangle is 7 centimeters and the length of the hypotenuse is 25 centimeters.

What is the length of the other side?

The triple 7, 24 and 25 is a Pythagorean triple, therefore the length of the second side is 24 centimeters.

a2+b2=c2
72+b2=252
b2=252-72
b2=625-49
b2=576
b=24
The length of the other side is 24 centimeters.

Consider the following example:

The length of the sides of a right triangle are 10 centimeters and 24 centimeters.

What is the length of the hypotenuse?

The triple 5, 12 and 13 is a Pythagorean triple, the given sides 10 and 24 are the triple sides 5 and 12 multiplied by 2. Therefore, the length of the hypotenuse is 13 multiplied by 2=26 centimeters.

a2+b2=c2
102+242=262
100+576=676
676=676

Calculating triangles side lengths with Pythagorean theorem and triangles similarity

Similar triangles have the same shape, but not the same size (they have the same angle measures, but not the same side lengths).

The corresponding side lengths of similar triangles are related by a constant ratio, which is called k.

Note that:
Similar triangles have the same angle measures.
If 2 lines are parallel, the triangles that are formed by them are similar.

Given parallel lines and a right triangle, we can calculate side lengths using triangles similarity combined with the Pythagorean theorem.

Consider the following example:

In the figure below the lines BE and CD are parallel. The angle measures and the side lengths (in centimeters) are given in the figure.

What are the lengths of the sides of triangle ACD?

Calculating the length of AB:

Since BE is parallel to CD, the triangles ABE and ACD are similar. Therefore, the angle ACD is equal to the angle ABE and equal to 90 degrees.

Since the angle ABE is equal to 90 degrees, the triangle ABE is a right triangle. Therefore, we can apply the Pythagorean theorem in the triangle ABE:

The sides 3, 4 and 5 are Pythagorean triple, or we can solve the Pythagorean theorem:
a2+b2=c2
32+42=c2
c2=9+16
c2=25
c=5
AB=5 centimeters

Calculating the length of CD and DE:
AC=BC+AB
AC=5+5=10 centimeters

Since the lines BE and CD are parallel, the triangles ACD and ABE are similar. Therefore, the ratio of the related sides is k=AC/AB=10/5=2 (the side of the big triangle divided by the side of the small triangle is equal to 2).

Since BE=3 and k=2, CD/3=2, CD=6 centimeters.

Checking by calculating Pythagorean theorem in the big triangle ACD:
a2+b2=c2
62+82=102
This is a Pythagorean triple 3, 4 and 5 multiplied by 2.

Trigonometric ratios (sine, cosine and tangent)

Trigonometric ratios (functions) represent connections between angle degrees and side lengths in a right triangle.

The sine of an angle (sin) in a right triangle is defined as the ratio of the length of the side that is opposite to the angle, to the length of the hypotenuse. In the figure below in a right triangle ABC, sine (A)= BC/AC.

The cosine of an angle (cos) in a right triangle is defined as the ratio of the length of the side that is adjacent to the angle, to the length of the hypotenuse. In the figure below in a right triangle ABC, cosine (A)= AB/AC.

The tangent of an angle (tan) in a right triangle is defined as the ratio of the length of the side that is opposite to the angle, side that is adjacent to the angle. In the figure below in a right triangle ABC, tangent (A)= BC/AB.

In the figure below we see a right triangle ABC.
sin (A)=opposite/hypotenuse=BC/AC

Sine and cosine of complementary angles

Complementary angles are two angles with the sum of 90 degrees.

Complementary angles in a right triangle: Since one angle in a right triangle is equal to 90 degrees and the sum of angles of a triangle is 180 degrees, the two other acute angles are complementary.

Sine of an angle in a right triangle is equal to cosine of its complementary angle. Meaning that given two complementary angles α and 90-α in a right triangle, sinα=cos(90-α).

Showing the connection between sine and cosine of complementary angles:
If we represent by α the measure of one angle in a right triangle, then the other angle measure is 90-α (complementary angles).

The figure below presents a right triangle ABC with complementary angles α and 90-α.

sinα=BC/AC

cos(90-α)=BC/AC

We see that sinα=cos(90-α)

Consider the following example:

What is sin (60) is equal to:
A. cos(30)
B. tan(60)
C. sin(30)
D. cos(60)

We know that sinα=cos(90-α), therefore, sin60=cos(90-60)=cos(30).

Calculation of side lengths with sine, cosine and tangent

To calculate a side length, we need to know the value of the trigonometric function (sine, cosine or tangent) and the value of the other side.

Consider the following example:

The figure above represents a right triangle ABC.

If sin(a)= 0.5 and AC=5 centimeters, what is the value of AC?

sin(A)=BC/AC

0.5=BC/5

BC=0.5*5=2.5 centimeters

Calculation of sine, cosine and tangent in similar triangles

Similar triangles have the same angle measures and their corresponding side lengths are related by a constant ratio therefore they also have similar sine, cosine and tangent.

We can calculate sine, cosine or tangent in one triangle and conclude that they are identical in all similar triangles.

Related corresponding side lengths lead to similar sine, cosine and tangent:
If the sides of triangle 1 are x, y and z and the sides of a similar triangle 2 are kx,ky and kz (multiplied by a factor of k) then:
The sine of an angle A in triangle 1 is sin A1=x/z
The sine of an angle A in triangle 2 is sin A2=kx/kx=x/z
The sine of triangle 1 is equal to the sine of triangle 2.

Consider the following example:

In the figure below the lines BE and CD are parallel. The angle measures and the side lengths (in centimeters) are given in the figure.

What is the cosine of angle E?

Special right triangles

Special right triangles are right triangles whose sides are in a particular ratio. Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.

The special right triangles with the side sizes length are given at the beginning of each SAT section.

30°, 60°, 90° triangle:
In a 30°, 60°, 90° right triangle the side opposite the 30° angle is half the length of the hypotenuse and the side opposite to 60° angle is equal to the length of a side opposite to 30° angle multiplied by √3.
In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.

45°, 45°, 90° triangle:
In a 45°, 45°, 90° right triangle the sides opposite the 45° angles are equal and the hypotenuse is equal to the side opposite to 45° angle multiplied by √2.
In 45°, 45°, 90° triangle the sides are s, s and s√2.

The following graphs present the special right triangles with the side sizes length.

Calculation of side lengths given angle measures in special triangles

Given the length of any side in a special right triangle, we can calculate the length of the two other sides.

If you identify that the angles of a right triangle have the measures of 30,60,90 or 45,45,90 you know the ratios between their side length according to the graph presented above:

In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.

In 45°, 45°, 90° triangle the sides are s, s and s√2.

Consider the following example:

In a right triangle the measures of the angles are 30°, 60°, 90° and the hypotenuse is equal to 8 centimeters.

What are the lengths of the other sides?

In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.

Given the hypotenuse is equal to 8 centimeters, we know that the side opposite to 30° angle is equal to 8/2=4 centimeters. The side opposite to 60° angle is equal to 4*√3=√16*√3=√48.

Checking with Pythagorean theorem:
a2+b2=c2
42+√482=82
16+48=64
64=64

Consider the following example:

In a right triangle the measures of the angles are 45°, 45°, 90° and the hypotenuse is equal to √40 centimeters.

What are the lengths of the sides?

In 45°, 45°, 90° triangle the sides are s, s and s√2.

We are given that the hypotenuse is equal to √40 centimeters, therefore s√2=√40. Continue solving:
s√2=√40
s=√40/√2
s=√20

Checking with Pythagorean theorem:
a2+b2=c2
√202+√202=√402
20+20=40
40=40

Calculation of angle measures given side lengths in special triangles

Given the side lengths of a special right triangle, we can calculate the angle measures.

If the ratio between a side and a hypotenuse in a right triangle is 1/2 we know that the triangle is a special right triangle with 30°, 60°, 90° angles (the side used in the ratio is located opposite to the 30° angle).

If the ratio between a side and a hypotenuse in a right triangle is √3/2 we know that the triangle is a special right triangle with 30°, 60°, 90° angles (the side used in the ratio is located opposite to the 60° angle).

If the ratio between two sides in a right triangle is √3 or 1/√3 we know that the triangle is a special right triangle with 30°, 60°, 90° angles (the bigger side which is multiplied by √3 is located opposite to the 60° angle).

If a right triangle is an isosceles triangle, it is a special right triangle with angles 45°, 45° and 90°.

If the ratio between a side and a hypotenuse in a right triangle is 1/√2 we know that the triangle is a special right triangle with 45°, 45°, 90° angles.

Consider the following example:

The ratio between two sides in a right triangle is √27/9, what are the measures of the angles of the triangle?

Simplifying the ratio √27/9 gives us:
√27/9=√27/√81
√27/√81=√27/(√27*√3)
√27/(√27*√3)=1/√3

If the ratio between two sides in a right triangle is 1/√3 we know that the triangle is a special right triangle with 30°, 60°, 90° angles (the bigger side which is multiplied by √3 is located opposite to the 60° angle).

You just finished studying right triangle trigonometry and word problems topic, the fourth topic of additional topics in math!

Continue studying the next additional topic in math- circle theorems.

Volume word problems on the SAT test

Studying volume word problems

On the SAT test volume word problems topic is part of additional topics in math that include 7 advanced topics (see the full topics list on the top menu).

Volume word problems topic is the second topic of additional topics in math. It is recommended to start learning additional topics in math with its first topic called complex numbers.

Volume word problems topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topics in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math).

Volume word problems- summary

Volume word problems require making calculations of volumes of three-dimensional shapes using volume formulas. To calculate the volume, plug the given dimension into the relevant volume formula.

Volume formulas of five basic shapes:
The volume formulas of five basic shapes are given at the beginning of the math sections of the SAT exam: right rectangular prism, right circular cylinder, sphere, right circular cone and rectangular pyramid.
Right rectangular prism volume formula is V= lwh.
Right circular cylinder volume formula is V=πr2h.
A sphere volume formula is V= 4/3 πr3.
A right circular cone volume formula is V= 1/3 πr2h.
A rectangular pyramid volume formula is V=1/3 lwh.

Calculating the effect of changes in dimensions on volume: The power of the dimension determines the size of the change in the volume value.

• If a dimension in the volume formula is raised to a first power, the volume changes by the same factor as the shape.
• If a dimension in the volume formula is raised to a second power, when the shape changes by a factor the volume changes by a square of the factor.
• If a dimension in the volume formula is raised to a third power, when the shape changes by a factor the volume changes by a third degree of the factor.

Comparing volumes of two shapes: In these questions we are given ratios between the dimensions of two shapes and we are required to compare their volumes. We need to calculate the total effect on the volume of all the ratios between the shapes.

Volume formulas worksheet

Volume formulas worksheet is given below, it includes 5 formulas:

Right rectangular prism volume formula is V= lwh.

Right circular cylinder volume formula is V=πr2h.

A sphere volume formula is V= 4/3 πr3.

A right circular cone volume formula is V= 1/3 πr2h.

A rectangular pyramid volume formula is V=1/3 lwh.

In the formulas below:
l=length, w=width, h=height, V=volume, A=area

The volume of a right rectangular prism

A right rectangular prism is a three-dimensional object with 6 faces, where all the 6 faces are rectangles. In a right rectangular prism, the angles between the base and sides are right angles.

The volume of a right rectangular prism is defined as the product of the area of one face (length*width) multiplied by its height.

Right rectangular prism volume formula is V= lwh.

Consider the following example:

Calculate the volume of a rectangular prism that has a height of 6 centimeters, a length of 3 centimeters and a width of 4 centimeters.

Right rectangular prism volume formula is V= lwh.

V=lwh=3*4*6=72 cubic centimeters.

The volume of a right circular cylinder

A right circular cylinder has 2 identical and parallel circular bases at the ends. The elements are perpendicular to the bases, therefore the cylinder is called right.

The volume of a right circular cylinder is defined as the product of the area of the circular base multiplied by the height of the cylinder.

Circular base area formula is A=πr2.

Right circular cylinder volume formula is V=πr2h.

Consider the following example:

Calculate the volume of a circular cylinder that has a radius of 5 centimeters and a height of 10 centimeters.

Right circular cylinder volume formula is V=πr2h.

V=πr2h=π*52*10=250π=785 cubic centimeters.

The volume of a sphere

A sphere is a three-dimensional object that has a surface of a ball, all the point on the surface of the sphere are lying at the same distance (the radius) from the center.

A sphere volume formula is V= 4/3 πr3.

Consider the following example:

Calculate the volume of a sphere that has a radius of 5 centimeters.

A sphere volume formula is 4/3 πr3.

V=4/3 πr3=4/3*π*53= 4/3*53*π=167π =524 cubic centimeters.

The volume of a right circular cone

A right circular cone is a cone in which the center point of the circular base is joined with the vertex of the cone and forms a right angle (the height is perpendicular to the radius or the circle).

A right circular cone volume formula is V= 1/3 πr2h.

Consider the following example:

A cone has a height of 10 centimeters and a circular base with a radius of 3 centimeters. What is the volume of the cone?

A right circular cone volume formula is V= 1/3 πr2h.

V= 1/3 πr2h=1/3*π*32*10=30π=30*3.14159=94 cubic centimeters.

The volume of a rectangular pyramid

A rectangular pyramid is a pyramid that has four-sided base and a vertex.

A rectangular pyramid volume is defined as the product of the area of the base multiplied by the height of the pyramid (the height is the distance from the center point of the base to the vertex) divided by 3.

A rectangular pyramid volume formula is V=1/3 lwh.

Consider the following example:

A pyramid has a height of 10 centimeters a length of 3 centimeters and a width of 5 centimeters. What is the volume of the pyramid?

A rectangular pyramid volume formula is V=1/3 lwh.

V=1/3 lwh= 1/3*3*5*10=50 cubic centimeters.

Calculating dimensions given the volumes values of the shapes

To calculate an unknown dimension, plug the given dimensions and the volume values into the volume formula and solve.

Consider the following example:

A sphere has a volume of 33.5 cubic centimeters.

What is the radius of the sphere?

A sphere volume formula is V= 4/3 πr3.
4/3 πr3=33.5
πr3=33.5*3/4
πr3=25.125
3.14159r3=25.125
r3=25.125/3.14159
r3=8
r=3√8
r=2
The radius of the sphere is 2 centimeters.

Consider the following example:

A pyramid has a square base with a height that is 3 timed bigger than the length of the base.

If the volume of the pyramid is 27 cubic centimeters, what is its height?

x= the length and the width of the square pyramid
3x= the height of the pyramid

A rectangular pyramid volume formula is V=1/3 lwh.
V=1/3 lwh= 1/3 x*x*3x
1/3 x*x*3x=27
x3=27
x=3√27
x=3

Calculating the effect of changes in dimensions on volume

The power of the dimension determines the size of the change in the volume value:

If a dimension in the volume formula is raised to a first power, the volume changes by the same factor as the shape.
For example: A rectangular pyramid volume formula is V=1/3 lwh. The length, width and height in the formula are raised to a first power. If we double the length or the width or the height (one of them), then the volume will also be doubled.

If a dimension in the volume formula is raised to a second power, when the shape changes by a factor the volume changes by a square of the factor.
For example: A right circular cone volume formula is V= 1/3 πr2h. The radius of the circle is raised to a second power. If we double the radius, the volume will change by 22=4. The height in the formula is raised to a first power. If we double the height, then the volume will also be doubled.

If a dimension in the volume formula is raised to a third power, when the shape changes by a factor the volume changes by a third degree of the factor.
For example: A sphere volume formula is 4/3 πr3. The radius of the sphere is raised to a third power. If we double the radius, the volume will change by 23=8.

Consider the following example:

A circular cylinder has a radius of 5 centimeters and a height of 10 centimeters. What is the change in the volume if we double the radius and halve the height?

Right circular cylinder volume formula is V=πr2h.

We see in the formula that the radius is raised to a second power and the height is raised to a first power.
If we doble the radius, the volume will be multiplied by 22=4.
If we halve the height, the volume will be multiplied by 0.51.
The total change in the volume is 4*0.5=2.

We can check the answer by calculating the volumes in the two scenarios:
V=πr2h=π*52*10= 250π cubic centimeters.
V=πr2h=π*102*5= 500π cubic centimeters.
The volume value is doubled.

Comparing volumes of two shapes

In these questions we are given ratios between the dimensions of two shapes and we are required to compare their volumes.

We need to calculate the total effect on the volume of all the ratios between the shapes.

Consider the following example:

A right circular cone has a volume of 200 cubic centimeters.
A right circular cone volume formula is V= 1/3 πr2h.

Which of the following cones has a higher volume than the given cone?
A. A cone with 3 times the radius and 1/10 times the height.
B. A cone with 2 times the radius and 1/4 times the height.
C. A cone with 1/2 times the radius and 3 times the height.
D. A cone with 1/3 times the radius and 10 times the height.

A right circular cone volume formula is V= 1/3 πr2h.

We see in the formula that the radius is raised to a second power and the height is raised to a first power.

Answer A: A cone with 3 times the radius and 1/10 times the height:
The total change is (32)*1/10=9/10 <1.
This cone has a smaller volume than the given cone.

Answer B: A cone with 2 times the radius and 1/4 times the height.
The total change is (22)*1/4=4/4=1
This cone has the same volume as the given cone.

Answer C: A cone with 1/2 times the radius and 3 times the height.
The total change is (1/2)2*3=(1/4)*3=2/4 <1.
This cone has a smaller volume than the given cone.

Answer D: A cone with 1/3 times the radius and 10 times the height.
The total change is (1/3)2*10=(1/9)*10=10/9 >1
This cone has a bigger volume than the given cone.

The correct answer is D– A cone with 1/3 times the radius and 10 times the height has a bigger volume than the given cone.

You just finished studying volume word problems topic, the second topic of additional topics in math!

Continue studying the next additional topic in math- congruence and similarity.