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HiSET: Math : Measurement and Geometry

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Example Questions

Example Question #81 : Measurement And Geometry

On the graph of the equation

$x^{2}+ y^{2} = 64$ ,

perform a dilation with center (0,4) and scale factor $\frac{1}{3}$ .

Give the equation of the resulting circle.

Possible Answers:

None of the other choices gives the correct response.

$x^{2}+\left (y- \frac{4}{3} \right )^{2} = \frac{16}{9}$

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{16}{9}$

$x^{2}+\left (y- \frac{4}{3} \right )^{2} = \frac{64}{9}$

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{64}{9}$

Correct answer:

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{64}{9}$

Explanation:

The graph of the equation

$(x-h)^{2}+(y-k)^{2} = r^{2}$

is a circle with center (h, k) and radius .

$x^{2}+ y^{2} = 64$ ,

$(x-0) ^{2}+ (y-0)^{2} = 8^{2}$

is a circle with center at origin (0,0) and radius 8.

A dilation of a circle with scale factor $\frac{1}{3}$ will result in multiplying that radius by $\frac{1}{3}$ , so the radius of the circle will be $8\cdot \frac{1}{3}= \frac{8}{3}$ .

To find the center of the image, note that the origin is 4 units below (0,4) . The center of the new circle must be $4 \cdot \frac{1}{3} = \frac{4}{3}$ units below (0,4) , so this center will be $\left ( 0,4- \frac{4}{3} \right )$ , or $\left ( 0, \frac{8}{3} \right )$ . See the figure below:

Substituting in the circle formula, this is

$(x-0)^{2}+\left (y- \frac{8}{3} \right )^{2} = \left (\frac{8}{3} \right )^{2}$ ,

$x^{2}+\left (y- \frac{8}{3} \right )^{2} = \frac{64}{9}$ .

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Example Question #82 : Measurement And Geometry

Obtuse triangle

On the above obtuse triangle perform a dilation of scale factor $\frac{1}{2}$ with the center of the dilation at the circumcenter of the triangle. Let the images of , , and be , , and C ' , respectively.

Which of the following correctly shows $\bigtriangleup A'B'C'$ relative to $\bigtriangleup ABC$ ?

Possible Answers:

Obtuse triangle

Correct answer:

Obtuse triangle

Explanation:

The circumcenter of a triangle can be located by finding the intersection of the perpendicular bisectors of the three sides of the triangle. The perpendicular bisectors are shown below, with point of intersection :

Obtuse triangle

Construct $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ . A dilation of scale factor $\frac{1}{2}$ with center can be performed by letting A ' , , and be the midpoints of $\overline{AO}$ , $\overline{BO}$ , and $\overline{CO}$ , respectively:

Obtuse triangle

Removing the perpendicular bisectors and , we see that the correct choice is the figure

Obtuse triangle

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Example Question #1 : Circumference

Find the circumference of a circle with the following diameter:

d=14

Possible Answers:

$15\pi$

$16\pi$

$14\pi$

$7\pi$

$8\pi$

Correct answer:

$14\pi$

Explanation:

A circle's circumference is calculated using the following formula:

$C=2\pi r$

In this equation, the variable, , is the circle's radius.

In our problem we are given the diameter. The diameter is related to the radius in the following manner:

d=2r

Let's rewrite our formula by substituting the diameter for the radius.

$C=d\pi$

Let's substitute and solve.

$C=14\pi$

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Example Question #2 : Equations Of Circles

Which of the following is closest to the length of a 270-degree arc of a circle with radius 20?

Possible Answers:

105

Correct answer:

Explanation:

A circle with radius 20 has as its circumference $2 \pi$ times this, or

$C = 2 \pi r = 2 \pi (20 )= 40 \pi$

There are 360 degrees in a circle, so a 270-degree arc is

$\frac{270}{360} = \frac{270 \div 90}{360 \div 90} = \frac{3}{4}$

of the circle. Therefore, the length of this arc is

$L= \frac{3}{4} C = \frac{3}{4} (40 \pi)= 30 \pi$

$\pi \approx 3.14159$ ,

so a reasonable estimate of the length of this arc is

$L \approx 30 (3.14159) \approx 94.2477$

Of the five choices, 95 comes closest to the correct length.

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Example Question #85 : Measurement And Geometry

A circle on the coordinate plane has a diameter with endpoints (3,4) and (9, 10) . Give its equation.

Possible Answers:

Insufficient information exists to determine its equation.

$(x-7)^{2} + (y-6)^{2}= 36$

$(x-7)^{2} + (y-6)^{2}= 18$

$(x-6)^{2} + (y-7)^{2}= 36$

$(x-6)^{2} + (y-7)^{2}= 18$

Correct answer:

$(x-6)^{2} + (y-7)^{2}= 18$

Explanation:

The equation, in standard form, of a circle on the coordinate plane with center (h, k) and radius is

$(x-h)^{2} + (y-k)^{2}= r^{2}$ .

It is necessary to find the center and the radius of the circle in order to determine its equation.

The center of the circle is the midpoint of a given diameter. The midpoint $\left ( x_{M}, y_{M} \right )$ of a segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be found by applying the midpoint formula:

$x_{M}= \frac{x_{1}+ x_{2} }{2}$

$y_{M}= \frac{y_{1}+ y_{2} }{2}$

Setting $x_{1}= 3, x_{2}= 9$ :

$x_{M}= \frac{3+9 }{2}= \frac{12}{2} = 6$

Setting $y_{1}= 4, y_{2}= 10$

$y_{M}= \frac{4+10 }{2} = \frac{14}{2} = 7$

The midpoint, and the center of the circle, is at (6, 7) .

The radius of the circle is the distance from the center to an endpoint, so we can use the distance formula

$r = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$ .

Substitute $x_{1}= 6, x_{2}= 9 , y_{1}= 7, y_{2}= 10$ :

$r = \sqrt{(9-6)^{2}+(10-7 )^{2} }$

$= \sqrt{3^{2}+3 ^{2} }$

$= \sqrt{9+9 }$

$= \sqrt{18 }$

and

$r^{2} = 18$

Now, set $r^{2} = 18 , h = 6, k= 7$ in the standard form of the equation of a circle, and the result is

$(x-6)^{2} + (y-7)^{2}= 18$

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Example Question #86 : Measurement And Geometry

A circle on the coordinate plane has a radius with endpoints (3,4) and (9, 10) . Give its equation.

Possible Answers:

$(x-3)^{2}+ (y-9)^{2} = 72$

$(x-3)^{2}+ (y-9)^{2} = 18$

$(x-4)^{2}+ (y-10)^{2} = 72$

$(x-4)^{2}+ (y-10)^{2} = 18$

Insufficient information exists to determine its equation.

Correct answer:

Insufficient information exists to determine its equation.

Explanation:

The equation, in standard form, of a circle on the coordinate plane with center (h, k) and radius is

$(x-h)^{2} + (y-k)^{2}= r^{2}$ .

It is necessary to find the center and the radius of the circle in order to determine its equation. The radius of the circle can be determined by applying the distance formula to the coordinates of the endpoints of the radius. However, it is not clear from the information given which endpoint is the center of the circle. Therefore, while can be determined, (h, k) cannot. The correct response is that insufficient information exists to determine its equation.

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Example Question #87 : Measurement And Geometry

What is the circumference of a circle with an area of $1369\pi$ ?

Possible Answers:

$70\pi$

$37\pi$

$35\pi$

$74\pi$

Correct answer:

$74\pi$

Explanation:

Step 1: Since we are given the area, let's first find the radius of the circle.

If the area of a circle is calculated by: $A=\pi r^2$ , we can plug in the values that are given to us. $A=1369\pi$ .

We have $1369\pi=\pi r^2$ . We will divide by $\pi$ .

1369=r^2 . To find the radius, we will take the square root of both sides.

So, r=37 .

Step 2: Now that we have the radius, we will now find the circumference.

The circumference formula by using the radius is $C=2\pi r$ .

Plug in all the information that we have...

$C=2\pi(37)\Rightarrow C=74\pi$

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Example Question #88 : Measurement And Geometry

A circle on the coordinate plane has center (5, -9) and circumference $50 \pi$ .

Give its equation.

Possible Answers:

$(x-5)^{2}+ (y+9)^{2}= 25$

$(x-5)^{2}+ (y+9)^{2}= 5$

$(x-5)^{2}+ (y+9)^{2}= 625$

$(x-5)^{2}+ (y+9)^{2}= 2,500$

$(x-5)^{2}+ (y+9)^{2}= 50$

Correct answer:

$(x-5)^{2}+ (y+9)^{2}= 625$

Explanation:

The standard form of the equation of a circle with center at (h, k) and radius is

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The circumference of the circle is $50 \pi$ ; to find the radius, divide this circumference by $2 \pi$ :

$r = \frac{C}{2\pi }= \frac{50 \pi }{2\pi } = 25$

Now, set h = 5, k = -9, r = 25 in the equation of the circle:

$(x-5)^{2}+ (y-(-9))^{2}= 25^{2}$

Simplifying:

$(x-5)^{2}+ (y+9)^{2}= 625$ ,

the correct equation.

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Example Question #89 : Measurement And Geometry

A circle on the coordinate plane has center (5, -9) and area $50 \pi$ .

Give its equation.

Possible Answers:

$(x-5)^{2}+ (y+9)^{2}= 25$

$(x-5)^{2}+ (y+9)^{2}= 2,500$

$(x-5)^{2}+ (y+9)^{2}= 50$

$(x-5)^{2}+ (y+9)^{2}= 625$

$(x-5)^{2}+ (y+9)^{2}= 5$

Correct answer:

$(x-5)^{2}+ (y+9)^{2}= 50$

Explanation:

The standard form of the equation of a circle with center at (h, k) and radius is

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The area of the circle is $50 \pi$ ; to find the radius, substitute in the formula:

$\pi r^{2}= A$

$\pi r^{2}= 50 \pi$

$\frac{ \pi r^{2}}{\pi} = \frac{50 \pi}{\pi}$

$r^{2} = 50$

We do not need to find . Set $h = 5, k = -9, r ^{2} = 50$ in the equation of the circle:

$(x-5)^{2}+ (y-(-9))^{2}= 50$

Simplifying:

$(x-5)^{2}+ (y+9)^{2}=50$ ,

the correct equation.

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Example Question #1 : Circumference

A circle on the coordinate plane has its center at (2,9 ) and passes through (10, 3) . Give its circumference.

Possible Answers:

$25 \pi$

$100 \pi$

$10 \pi$

$20 \pi$

The information given is insufficient to answer the question

Correct answer:

$20 \pi$

Explanation:

The radius of the circle is the distance between its center and a point that it passes through. This radius can be calculated using the distance formula:

$r = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$ .

Setting $x_{1} = 2, x_{2} = 10, y_{1}=9, y_{2} =3$ :

$r = \sqrt{(10-2)^{2}+(3-9)^{2} }$

$= \sqrt{8^{2}+(-6)^{2} }$

$= \sqrt{64+36}$

$= \sqrt{100}$

=10

The circumference of a circle is equal to $2 \pi$ multiplied by the radius, so

$C= 2 \pi r = 2 \pi (10) = 20 \pi$ ,

the correct response.

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HiSET: Math : Measurement and Geometry

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #81 : Measurement And Geometry

Example Question #82 : Measurement And Geometry

Example Question #1 : Circumference

Example Question #2 : Equations Of Circles

Example Question #85 : Measurement And Geometry

Example Question #86 : Measurement And Geometry

Example Question #87 : Measurement And Geometry

Example Question #88 : Measurement And Geometry

Example Question #89 : Measurement And Geometry

Example Question #1 : Circumference

All HiSET: Math Resources

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