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HiSET: Math : Measurement and Geometry

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125 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Equations Of Circles

Give the circumference of the circle on the coordinate plane with equation

$x^{2}+y^{2}+ 6x-20y+ 9=0$ .

Possible Answers:

$40 \pi$

$100 \pi$

$10 \pi$

$50 \pi$

$20 \pi$

Correct answer:

$20 \pi$

Explanation:

We must first rewrite the equation of the circle in standard form

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

will be the radius.

$x^{2}+y^{2}+ 6x-20y+ 9=0$

Subtract 9 from both sides, and rearrange the terms remaining on the left as follows:

$x^{2}+y^{2}+ 6x-20y+ 9- 9 =0- 9$

$x^{2}+ 6x + \underline{\; \; \; \; \; \; }+y^{2}-20y+ \underline{\; \; \; \; \; \; } = - 9$

Note that blanks have been inserted after the linear terms. In these blanks, complete two perfect square trinomials by dividing linear coefficients by 2 and squaring:

$(6 \div 2)^{2} = 3^{2} = 9$

$(20 \div 2)^{2} = 10^{2} = 100$

Add to both sides, as follows:

$x^{2}+ 6x + 9+y^{2}-20y+ 100 = - 9+9+ 100$

$x^{2}+ 6x + 9+y^{2}-20y+ 100 = 100$

Rewrite the expression on the left as the sum of the squares of two binomials.

$(x+3)^{2} + (y-10)^{2} = 100$

The equation is now in standard form.

$r^{2} = 100$

The radius is the square root of this, or

r = 10

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

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

the correct response.

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Example Question #141 : Hi Set: High School Equivalency Test: Math

Find the area of a circle with the following radius:

r=6

Possible Answers:

$12\pi$

$36\pi$

$18\pi$

$6\pi$

$66\pi$

Correct answer:

$36\pi$

Explanation:

The area of a circle is found using the following formula:

$A=\pi r^2$

In this formula the variable, , is the radius. Let's substitute in our known values and solve for the area.

$A=\pi 6^2$

$A=36\pi$

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Example Question #151 : Hi Set: High School Equivalency Test: Math

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

Possible Answers:

$10 \pi$

$20 \pi$

$100 \pi$

$25 \pi$

The information given is insufficient to answer the question

Correct answer:

$100 \pi$

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 area of a circle can be calculated by substituting 10 for in the equation

$A = \pi r^{2}$

$= \pi (10)^{2}$

$= 100 \pi$ ,

the correct response.

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Example Question #152 : Hi Set: High School Equivalency Test: Math

Give the area of the circle on the coordinate plane with equation

$x^{2}+y^{2}+ 6x-20y+ 9=0$ .

Possible Answers:

$20 \pi$

$40 \pi$

$100 \pi$

$10 \pi$

$50 \pi$

Correct answer:

$100 \pi$

Explanation:

We must first rewrite the equation of the circle in standard form

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

will be the radius.

$x^{2}+y^{2}+ 6x-20y+ 9=0$

Subtract 9 from both sides, and rearrange the terms remaining on the left as follows:

$x^{2}+y^{2}+ 6x-20y+ 9- 9 =0- 9$

$x^{2}+ 6x + \underline{\; \; \; \; \; \; }+y^{2}-20y+ \underline{\; \; \; \; \; \; } = - 9$

Note that blanks have been inserted after the linear terms. In these blanks, complete two perfect square trinomials by dividing linear coefficients by 2 and squaring:

$(6 \div 2)^{2} = 3^{2} = 9$

$(20 \div 2)^{2} = 10^{2} = 100$

Add to both sides, as follows:

$x^{2}+ 6x + 9+y^{2}-20y+ 100 = - 9+9+ 100$

$x^{2}+ 6x + 9+y^{2}-20y+ 100 = 100$

Rewrite the expression on the left as the sum of the squares of two binomials.

$(x+3)^{2} + (y-10)^{2} = 100$

The equation is now in standard form.

$r^{2} = 100$

The area of this circle can be found using this formula:

$A = \pi r^{2} = \pi (100) = 100 \pi$ ,

the correct response.

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Example Question #1 : Midpoint Of Line Segments

Give the coordinates of the midpoint of the line segment whose endpoints are the intercepts of the line of the equation

2x+ 5y = 40 .

Possible Answers:

(-4, -14)

(4, 14)

(-14, -4)

None of the other choices gives the correct response.

(14, 4)

Correct answer:

(14, 4)

Explanation:

The x-intercept and y-intercept of the line of an equation can be found by substituting 0 for and , respectively, as follows:

x-intercept:

2x+ 5y = 40

Substitute and simplify:

2x+ 5(0)= 40

2x+ 0 = 40

2x= 40

Solve for by dividing by 2 on both sides:

$2x \div 2 = 40 \div 2$

x = 20

The x-intercept of the line is located at (20,0)

The y-intercept can be found by setting x= 0 and solving for in a similar fashion:

2x+ 5y = 40

2 (0)+ 5y = 40

0 + 5y = 40

5y = 40

$5y \div 5 = 40 \div 5$

y = 8

The y-intercept is located at (0,8) .

The midpoint of a line segment, given its endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , is located at

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$ ;

substituting accordingly,

$\frac{x_{1}+x_{2}}{2} =\frac{20+8}{2} = \frac{28}{2} = 14$

$\frac{y_{1}+y_{2}}{2} = \frac{ 0+8}{2} = \frac{8}{2} = 4$

The midpoint is located at (14, 4) .

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Example Question #1 : Midpoint Of Line Segments

What is the midpoint of the line segment created by the points (-2,16) and (3,7) ?

Possible Answers:

$(-\frac{5}{2},-\frac{13}{2})$

$(\frac{3}{2},\frac{3}{2})$

$(\frac{1}{2},\frac{23}{2})$

$(\frac{9}{2},-\frac{17}{2})$

$(\frac{23}{2},\frac{3}{2})$

Correct answer:

$(\frac{1}{2},\frac{23}{2})$

Explanation:

The midpoint of the line segment between two points

$(x_{0},y_{0})$ and $(x_{1},y_{1})$

is given by the formula

$(\frac{x_{0}+x_{1}}{2},\frac{y_{0}+y_{1}}{2})$ .

Thus, since the line segment is generated by

(-2,16) and (3,7)

applying the midpoint formula we have

$(\frac{-2+3}{2},\frac{16+7}{2})= (\frac{1}{2},\frac{23}{2})$

as the midpoint.

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Example Question #1 : Midpoint Of Line Segments

What is the midpoint between A(-12,10) and B(14,-6) ?

Possible Answers:

(-1,2)

(2,1)

(-2,-1)

(1,2)

Correct answer:

(1,2)

Explanation:

Step 1: Recall the midpoint Formula....

$(\dfrac {x_1+x_2}{2},\dfrac {y_1+y_2}{2})$ , where (x_1,y_1) is the first point and (x_2,y_2) is the second point

Step 2: Identify the points and their specific terms...

x_1=-12, y_1=10, x_2=14, y_2=-6

Step 3: Plug in the values into the midpoint formula...

$(\dfrac {-12+14}{2},\dfrac {10-6}{2})$

Simplify...

$(\dfrac {2}{2},\dfrac {4}{2})=(1,2)$

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Example Question #151 : Hi Set: High School Equivalency Test: Math

$\tan N = \frac{\sqrt{y}}{5}$

Evaluate $\sin N$ in terms of .

Possible Answers:

$\frac{5} {\sqrt{y}}$

$\frac{ \sqrt{y+25}}{5}$

$\frac{5}{ \sqrt{y+25}}$

$\sqrt{ \frac{y}{y+25}}$

$\sqrt{ \frac{y+25}{y}}$

Correct answer:

$\sqrt{ \frac{y}{y+25}}$

Explanation:

Suppose we allow O, A, H be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring .

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, so

$\tan N = \frac{O}{A} =\frac{\sqrt{y}}{5}$

We can set the lengths of the opposite and adjacent legs to $\sqrt{y}$ and 5, respectively. The length of the hypotenuse can be determined using the Pythagorean Theorem:

$H = \sqrt{O^{2}+A^{2}} = \sqrt{( \sqrt{y})^{2}+5^{2}} = \sqrt{y+25}$

The sine of the angle is equal to the ratio of the length opposite leg to that of the hypotenuse, so

$\sin N = \frac{O}{H} = \frac{\sqrt{y }}{ \sqrt{y+25}} =\sqrt{ \frac{y}{y+25}}$ .

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Example Question #2 : Problems Involving Right Triangle Trigonometry

$\cos N = \frac{x}{3}$

Evaluate $\sin N$ in terms of .

Possible Answers:

$\frac{x}{\sqrt{x^{2}-9}}$

$\frac{3}{x}$

$\frac{\sqrt{x^{2}-9}}{3}$

$\frac{3}{\sqrt{x^{2}-9}}$

$\frac{\sqrt{x^{2}-9}}{x}$

Correct answer:

$\frac{\sqrt{x^{2}-9}}{3}$

Explanation:

Suppose we allow O, A, H be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring . The cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = \frac{A}{H} = \frac{x}{3}$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = \sqrt{H^{2}- A^{2}} = \sqrt{x^{2}-3^{2}} = \sqrt{x^{2}-9 }$

The sine of the angle is equal to the ratio of the length of the opposite leg to that of the hypotenuse, so

$\sin N = \frac{O}{H} = \frac{\sqrt{x^{2}-9 }}{3}$ .

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Example Question #1 : Problems Involving Right Triangle Trigonometry

$\tan N = \frac{\sqrt{y}}{5}$

Evaluate $\cos N$ in terms of .

Possible Answers:

$\frac{5}{ \sqrt{y+25}}$

$\frac{ \sqrt{y+25}}{5}$

$\sqrt{ \frac{y+25}{y}}$

$\sqrt{ \frac{y}{y+25}}$

$\frac{5} {\sqrt{y}}$

Correct answer:

$\frac{5}{ \sqrt{y+25}}$

Explanation:

Suppose we allow O, A, H be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring .

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, so

$\tan N = \frac{O}{A} =\frac{\sqrt{y}}{5}$

We can set the lengths of the opposite and adjacent legs to $\sqrt{y}$ and 5, respectively. The length of the hypotenuse can be determined using the Pythagorean Theorem:

$H = \sqrt{O^{2}+A^{2}} = \sqrt{( \sqrt{y})^{2}+5^{2}} = \sqrt{y+25}$

The cosine of the angle is equal to the ratio of the length of the adjacent leg to that of the hypotenuse, so

$\cos N = \frac{A}{H} = \frac{5}{ \sqrt{y+25}}$ .

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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 #1 : Equations Of Circles

Example Question #141 : Hi Set: High School Equivalency Test: Math

Example Question #151 : Hi Set: High School Equivalency Test: Math

Example Question #152 : Hi Set: High School Equivalency Test: Math

Example Question #1 : Midpoint Of Line Segments

Example Question #1 : Midpoint Of Line Segments

Example Question #1 : Midpoint Of Line Segments

Example Question #151 : Hi Set: High School Equivalency Test: Math

Example Question #2 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Problems Involving Right Triangle Trigonometry

All HiSET: Math Resources

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