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HSPT Math : Concepts

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Example Questions

Example Question #561 : Algebra

Axes 4

Refer to the above diagram. Which line is perpendicular to the line of the equation 8x - 12y = 71 ?

Possible Answers:

Correct answer:

Explanation:

Rewrite the equation of the perpendicular line in slope-intercept form y = mx+b as follows:

8x - 12y = 71

8x - 12y- 8x = 71 - 8x

- 12y =- 8x+ 71

$-\frac{1}{12} \cdot (- 12y )= -\frac{1}{12} \cdot (- 8x+ 71)$

$y = \frac{8}{12}x - \frac{71}{12}$

$y = \frac{2}{3}x - \frac{71}{12}$

The slope of the line is the coefficient of , which is $\frac{2}{3}$ . The slope of a line perpendicular to this is the opposite of the reciprocal of $\frac{2}{3}$ , which is $-\frac{3}{2}$ .

The slope of a line is its ratio of rise to run. The slopes of the four lines in the diagram are given below, with their justification shown.

Axes 4

Of the four lines, the line marked has slope $-\frac{3}{2}$ .

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Example Question #562 : Algebra

Axes 4

Refer to the above diagram. Which of the above points is on the line of the equation y = 2x+ 7 ?

Possible Answers:

Correct answer:

Explanation:

All four points on the diagram have -coordinate , so set x= -1 in the equation to find the corresponding -coordinate:

y = 2x+ 7

y = 2(-1)+ 7 = -2 + 7 = 5

The correct point is the one with coordinates (-1, 5) . This point is one unit left and five units up from the orgin - the point at this location is .

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Example Question #1181 : Concepts

Axes 4

Refer to the above diagram. Which line is parallel to the line of the equation 8x - 12y = 55 ?

Possible Answers:

Correct answer:

Explanation:

Rewrite the equation of the parallel line in slope-intercept form y = mx+b as follows:

8x - 12y = 55

8x - 12y- 8x = 55 - 8x

- 12y =- 8x+ 55

$-\frac{1}{12} \cdot (- 12y )= -\frac{1}{12} \cdot (- 8x+ 55)$

$y = \frac{8}{12}x - \frac{55}{12}$

$y = \frac{2}{3}x - \frac{55}{12}$

The slope of the line is the coefficient of , which is $\frac{2}{3}$ . The slope of a line parallel to this is also $\frac{2}{3}$ .

The slope of a line is its ratio of rise to run. The slopes of the four lines in the diagram are given below, with their justification shown.

Axes 4

Of the four lines, the line marked also has slope $\frac{2}{3}$ .

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Example Question #1182 : Concepts

Solve for :

$10=\frac{x}{3}+5$

Possible Answers:

x=10

x=30

x=15

x=5

Correct answer:

x=15

Explanation:

To get by itself, you must first get rid of the constant.

To undo the positive , you must subtract it from both sides leaving us with,

$5=\frac{x}{3}$ .

Then you must multiple both sides by and that means,

x=5*3=15 .

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Example Question #1183 : Concepts

What is the area of a triangle with a base of 5 and a height of 20?

Possible Answers:

100

Correct answer:

Explanation:

When searching for the area of a triangle we are looking for the amount of the space enclosed by the triangle.

The equation for area of a triangle is $\frac{1}{2}(base)(height)= Area\: of\: Triangle$

Plug in the numbers for base and height into the equation yielding $\frac{1}{2}(5)(20)= Area\: of\: Triangle$

Then multiply the numbers together to arrive at the answer which is .

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Example Question #2 : Geometry

A trapezoid has height 32 inches and bases 25 inches and 55 inches. What is its area?

Possible Answers:

$800\; \textrm{in}^{2}$

$1,760\; \textrm{in}^{2}$

$1,920\; \textrm{in}^{2}$

$1,280 \; \textrm{in}^{2}$

$1,375\; \textrm{in}^{2}$

Correct answer:

$1,280 \; \textrm{in}^{2}$

Explanation:

Use the following formula, with B = 55,b = 25,h=32 :

$A = \frac{1}{2} (B+b)h = \frac{1}{2} (55+25) \cdot 32 = 1,280$

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Example Question #3 : Geometry

What is the area of a triangle with a base of 22 cm and a height of 9 cm?

Possible Answers:

$45\ cm^{2}$

$99\ cm^{2}$

$198\ cm^{2}$

$98\ cm^{2}$

Correct answer:

$99\ cm^{2}$

Explanation:

Use the area of a triangle formula

$A=\frac{1}{2}bh$

Plug in the base and height. This gives you $99\ cm^{2}$ .

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Example Question #1 : How To Find The Area Of A Right Triangle

Right Triangle A has hypotenuse 25 inches and one leg of length 24 inches; Right Triangle B has hypotenuse 15 inches and one leg of length 9 inches; Rectangle C has length 16 inches. The area of Rectangle C is the sum of the areas of the two right triangles. What is the width of Rectangle C?

Possible Answers:

$17 \frac{1}{4} \textrm{ in}$

$6 \frac{7}{8} \textrm{ in}$

$10 \frac{1}{2} \textrm{ in}$

$8 \frac{5}{8} \textrm{ in}$

$14\frac{7}{8} \textrm{ in}$

Correct answer:

$8 \frac{5}{8} \textrm{ in}$

Explanation:

The area of a right triangle is half the product of its legs. In each case, we know the length of one leg and the hypotenuse, so we need to apply the Pythagorean Theorem to find the second leg, then take half the product of the legs:

Right Triangle A:

The length of the second leg is

$\sqrt{25^{2} -24 ^{2}} = \sqrt{625 - 576} = \sqrt{49} = 7$ inches.

The area is

$\frac{1}{2} \times 7 \times 24 = 84$ square inches.

Right Triangle B:

The length of the second leg is

$\sqrt{15^{2} -9 ^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12$ inches.

The area is

$\frac{1}{2} \times 9 \times 12 = 54$ square inches.

The sum of the areas is 84 + 54 = 138 square inches.

The area of a rectangle is the product of its length and its height. Therefore, the height is the quotient of the area and the length, which, for Rectangle C, is $138 \div 16 = 8 \frac{5}{8}$ inches.

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Example Question #4 : Geometry

Right Triangle A has legs of lengths 10 inches and 14 inches; Right Triangle B has legs of length 20 inches and 13 inches; Rectangle C has length 30 inches. The area of Rectangle C is the sum of the areas of the two right triangles. What is the height of Rectangle C?

Possible Answers:

Insufficient information is given to determine the height.

$6 \frac{2}{3} \textrm{ in}$

$10 \textrm{ in}$

$13\frac{1}{3} \textrm{ in}$

$5 \textrm{ in}$

Correct answer:

$6 \frac{2}{3} \textrm{ in}$

Explanation:

The area of a right triangle is half the product of its legs. The area of Right Triangle A is equal to $\frac{1}{2} \times 10 \times 14 = 70$ square inches; that of Right Triangle B is equal to $\frac{1}{2} \times 20 \times 13 = 130$ square inches. The sum of the areas is 70 + 130 = 200 square inches, which is the area of Rectangle C.

The area of a rectangle is the product of its length and its height. Therefore, the height is the quotient of the area and the length, which, for Rectangle C, is $200 \div 30 = 6 \frac{2}{3}$ inches.

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Example Question #5 : Geometry

Heron

Give the area of the above triangle.

Possible Answers:

1,125

450

900

800

675

Correct answer:

900

Explanation:

By Heron's formula, the area of a triangle given its sidelengths is

$A =\sqrt{ s (s-a) (s-b) (s-c)}$

Where a, b, c are the sidelengths and

$s = \frac{1}{2} (a+b+c)$ ,

or half the perimeter.

Setting a = 50, b = 45, c=85 ,

$s = \frac{1}{2} (a+b+c) = \frac{1}{2} (50+45+85) = 90$

Therefore,

$A =\sqrt{ s (s-a) (s-b) (s-c)}$

$A =\sqrt{90 (90 -50) (90 -45) (90 -85)}$

$A =\sqrt{90 \cdot 40 \cdot 45 \cdot 5 }$

$A =\sqrt{810,000 }$

A =900

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HSPT Math : Concepts

Study concepts, example questions & explanations for HSPT Math

All HSPT Math Resources

Example Questions

Example Question #561 : Algebra

Example Question #562 : Algebra

Example Question #1181 : Concepts

Example Question #1182 : Concepts

Example Question #1183 : Concepts

Example Question #2 : Geometry

Example Question #3 : Geometry

Example Question #1 : How To Find The Area Of A Right Triangle

Example Question #4 : Geometry

Example Question #5 : Geometry

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