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HSPT Math : Geometry

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Example Questions

Example Question #1 : Geometry

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:

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

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

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

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

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

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:

$98\ cm^{2}$

$99\ cm^{2}$

$45\ cm^{2}$

$198\ cm^{2}$

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 #4 : Geometry

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:

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

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

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

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

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

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 #5 : 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:

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

$5 \textrm{ in}$

Insufficient information is given to determine the height.

$10 \textrm{ in}$

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

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 #4 : Geometry

Heron

Give the area of the above triangle.

Possible Answers:

675

800

1,125

450

900

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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Example Question #5 : Geometry

Heron

Give the area of the above triangle.

Possible Answers:

240

$30 \sqrt{ 14 }$

160

216

$60 \sqrt{7 }$

Correct answer:

$60 \sqrt{7 }$

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 = 16, b = 20, c=24 ,

$s = \frac{1}{2} (a+b+c) = \frac{1}{2} (16+20+24) = 30$ .

Therefore,

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

$A =\sqrt{ 30 (30-16) (30-20) (30-24)}$

$A =\sqrt{ 30 \cdot 14 \cdot 10 \cdot 6 }$

$A = \sqrt{25,200 }$

$A = \sqrt{3,600 } \cdot\sqrt{7}$

$A = 60 \sqrt{7 }$

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Example Question #172 : Quadrilaterals

Find the area of the following parallelogram:

Note: The formula for the area of a parallelogram is $A=b\times h$ .

Possible Answers:

$32\: in^2$

$30\: in^2$

$60\: in^2$

$50\: in^2$

Correct answer:

$50\: in^2$

Explanation:

The base of the parallelogram is 10, while the height is 5.

$A=b\times h$

$A=10\times5=50\: in^2$

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Example Question #173 : Quadrilaterals

Find the area:

Question_5

Possible Answers:

$\small 24$

$\small 32$

$\small 16$

$\small 12$

Correct answer:

$\small 24$

Explanation:

The area of a parallelogram can be determined using the following equation:

$\small A=bh$

Therefore,

$\small A=8\times3=24$

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Example Question #6 : Geometry

What is the area of a rectangle with length $14\ cm$ and width $12\ cm$ ?

Possible Answers:

$148\; cm^{2}$

$158\;cm^{2}$

$144\; cm^{2}$

$198\;cm^{2}$

$168\;cm^{2}$

Correct answer:

$168\;cm^{2}$

Explanation:

The formula for the area, , of a rectangle when we are given its length, , and width, , is A = L * W .

To calculate this area, just multiply the two terms.

$A = 14\ cm * 12\ cm = 168\; cm^{2}$

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HSPT Math : Geometry

Study concepts, example questions & explanations for HSPT Math

All HSPT Math Resources

Example Questions

Example Question #1 : Geometry

Example Question #2 : Geometry

Example Question #3 : Geometry

Example Question #4 : Geometry

Example Question #5 : Geometry

Example Question #4 : Geometry

Example Question #5 : Geometry

Example Question #172 : Quadrilaterals

Example Question #173 : Quadrilaterals

Example Question #6 : Geometry

All HSPT Math Resources

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