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Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

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All Intermediate Geometry Resources

8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : How To Find The Length Of The Hypotenuse Of An Acute / Obtuse Triangle

An isosceles right triangle has two sides with a length of inches each. Find the length of the hypotenuse.

Possible Answers:

$20\sqrt{2}in$

25in

40in

$\sqrt{20}in$

Correct answer:

$20\sqrt{2}in$

Explanation:

To find the length of the hypotenuse, apply the Pythagorean Theorem: a^2+b^2=c^2 , where the length of the hypotenuse.

Thus, the solution is:

20^2+20^2=c^2

400+400=c^2

c^2=800

$c=\sqrt{800}=\sqrt{400\times 2}=\sqrt{400}\sqrt{2}=20\sqrt{2}$

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Example Question #1 : How To Find The Length Of The Hypotenuse Of An Acute / Obtuse Triangle

A scalene right triangle has one side length of inches and another side length of inches. Find the length of the hypotenuse.

Possible Answers:

49in

$\sqrt{74} in$

74in

$\sqrt{49}in$

Correct answer:

$\sqrt{74} in$

Explanation:

To find the length of the hypotenuse, apply the Pythagorean Theorem: a^2+b^2=c^2 , where the length of the hypotenuse.

Thus, the solution is:

5^2+7^2=c^2

c^2=25+49=74

$c=\sqrt{74}$

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Example Question #561 : Intermediate Geometry

In ΔABC: a = 8, b = 13, c = 9.

Find the area of ΔABC (to the nearest tenth).

Possible Answers:

29.6

33.9

40.4

35.5

36.1

Correct answer:

35.5

Explanation:

In order to determine the area of a non-right triangle, we can use Heron's formula:

$Area =\sqrt{s(s-a)(s-b)(s-c)}$
$s=\frac{a+b+c}{2}$

Using the information from the question, we obtain:
$s=\frac{8+13+9}{2}=15$
$Area =\sqrt{15(15-8)(15-13)(15-9)}=35.5$

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Example Question #562 : Intermediate Geometry

In ΔABC: a = 16, b = 11, c = 19.

Find the area of ΔABC (to the nearest tenth).

Possible Answers:

75.5

78.8

87.9

85.2

80.1

Correct answer:

87.9

Explanation:

In order to determine the area of a non-right triangle, we can use Heron's formula:

$Area =\sqrt{s(s-a)(s-b)(s-c)}$
$s=\frac{a+b+c}{2}$

Using the information from the question, we obtain:
$s=\frac{16+11+19}{2}=23$
$Area =\sqrt{23(23-16)(23-11)(23-19)}=87.9$

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Example Question #563 : Intermediate Geometry

Find the height of a triangle if its base is $12\:cm$ long and its area is $24\:cm^2$ .

Possible Answers:

$2\:cm$

$6\:cm$

$12\:cm$

$4\:cm$

Correct answer:

$4\:cm$

Explanation:

The formula to find the area of a triangle is

$\text{Area}=\frac{\text{base }\times\text{height}}{2}$

Substitute in the given values for area and base to solve for the height, :

$24=\frac{12h}{2}$

12h=48

$h=4\:cm$

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Example Question #4 : How To Find The Area Of An Acute / Obtuse Triangle

In terms of , what is the area of a triangle with a height of $2a\:units$ and a base of $5a\:units$ ?

Possible Answers:

$15a^2\:units^2$

$20a^2\:units^2$

$5a^2\:units^2$

$10a^2\:units^2$

Correct answer:

$5a^2\:units^2$

Explanation:

The formula to find the area of a triangle is

$\text{Area}=\frac{base\times height}{2}$

Substitute in the given values for the base and the height to find the area.

$\text{Area}=\frac{5a\times 2a}{2}=\frac{10a^2}{2}=5a^2\:units^2$

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Example Question #1 : How To Find The Area Of An Acute / Obtuse Triangle

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$191.2\:units^2$

$207.7\:units^2$

$134.8\:units^2$

$201.4\:units^2$

Correct answer:

$207.7\:units^2$

Explanation:

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side, the triangle's height, and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta=\frac{adjacent}{hypotenuse}$

$\cos(33^{\circ})=\frac{height}{11}$

Now, solve for the height.

$\text{Height}=11\cos(33^{\circ})=9.23$

Now you can find the area of the triangle:

$\text{Area}=\frac{45\times9.23}{2}=207.7\:units^2$

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Example Question #3 : How To Find The Area Of An Acute / Obtuse Triangle

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$6.8\:units^2$

$8.2\:units^2$

$12.4\:units^2$

$18.8\:units^2$

Correct answer:

$18.8\:units^2$

Explanation:

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side—the triangle's height—and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta =\frac{adjacent}{hypotenuse}$

$\cos(20^{\circ})=\frac{height}{5}$

Now, solve for the height.

$\text{Height}=5\cos(20^{\circ})=4.70$

Now you can find the area.

$\text{Area}=\frac{8\times 4.70}{2}=18.8\:units^2$

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Example Question #1 : How To Find The Area Of An Acute / Obtuse Triangle

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$226.1\:units^2$

$100.7\:units^2$

$199.6\:units^2$

$159.2\:units^2$

Correct answer:

$226.1\:units^2$

Explanation:

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Remember that $\text{cosine }\theta=\frac{adjacent}{hypotenuse}$

$\cos(24^{\circ})=\frac{height}{34}$

Now, solve for the height.

$\text{Height}=11\cos(24^{\circ})=10.05$

Now you can find the area.

$\text{Area}=\frac{45\times 10.05}{2}=226.1\:units^2$

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Example Question #122 : Triangles

Find the area of the triangle below. Round to the nearest tenths place.

Possible Answers:

$7.6\:units^2$

$4.5\:units^2$

$9.2\:units^2$

$10.0\:units^2$

Correct answer:

$10.0\:units^2$

Explanation:

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side—the triangle's height—and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta=\frac{adjacent}{hypotenuse}$

$\cos(18^{\circ})=\frac{height}{3}$

Now, solve for the height.

$\text{Height}=3\cos(18^{\circ})=2.85$

Now you can find the area.

$\text{Area}=\frac{7\times 2.85}{2}=10.0\:units^2$

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All Intermediate Geometry Resources

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Intermediate Geometry : Intermediate Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #1 : How To Find The Length Of The Hypotenuse Of An Acute / Obtuse Triangle

Example Question #1 : How To Find The Length Of The Hypotenuse Of An Acute / Obtuse Triangle

Example Question #561 : Intermediate Geometry

Example Question #562 : Intermediate Geometry

Example Question #563 : Intermediate Geometry

Example Question #4 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #1 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #3 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #1 : How To Find The Area Of An Acute / Obtuse Triangle

Example Question #122 : Triangles

All Intermediate Geometry Resources

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