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SSAT Upper Level Math : How to find the perimeter of a right triangle

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Example Questions

Example Question #1 :一个三角形的周长

What is the perimeter of a right triangle with hypotenuseand a leg of length?

Possible Answers:

210

119

126

175

It cannot be determined from the information given.

Correct answer:

210

Explanation:

Using the Pythagorean Theorem, the length of the second leg can be determined.

a^2+b^2=c^2

We are given the length of the hypotenuse and one leg.

a=35, c=91

$b^2=c^2-a^2\rightarrow b=\sqrt{c^2-a^2}$

$\sqrt{91^{2}-35^{2}} = \sqrt{8,281-1,225} = \sqrt{7,056} = 84$

The perimeter of the triangle is the sum of the lengths of the sides.

P=35+84+91 = 210

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

Which of these polygons has the same perimeter as a right triangle with legs 6 feet and 8 feet?

Possible Answers:

A regular decagon with sidelength one yard.

None of the other responses is correct.

A regular hexagon with sidelength one yard.

A regular octagon with sidelength one yard.

A regular pentagon with sidelength one yard.

Correct answer:

A regular octagon with sidelength one yard.

Explanation:

A right triangle with legs 6 feet and 8 feet has hypotentuse 10 feet, as this is a right triangle that confirms to the well-known Pythagorean triple 6-8-10. The perimeter is therefore 6 + 8 + 10 = 24 feet, or 8 yards.

We are looking for a polygon with this perimeter. Each choice is a polygon with all sides one yard long, so we want the polygon with eight sides - the regular octagon is the correct choice.

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

The lengths of the legs of a right triangle areunits andunits. What is the perimeter of this right triangle?

Possible Answers:

units

112 units

units

Correct answer:

112 units

Explanation:

First, we need to use the Pythagorean Theorem to find the hypotenuse of the triangle.

$14^2+48^2=\text{Hypotenuse}^2$

$196+2304=\text{Hypotenuse}^2$

$\text{Hypotenuse}^2=2500$

$\text{Hypotenuse}=\sqrt{2500}=50$

Now, add up all three side lengths to find the perimeter of the triangle.

50+14+48=112

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Example Question #2 :Right Triangles

A right triangle has leg lengths of $7\:cm$ and $8\:cm$ . Find the perimeter of this triangle.

Possible Answers:

$164\:cm$

$\sqrt{113}\:cm$

$15+\sqrt{113}\:cm$

$15\:cm$

Correct answer:

$15+\sqrt{113}\:cm$

Explanation:

First, use the Pythagorean Theorem to find the length of the hypotenuse.

a^2+b^2=c^2

Substituting inandforand(the lengths of the triangle's legs), we get:

7^2+8^2=c^2

49+64=c^2

113=c^2

$\sqrt{113}=\sqrt{c^2}$

$\sqrt{113}=c$

Now, add up the three sides to find the perimeter:

$7+8+\sqrt{113}=15+\sqrt{113}$

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Example Question #3 :Right Triangles

What is the perimeter of a right triangle with legs of lengthand, respectively?

Possible Answers:

$\sqrt{74}$

$7+\sqrt{74}$

$5+\sqrt{74}$

$12+\sqrt{74}$

Correct answer:

$12+\sqrt{74}$

Explanation:

In order to find the perimeterof the right triangle, we need to first find the missing length of the hypotenuse. In order to find the hypotenuse, use the Pythagorean Theorem:

$a^{2}+ b^{2}=c^{2}$ , whereandare the lengths of the legs of the triangle, andis the length of the hypotenuse.

Substituting in our known values:

$5^{2}+ 7^{2}=c^{2}$

$25+ 49=c^{2}$

$74=c^{2}$

$\sqrt{74}=c$

Now that we have the lengths of all sides of the right triangle, we can calculate the perimeter:

$P=5+7+\sqrt{74}$

$P=12+\sqrt{74}$

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Example Question #4 :Right Triangles

What is the perimeter of a right triangle with a hypotenuse of lengthand a leg of length?

Possible Answers:

Not enough information provided

$22+ 2\sqrt{33}$

$14+ 2\sqrt{33}$

$2\sqrt{33}$

$8+ 2\sqrt{33}$

Correct answer:

$22+ 2\sqrt{33}$

Explanation:

In order to find the perimeterof the right triangle, we need to first find the missing length of the second leg. In order to find the second leg, use the Pythagorean Theorem:

$a^{2}+ b^{2}=c^{2}$ , whereandare the lengths of the legs of the triangle, andis the length of the hypotenuse.

Substituting in our known values:

$8^{2}+ b^{2}=14^{2}$

Subtracting 8^2 from each side of the equation lets us isolate the variable for which we are solving:

$b^{2}=14^{2}-8^{2}$

$b^{2}=196-64$

$b^{2}=132$

$b=\sqrt{132}$

$b=\sqrt{33\times 4}$

$b=2\sqrt{33}$

Now that we have the lengths of all three sides of the right triangle, we can calculate the perimeter:

$P=14+8+2\sqrt{33}$

$P=22+2\sqrt{33}$

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Example Question #5 :Right Triangles

Find the perimeterof a right triangle with two legs of lengthand, respectively.

Possible Answers:

$12+4\sqrt{13}$

$4\sqrt{13}$

Not enough information provided

$8+4\sqrt{13}$

$20+4\sqrt{13}$

Correct answer:

$20+4\sqrt{13}$

Explanation:

In order to find the perimeterof the right triangle, we need to first find the missing length of the hypotenuse. In order to find the length of the hypotenuse, use the Pythagorean Theorem:

$a^{2}+ b^{2}=c^{2}$ , whereandare the lengths of the legs of the triangle, andis the length of the hypotenuse.

Substituting in our known values:

$8^{2}+ 12^{2}=c^{2}$

$64+144=c^{2}$

$208=c^{2}$

$\sqrt{208}=c$

$\sqrt{13\times 16}=c$

$4\sqrt{13}=c$

Now that we have all three sides of the right triangle, we can calculate the perimeter:

$P=8+12+4\sqrt{13}$

$P=20+4\sqrt{13}$

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SSAT Upper Level Math : How to find the perimeter of a right triangle

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #1 :一个三角形的周长

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

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

Example Question #2 :Right Triangles

Example Question #3 :Right Triangles

Example Question #4 :Right Triangles

Example Question #5 :Right Triangles

All SSAT Upper Level Math Resources

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