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

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Example Questions

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

One angle of a right triangle measures 45, and the hypotenuse length is 6 centimeters. Give the perimeter of the triangle.

Possible Answers:

$6(2+\sqrt{2})\ cm$

$12\ cm$

$4(1+\sqrt{2})\ cm$

$6\ cm$

$6(\sqrt{2}+1)\ cm$

Correct answer:

$6(\sqrt{2}+1)\ cm$

Explanation:

This triangle has two angles of 45 and 90 degrees, so the third angle must measure 45 degrees; this is therefore an isosceles right triangle.

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let hypotenuse and side length.

$c^2=s^2+s^2\Rightarrow c^2=2s^2\Rightarrow 6^2=2s^2\Rightarrow s^2=18$

$\Rightarrow s=\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2}\ cm$

We can now find the perimeter since we have all three sides:

$Perimeter=s+s+c=2s+c=2\times 3\sqrt{2}+6=6\sqrt{2}+6=6(\sqrt{2}+1)\ cm$

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Example Question #2 : How To Find The Perimeter Of A Right Triangle

One leg of a right triangle measures $3\frac{1}{3}$ ; the hypotenuse measures $8 \frac{2}{3}$ . What is the perimeter of the triangle?

Possible Answers:

$20 \frac{2}{3}$

$19 \frac{1}{3}$

Correct answer:

Explanation:

The length of the second leg can be calculated using the Pythagorean Theorem. Set $c = 8 \frac{2}{3} = \frac{26}{3}, a =3\frac{1}{3} = \frac{10}{3}$ :

$b ^{2} = c ^{2}-a ^{2}$

$b ^{2} = \left ( \frac{26}{3} \right ) ^{2}- \left ( \frac{10}{3} \right )^{2}$

$b ^{2} = \frac{676}{9}- \frac{100}{9}$

$b ^{2} = \frac{576}{9}$

$b ^{2} =64$

$b = \sqrt{64} = 8$

Add the three sidelengths to get the perimeter:

$P = 3\frac{1}{3} + 8 + 8 \frac{2}{3} = 20$

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Example Question #2 : How To Find The Perimeter Of A Right Triangle

Right triangle 2

$\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . If $BX = \sqrt{6}$ , give the perimeter of $\bigtriangleup ABC$ .

Possible Answers:

$8 \sqrt{2}$

$\sqrt{6} + 3\sqrt{2}$

$4 \sqrt{6} + 2\sqrt{2}$

$2 \sqrt{6} + 6\sqrt{2}$

Correct answer:

$2 \sqrt{6} + 6\sqrt{2}$

Explanation:

An altitude of a right triangle, which here is $\overline{BX}$ , divides the triangle into two triangles similar to each other and to the large triangle. Therefore, their corresponding angles are equal.

$\angle ABC$ is right - and has degree measure 90 - and $\angle C$ has degree measure 30, so $\angle A$ has degree measure 60, making the triangle a 30-60-90 triangle. Therefore, the smaller triangles are as well.

$\overline{BX}$ is the short leg of $\bigtriangleup BXC$ , so the hypotenuse, $\overline{BC}$ , has twice the length of that leg, or $2 \sqrt{6}$ .

$\overline{BC}$ is the long leg of $\bigtriangleup ABC$ ; short leg $\overline{AB}$ has as its length the length of $\overline{BC}$ divided by $\sqrt{3}$ , or $\frac{2 \sqrt{6}}{\sqrt{3}}= 2\sqrt{\frac{6}{3}} = 2\sqrt{2}$ .

The hypotenuse of $\bigtriangleup ABC$ , $\overline{AC}$ , has as its length twice the length of short leg $\overline{AB}$ , which is $2 \cdot 2\sqrt{2} = 4 \sqrt{2}$ .

Add the lengths of the sides of $\bigtriangleup ABC$ :

$2 \sqrt{6} + 2\sqrt{2} + 4\sqrt{2}= 2 \sqrt{6} + 6\sqrt{2}$

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

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

All ISEE Upper Level Math Resources

Example Questions

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

Example Question #2 : How To Find The Perimeter Of A Right Triangle

Example Question #2 : How To Find The Perimeter Of A Right Triangle

All ISEE Upper Level Math Resources

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