How to find the perimeter of a right triangle - ISEE Upper Level Quantitative Reasoning

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Question

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

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Answer

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.

$\Rightarrow s=$\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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Question

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?

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Answer

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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Question

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

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Answer

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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Question

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

Tap to reveal answer

Answer

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.

$\Rightarrow s=$\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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Question

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?

Tap to reveal answer

Answer

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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Question

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

Tap to reveal answer

Answer

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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Question

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

Tap to reveal answer

Answer

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.

$\Rightarrow s=$\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}$+1)\ cm$

← Didn't Know|Knew It →

Question

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?

Tap to reveal answer

Answer

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$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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}$$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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.

$\Rightarrow s=$\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}$+1)\ cm$

← Didn't Know|Knew It →

Question

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?

Tap to reveal answer

Answer

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$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

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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