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

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Example Questions

Example Question #31 : Triangles

One angle of a right triangle measures 45 degrees, and the hypotenuse measures 8 centimeters. Give the area of the triangle.

Possible Answers:

$4\ cm^2$ $\displaystyle 4\ cm^2$

$32\ cm^2$ $\displaystyle 32\ cm^2$

$8\ cm^2$ $\displaystyle 8\ cm^2$

$64\ cm^2$ $\displaystyle 64\ cm^2$

$16\ cm^2$ $\displaystyle 16\ cm^2$

Correct answer:

$16\ cm^2$ $\displaystyle 16\ cm^2$

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 $\displaystyle c=$ hypotenuse and $\displaystyle s=$ side length.

$c^2=s^2+s^2\Rightarrow c^2=2s^2\Rightarrow 8^2=2s^2\Rightarrow s^2=32$ $\displaystyle c^2=s^2+s^2\Rightarrow c^2=2s^2\Rightarrow 8^2=2s^2\Rightarrow s^2=32$

$\Rightarrow s=\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}\ cm$ $\displaystyle \Rightarrow s=\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}\ cm$

We can then plug this side length into the formula for area.

$area=\frac{s^2}{2}=\frac{(4\sqrt{2})^2}{2}=\frac{32}{2}=16\ cm^2$ $\displaystyle area=\frac{s^2}{2}=\frac{(4\sqrt{2})^2}{2}=\frac{32}{2}=16\ cm^2$

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

The legs of a right triangle measure $2 \frac{2}{5}$ $\displaystyle 2 \frac{2}{5}$ and $3 \frac{1}{5}$ $\displaystyle 3 \frac{1}{5}$. What is its perimeter?

Possible Answers:

$8 \frac{3}{5}$ $\displaystyle 8 \frac{3}{5}$

$9 \frac{1}{5}$ $\displaystyle 9 \frac{1}{5}$

$9 \frac{3}{5}$ $\displaystyle 9 \frac{3}{5}$

$10\frac{1}{5}$ $\displaystyle 10\frac{1}{5}$

Correct answer:

$9 \frac{3}{5}$ $\displaystyle 9 \frac{3}{5}$

Explanation:

The hypotenuse of the triangle can be calculated using the Pythagorean Theorem. Set $a = 2 \frac{2}{5} = \frac{12}{5} , b=3 \frac{1}{5}= \frac{16}{5}$ $\displaystyle a = 2 \frac{2}{5} = \frac{12}{5} , b=3 \frac{1}{5}= \frac{16}{5}$:

$c ^{2} = a ^{2} + b ^{2}$ $\displaystyle c ^{2} = a ^{2} + b ^{2}$

$c ^{2} = \left ( \frac{12}{5} \right ) ^{2} + \left ( \frac{16}{5} \right ) ^{2}$ $\displaystyle c ^{2} = \left ( \frac{12}{5} \right ) ^{2} + \left ( \frac{16}{5} \right ) ^{2}$

$c ^{2} = \frac{144}{25} + \frac{256}{25}$ $\displaystyle c ^{2} = \frac{144}{25} + \frac{256}{25}$

$c ^{2} = \frac{400}{25} = 16$ $\displaystyle c ^{2} = \frac{400}{25} = 16$

$c = \sqrt{16} = 4$ $\displaystyle c = \sqrt{16} = 4$

Add the three sidelengths:

$P =2 \frac{2}{5} + 3 \frac{1}{5} + 4 = 9 \frac{3}{5}$ $\displaystyle P =2 \frac{2}{5} + 3 \frac{1}{5} + 4 = 9 \frac{3}{5}$

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

Right triangle

Figure NOT drawn to scale

$\bigtriangleup ABC$ $\displaystyle \bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ $\displaystyle \overline{BX}$. Give the ratio of the area of $\bigtriangleup BXC$ $\displaystyle \bigtriangleup BXC$ to that of $\bigtriangleup AXB$ $\displaystyle \bigtriangleup AXB$.

Possible Answers:

16:9 $\displaystyle 16:9$

25:9 $\displaystyle 25:9$

25:16 $\displaystyle 25:16$

4:3 $\displaystyle 4:3$

Correct answer:

16:9 $\displaystyle 16:9$

Explanation:

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ $\displaystyle \overline{BX}$ - divides the triangle into two triangles similar to each other. The ratio of the hypotenuse of the white triangle to that of the gray triangle (which are corresponding sides) is

$\frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}$ $\displaystyle \frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}$ ,

making this the similarity ratio. The ratio of the areas of two similar triangles is the square of their similarity ratio, which here is

$\left ( \frac{4}{3} \right )^{2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$ $\displaystyle \left ( \frac{4}{3} \right )^{2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$, or 16: 9 $\displaystyle 16: 9$.

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Example Question #31 : Right Triangles

Find the area of a right triangle with a base of 7cm and a height of 20cm.

Possible Answers:

$120\text{cm}^2$ $\displaystyle 120\text{cm}^2$

$80\text{cm}^2$ $\displaystyle 80\text{cm}^2$

$140\text{cm}^2$ $\displaystyle 140\text{cm}^2$

$70\text{cm}^2$ $\displaystyle 70\text{cm}^2$

$110\text{cm}^2$ $\displaystyle 110\text{cm}^2$

Correct answer:

$70\text{cm}^2$ $\displaystyle 70\text{cm}^2$

Explanation:

To find the area of a right triangle, we will use the following formula:

$A = \frac{1}{2} \cdot b \cdot h$ $\displaystyle A = \frac{1}{2} \cdot b \cdot h$

where b is the base and h is the height of the triangle.

We know the base of the triangle is 7cm. We know the height of the triangle is 20cm. Knowing this, we can substitute into the formula. We get

$A = \frac{1}{2} \cdot 7\text{cm} \cdot 20\text{cm}$ $\displaystyle A = \frac{1}{2} \cdot 7\text{cm} \cdot 20\text{cm}$

$A = 7\text{cm} \cdot 10\text{cm}$ $\displaystyle A = 7\text{cm} \cdot 10\text{cm}$

$A = 70\text{cm}^2$ $\displaystyle A = 70\text{cm}^2$

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

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #31 : Triangles

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

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

Example Question #31 : Right Triangles

All ISEE Upper Level Math Resources

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