A right triangle has sides A, B, and C. Side A is the hypotenuse and side B is also a side of a rectangle. Sides A, C, and the side of the rectangle adjacent to side B have lengths of $10$ $10$ , $8$ $8$ , and $16$ $16$ , respectively. What is the rectangle's area?

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A right triangle has sides A, B, and C. Side A is the hypotenuse and side B is also a side of a rectangle. Sides A, C, and the side of the rectangle adjacent to side B have lengths of $10$ $10$ , $8$ $8$ , and $16$ $16$ , respectively. What is the rectangle's area?

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Answer 1 · 2016-03-04T09:22:54

The area of the rectangle is $96 {units}^{2}$ $96 " units"^2$ .

Explanation:

Your triangle and rectangle should look like this:

enter image source here

Let's compute the length of the side $B$ $B$ first.

As it is one of the sides of a right triangle, this can be done with Pythagorean Theorem:

$C^{2} + B^{2} = A^{2}$ $C^2 + B^2 = A^2$

$\Rightarrow 8^{2} + B^{2} = 10^{2}$ $=> 8^2 + B^2 = 10^2$

$\Rightarrow 64 + B^{2} = 100$ $=> 64 + B^2 = 100$

$\Rightarrow B^{2} = 36$ $=> B^2 = 36$

$\Rightarrow B = 6$ $=> B = 6$

So now we know the length of two adjacent sides of a rectangle $B = 6$ $B = 6$ and $D = 16$ $D = 16$ and we can compute the area of the rectangle:

$Area = B \cdot D = 6 \cdot 16 = 96 {units}^{2}$ $"Area" = B * D = 6 * 16 = 96 " units"^2$

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A right triangle has sides A, B, and C. Side A is the hypotenuse and side B is also a side of a rectangle. Sides A, C, and the side of the rectangle adjacent to side B have lengths of $10$ $10$ , $8$ $8$ , and $16$ $16$ , respectively. What is the rectangle's area?

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A right triangle has sides A, B, and C. Side A is the hypotenuse and side B is also a side of a rectangle. Sides A, C, and the side of the rectangle adjacent to side B have lengths of $10$ $10$ , $8$ $8$ , and $16$ $16$ , respectively. What is the rectangle's area?