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 $2$ $2$ , $\frac{1}{3}$ $1/3$ , and $3$ $3$ , 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 $2$ $2$ , $\frac{1}{3}$ $1/3$ , and $3$ $3$ , respectively. What is the rectangle's area?

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Answer 1 · 2016-03-15T21:09:15

$\sqrt{35} \approx 5.916 square units$ $sqrt35 ≈ 5.916" square units "$

Explanation:

The area of the rectangle = 3B . To calculate this area , require to find length of B.
Since this is a right triangle , B can be found using $Pythagoras' theorem$ $color(blue)" Pythagoras' theorem "$

A is the hypotenuse hence $A^{2} = B^{2} + C^{2}$ $A^2 = B^2 + C^2$

thus $B^{2} = A^{2} - C^{2}$ $B^2 = A^2 - C^2$

here A = 3 and C $= \frac{1}{3}$ $= 1/3$

hence : $B^{2} = 2^{2} - {(\frac{1}{3})}^{2} = 4 - \frac{1}{9} = \frac{35}{9}$ $B^2 = 2^2 - (1/3)^2 = 4 - 1/9 = 35/9$

now $B^{2} = \frac{35}{9} \Rightarrow B = \sqrt{\frac{35}{9}} = \frac{1}{3} \sqrt{35}$ $B^2 = 35/9 rArr B = sqrt(35/9) = 1/3sqrt35$

area of rectangle = 3B = $3 \times \frac{1}{3} \sqrt{35} = \sqrt{35} \approx 5.916$ $3xx1/3 sqrt35 = sqrt35 ≈ 5.916$

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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 $2$ $2$ , $\frac{1}{3}$ $1/3$ , and $3$ $3$ , 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 $2$ $2$ , $\frac{1}{3}$ $1/3$ , and $3$ $3$ , respectively. What is the rectangle's area?