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 $13$ $13$ , $4$ $4$ , and $3$ $3$ , respectively. What is the rectangle's area?

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Answer 1 · 2016-04-11T05:05:45

$9 \sqrt{17}$ $9sqrt17$

Consider the diagram

enter image source here

We can find the length of $b$ $b$ using the Pythagoras-theorem

$a^{2} + b^{2} = c^{2}$ $color(blue)(a^2+b^2=c^2$

Where,

$a and b$ $aand b$ are the right-containing sides and $c$ $c$ is the Hypotenuse (longest side of a right-triangle)

But,

In this case, $a$ $a$ is the Hypotenuse and $b and c$ $band c$ are the right-containing sides

So,

$\to b^{2} + 4^{2} = 13^{2}$ $rarrb^2+4^2=13^2$

$\to b^{2} + 16 = 169$ $rarrb^2+16=169$

$\to b^{2} = 169 - 16$ $rarrb^2=169-16$

$\to b^{2} = 153$ $rarrb^2=153$

$\to b = \sqrt{153}$ $rarrb=sqrt153$

$\Rightarrow b = \sqrt{153} = \sqrt{9 \cdot 17} = 3 \sqrt{17}$ $color(green)(rArrb=sqrt153=sqrt(9*17)=3sqrt17$

Now we need to find the area of the rectangle

Area= $l \cdot b$ $color(purple)(l*b$

Where,

$l = l$ $l=l$ $e n g$ $eng$ $t h, b = b r e a d t h$ $th,b=breadth$

$l = 3 \sqrt{17}, b = 3$ $color(violet)(l=3sqrt17,b=3$

$\to A r e a = 3 \sqrt{17} \cdot 3$ $rarrArea=3sqrt17*3$

$\Rightarrow 9 \sqrt{17}$ $color(green)(rArr9sqrt17$

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 13 1313 , 4 44 , and 3 33 , respectively. What is the rectangle's area?