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Find the perimeter of the rectangle whose length is $40\ cm$ and a diagonal is $41\ cm$.
Given: A rectangle whose length is $40\ cm$ and a diagonal is $41\ cm$.
To do: To find the perimeter of the rectangle.
Solution:
$PR=41\ cm$, $PQ=40\ cm$
Let breadth $(QR)$ be $x\ cm$
Now, in right-angled triangle PQR
$(PR)^2=(RQ)^2+(PQ)^2$
$\Rightarrow (41)^2=x^2+(40)^2$
$\Rightarrow 1681=x^2+1600$
$\Rightarrow 1681-1600=x^2$
$\Rightarrow \sqrt{81}=x$
$\Rightarrow 9=x$
Therefore, the breadth of the rectangle$=9\ cm$
The perimeter of the rectange$=2(l+b)$
$=2(40+9)$
$=2(49)$
$=98\ cm$.
Updated on: 10-Oct-2022
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