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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$=2(l+b)$

$=2(40+9)$

$=2(49)$

$=98\ cm$.

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