Question
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
✓
Answer
Let $A B C D$ be a rectangle whose length, $A B=40 cm$
and diagonal $A C=41 cm$
In right angled $\triangle A B C$, using
Pythagoras property,
$
\begin{array}{l}
A C^2=A B^2+B C^2 \\
\Rightarrow \quad B C^2=A C^2-A B^2 \\
\Rightarrow \quad B C^2=(41)^2-(40)^2 \\
\Rightarrow \quad B C^2=1681-1600 \\
\Rightarrow \quad B C^2=81 \\
\Rightarrow \quad B C=\sqrt{81}=9 cm
\end{array}
$
Now, perimeter of the rectangle
$
\begin{array}{l}
=2(A B+B C) \\
\quad[\because \text { Perimeter of a rectangle }=2(l+b)] \\
=2(40+9) \\
=2 \times 49=98 cm
\end{array}
$
Hence, the perimeter of the rectangle is 98 cm .
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