Question
Find the perimeter of the rectangle whose length is $40\ cm$ and a diagonal is $41\ cm.$
✓
Answer
Given,
$A B C D$ is a rectangle and $A C$ is its diagonal.
$\mathrm{AB}=40 \mathrm{~cm} \text { and } \mathrm{AC}=41 \mathrm{~cm}$
Now in right $\triangle \mathrm{ABC}$
$A C^2=A B^2+B C^2(B y \text { Pythagoras Theorom) }$
$\Rightarrow(41)^2=(40)^2+B C^2$
$\Rightarrow 1681=1600+B C^2$
$\Rightarrow B C^2=1681-1600$
$\Rightarrow B C^2=81$
$\Rightarrow B C^2=(9)^2$
$B C=9 \mathrm{~cm}$
Now perimeter of rectangle $A B C D=2(A B+B C)=2(40+9)=2 \times 49=98 \mathrm{~cm}$.
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