Question

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Answer 1

AB = 40 cm
Diagonal, AC = 41 cm

Then, by Pythagoras theorem in right $△ABC$:
$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2 \\ \Rightarrow {\mathrm{BC}}^2={41}^2-{40}^2 \\ \Rightarrow {\mathrm{BC}}^2=1681-1600 \\ \Rightarrow {\mathrm{BC}}^2=81 \\ \Rightarrow {\mathrm{BC}}^2=9^2 \\ \Rightarrow \mathrm{BC}=9\mathrm{cm} \end{gathered}$$


∴ Length = 40 cm
Breadth = 9 cm

∴ Perimeter of the rectangle = 2(length + breadth)
= 2(40+9)
= 98 cm