Question

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm.

Answer 1

It is given that, area of rectangle is 192 sq.cm.

$$\begin{gathered} \mathrm{Area}=\mathrm{Length}\times \mathrm{Breadth} \\ \Rightarrow 192=16\times \mathrm{BC} \\ \Rightarrow \mathrm{BC}=\frac{192}{16} \\ \Rightarrow \mathrm{BC}=12\mathrm{cm}...\left(1\right) \end{gathered}$$

According to Pythagoras theorem,

In ∆ABC

$$\begin{gathered} {\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2 \\ \Rightarrow {\left(16\right)}^2+{\left(12\right)}^2={\mathrm{AC}}^2 \\ \Rightarrow 256+144={\mathrm{AC}}^2 \\ \Rightarrow {\mathrm{AC}}^2=400 \\ \Rightarrow \mathrm{AC}=20\mathrm{cm} \end{gathered}$$

Hence, the length of a diagonal of the rectangle is 20 cm.