Question

11. The length of a rectangle is twice its width. If the
length of its diagonal is 16√5 cm, find its area.​

Answer 1

To find : area of rectangle

Given :  length of a rectangle is twice its width.

         length of its diagonal is $16\sqrt{5} cm$

Solution :

• Let , width of rectangle be $x$ then length of rectangle be $2x$ .
• Given , diagonal = $16\sqrt{5} cm$  
• Using pythagoras theorem we get ,

         $(16\sqrt{5} )^{2} = (x)^{2} + (2x)^{2}$  

        $\begin{array}{l}\sqrt{\left(4 x^{2}+x^{2}\right)}=\sqrt{5 x} \\\sqrt{5 x}=16 \sqrt{5} \\x=16\end{array}$

• Therefore , width = $x$ = $16\ cm$

                          length =  $2x$  = $2\times 16 = 32\ cm$

          $\begin{array}{l}\text { Area }=\text { length } \times \text { breadth } \\=(16 \times 32) \mathrm{cm}^{2} \\=512 \mathrm{~cm}^{2}\end{array}$

Hence , area of rectangle is $512\ m^{2}$ .