Question

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

Answer 1

Question :-

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

Answer :-

Given :-

• Length of rectangle is twice it's width
• Length of diagnol = 16√5 cm

To Find :-

• Area

Solution :-

Let the length be 2x

Breadth = x

$\sf Diagnol = \sqrt{l^2 + b^2}$

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

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

⇒ $\sf 16^2 \times 5 = x^2 + 4x^2$

⇒ $\sf 1280 = 5x^2$

⇒ $\sf x^2 = \dfrac{1280}{5}$

⇒ $\sf x^2 = 256$

⇒ $\sf x = \sqrt{256}$

⇒ $\sf x = 16$

• Breadth = 16 cm
• Length = 16 × 2 = 32 cm

Area of rectangle = length × breadth

⇒ Area = 16 × 32

⇒ Area = 512

Area of rectangle = 512 cm²

Answer 2

Given Question :-

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

$\large\underline\green{\bold{ \sf \: ANSWER}}$

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{diagonal \: of \: rectangle = 16 \sqrt{5} \: cm} \\ &\sf{length \: = \: 2(breadth)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: find - \begin{cases} &\sf{Area_{(rectangle)}}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: Formula \: Used - \begin{cases} &\rm{ {d}^{2} = {l}^{2} + {b}^{2} } \\ &\rm{Area_{(rectangle)} = l \times b} \end{cases}\end{gathered}\end{gathered}$

where,

• d represents diagonal of a rectangle.

• l represents length of a rectangle.

• b represents breadth of a rectangle.

$\large\underline\purple{\bold{Solution :- }}$

Given that

• Length of a rectangle = twice the Breadth of rectangle

So,

$\begin{gathered}\begin{gathered}\bf \: Let - \begin{cases} &\sf{breadth_{(rectangle)} (b)\: = \: x \: cm} \\ &\sf{length_{(rectangle)} (l)= \: 2x \: cm} \end{cases}\end{gathered}\end{gathered}$

Also,

Given that

$\bullet \: \sf \: diagonal_{(rectangle)} (d)\: = 16 \sqrt{5} \: cm$

Now,

• Using the formula,

$\rm :\implies\: \boxed{ \pink{ \bf \: {d}^{2} \: = \tt \: {l}^{2} + {b}^{2} }}$

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

$\rm :\implies\:256 \times 5 = {x}^{2} + 4 {x}^{2}$

$\rm :\implies\:256 \times 5 = 5 {x}^{2}$

$\rm :\implies\: {x}^{2} = 256$

$\rm :\implies\: \boxed{ \pink{ \bf \: x \: = \tt \:16 }}$

$\begin{gathered}\begin{gathered}\bf \: So - \begin{cases} &\sf{breadth_{(rectangle)} (b)\: = \: x \: = 16 \: cm} \\ &\sf{length_{(rectangle)} (l)= \: 2x \: = 32\: cm} \end{cases}\end{gathered}\end{gathered}$

Now,

• Using the formula,

$\rm :\implies\: \boxed{ \pink{ \bf \: Area_{(rectangle)} \: = \tt \: length \times breadth}}$

$\rm :\implies\:Area_{(rectangle)} = 32 \times 16$

$\rm :\implies\: \boxed{ \pink{ \bf \: Area_{(rectangle)} \: = \tt \:512 \: {cm}^{2} }}$