Question : The sum of the length and breadth of a rectangle is 6 cm. A square is constructed such that one of its sides is equal to a diagonal of the rectangle. If the ratio of areas of the square and rectangle is 5 : 2, the area of the square (in cm2) is:
Option 1: $20$
Option 2: $10$
Option 3: $4\sqrt{5}$
Option 4: $10\sqrt{2}$
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Correct Answer: $20$
Solution :
Let the length of the rectangle as $l$ and the breadth as $b$.
Given that the sum of the length and breadth of the rectangle is 6 cm.
$l + b = 6 \tag{1}$____(1)
The diagonal $d$ of the rectangle can be found using the Pythagorean theorem:
$d = \sqrt{l^2 + b^2} \tag{2}$____(2)
Given that the ratio of the areas of the square and the rectangle is 5 : 2,
$\frac{A_{\text{sq}}}{A_{\text{rect}}} = \frac{5}{2} \Rightarrow \frac{d^2}{l \times b} = \frac{5}{2} \tag{3}$____(3)
Substituting equation (2) into equation (3),
$\frac{l^2 + b^2}{l \times b} = \frac{5}{2} $
$⇒ 2l^2 + 2b^2 = 5lb \tag{4}$____(4)
Solving equations (1) and (4) simultaneously,
$2(6-b)^2+2b^2=5b(6-b)$
⇒ $b^2-6b+8=0$
⇒ $(b-2)(b-4)=0$
$b=2 $ or $4$
Taking
$l = 2\;\mathrm{cm}$ and $b = 4\;\mathrm{cm}$
Substituting these values into equation (2),
$d = \sqrt{2^2 + 4^2}$
$d = 2\sqrt{5}\;\mathrm{cm}$
The area of the square,
$ = d^2 = (2\sqrt{5})^2 = 20 \text{ cm}^2$
Hence, the correct answer is $20$.
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