Sum of area of two square is ${468 m}^{2}$ and difference in the perimeter square is $24$ . What is the measure of the their sides?

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Solution

$\begin{matrix} L e t s i d e o f t h e f i r s t s q u a r e b e x a n d s i d e o f sec o n d s q u a r e i s y, t h e n, f r o m t h e q u e s t i o n x^{2} + y^{2} = 468 m^{2} - - - (1) P e r i m e t e r o f s q u a r e = 4 x a n d 4 y ∴ 4 x - 4 y = 24 \Rightarrow x - y = 6 ∴ y = x - 6 - - - (2) P u t t i n g (2) i n e q u^{n} (1) w e g e t x^{2} + {(x - 6)}^{2} = 468 \Rightarrow x^{2} + x^{2} + 36 - 12 x = 468 \Rightarrow 2 x^{2} - 12 x - 432 = 0 \Rightarrow x^{2} - 6 x - 216 = 0 ∴ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- (- b) \pm \sqrt{900}}{2 \times 1} = \frac{6 \pm 30}{2} ∴ x = \frac{36}{2} = 18 a n d x = \frac{- 24}{2} = - 12 [d o e s n o t e x i s t] s o, y = x - 6 = 12 H e n c e, t h e s i d e s o f t h e s q u a r e a r e 18 a n d 12 \end{matrix} .$

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