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# Sum of the Area of Two Squares is 400 Cm2. If the Difference of Their Perimeters is 16 Cm, Find the Sides of Two Squares. - Mathematics

Answer in Brief

Sum of the area of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of two squares.

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#### Solution

Let the side of the smaller square be cm.
Perimeter of any square = (4 × side of the square) cm.

It is given that the difference of the perimeters of two squares is 16 cm.
Then side of the bigger square = $\frac{16 + 4x}{4} = \left( 4 + x \right)$ cm.

According to the question,

$x^2 + \left( 4 + x \right)^2 = 400$

$\Rightarrow x^2 + 16 + x^2 + 8x = 400$

$\Rightarrow 2 x^2 + 8x - 384 = 0$

$\Rightarrow x^2 + 4x - 192 = 0$

$\Rightarrow x^2 + 16x - 12x - 192 = 0$

$\Rightarrow x(x + 16) - 12(x + 16) = 0$

$\Rightarrow (x - 12)(x + 16) = 0$

$\Rightarrow x - 12 = 0 \text { or } x + 16 = 0$

$\Rightarrow x = 12 \text { or } x = - 16$

Since, side of the square cannot be negative
Thus, the side of the smaller square is 12 cm.
and the side of the bigger square is (4 + 12) = 16 cm.

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 4 Quadratic Equations
Q 8 | Page 71
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