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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. - CBSE Class 10 - Mathematics

ConceptSolutions of Quadratic Equations by Completing the Square

Question

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.

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.

Is there an error in this question or solution?

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Solution 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. Concept: Solutions of Quadratic Equations by Completing the Square.
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