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Sum of the Areas of Two Squares is 640 M2. If the Difference of Their Perimeters is 64 M. Find the Sides of the Two Squares. - CBSE Class 10 - Mathematics

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Question

Sum of the areas of two squares is 640 m2. If the difference of their perimeters is 64 m. Find the sides of the two squares.

Solution

Let the sides of the squares are m and = y m.Then

According to question,

Sum of the difference of their perimeter=64 m

4x - 4y = 64

x - y = 16

y = x - 16        .................... (1)

And sum of the areas of square = 640 m2

x2 + y2 = 640               ............ (2)

Putting the value of x in equation (2) from equation (1)

x2 + (x - 16)2 = 640

x2 + x2 - 32x + 256 = 640

2x2 - 32x + 256 - 640 = 0

2x2 - 32x - 384 = 0

2(x2 - 16x - 192) = 0

x2 - 16x - 192 = 0

x2 - 24x + 8x - 192 = 0

x(x - 24) + 8(x - 24) = 0

(x - 24)(x + 8) = 0

x - 24 = 0

x = 24

or

x + 8 = 0

x = -8

Sides of the square never are negative.

Therefore, putting the value of x in equation (1)

y = x - 16 = 24 - 16 = 8

Hence, sides of the square be 24m and 8m respectively.

  Is there an error in this question or solution?

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Solution Sum of the Areas of Two Squares is 640 M2. If the Difference of Their Perimeters is 64 M. Find the Sides of the Two Squares. Concept: Solutions of Quadratic Equations by Factorization.
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