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

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Question

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

Sum

Solution

Let the sides of two squares by a and b respectively.
Then, area of one square, S1 = a2
And, area of second square, S2 = b2
Given, S1 + S2 = 400 cm2
⇒ a2 + b2 = 400 cm2 …..(1)
Also, difference in perimeter = 16 cm
⇒ 4a - 4b = 16 cm
⇒ a - b = 4
⇒ a = (4 + b)

Substituting the value of 'a' in (1), we get
(4 + b)2 + b2 = 400
⇒ 16 + 8b + b2 + b2 = 400
⇒ 2b2 + 8b - 384 = 0
⇒ b2 + 4b - 192 = 0
⇒ b2 + 16b - 12b - 192 = 0
⇒ b(b + 16) - 12(b + 16) = 0
⇒ (b +16)(b - 12) = 0
⇒ b + 16 = 0 or b - 12 = 0
⇒ b = - 16 or b = 12

Since, the side of a square cannot be negative, we reject - 16.
Thus, b = 12
⇒ a = 4 + b = 4 + 12 = 16

Hence, the sides of a square are 16 cm and 12 cm respectively.

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Chapter 20: Area and Perimeter of Plane Figures - Exercise 20 (D) [Page 263]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 20 Area and Perimeter of Plane Figures
Exercise 20 (D) | Q 4 | Page 263

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