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Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares. - Mathematics

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Question

Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Sum

Solution

Let the side of the first square be 'a' m and that of the second be ′A′ m.

Area of the first square = a2 sq m.

Area of the second square = A2 sq m.

Their perimeters would be 4a and 4A respectively.

Given 4A - 4a = 24
A - a = 6              ......(1)
A2 + a2 = 468      ......(2)
From (1), A = a + 6
Substituting for A in (2), we get
(a + 6)2 + a2 = 468
a2 + 12a + 36 + a2 = 468
2a2 + 12a + 36 = 468
a2 + 6a + 18 = 234
a2 + 6a - 216 = 0
a2 + 18a - 12a - 216 = 0
a(a + 18) - 12(a + 18) = 0
(a - 12)(a + 18) = 0
a = 12, - 18
So, the side of the first square is 12 m. and the side of the second square is 18 m.
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Chapter 4: Quadratic Equations - Exercise 4.3 [Page 88]

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NCERT Mathematics [English] Class 10
Chapter 4 Quadratic Equations
Exercise 4.3 | Q 11 | Page 88
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