Sum of squars of two munden in 160. Eind then product, if there som ا
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>>Sum of the areas of two squares is 468 m
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Sum of the areas of two squares is 468m
2
. If the difference of their perimeters is 24 m, find the sides of the two squares.
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Solution
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Let the side of the first square be 'a'm and that of the second be
′
A
′
m.
Area of the first square =a
2
sq m.
Area of the second square =A
2
sq m.
Their perimeters would be 4a and 4A respectively.
Given 4A−4a=24
A−a=6 --(1)
A
2
+a
2
=468 --(2)
From (1), A=a+6
Substituting for A in (2), we get
(a+6)
2
+a
2
=468
a
2
+12a+36+a
2
=468
2a
2
+12a+36=468
a
2
+6a+18=234
a
2
+6a−216=0
a
2
+18a−12a−216=0
a(a+18)−12(a+18)=0
(a−12)(a+18)=0
a=12,−18