sum of the area of two squares is 468m² if the difference of their perimeters is 24m , formulate the quadra equation to find the sides of the two squares.
Answers
Answer:
If the difference of their perimeters is 24 m, find the sides of the two squares. According to question, (X)² + (Y)² = 468 m² ——(1). Side of second square = Y = 12 m.
Explanation:
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Answer:
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
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So, the side of the first square is 12 m. and the side of the second square is 18 m.