sum of the areas of two squares is 468 m2.if the difference of their perimetres is 24 m ,find the slides of the two squares by quadratic formula by quadratic formula
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Answered by
1
Step-by-step explanation:
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
So, the side of the first square is 12 m. and the side of the second square is 18 m.
Answered by
1
Step-by-step explanation:
Sum of the areas of two squares = 468 m2
Let a and b be the sides of the two squares.
⇒a2 + b2 = 468…(1)
Also given that,
the difference of their perimeters = 24m
⇒4a - 4b = 24
⇒a - b = 6
⇒a = b + 6…(2)
We need to find the sides of the two squares.
Substituting the value of a from equation (2) in equation (1), we have,
(b + 6)2 + b2 = 468
⇒b2 + 62 + 2 × b × 6 + b2 = 468
⇒2b2 + 36 + 12b = 468
⇒2b2 + 36 + 12b - 468 = 0
⇒2b2 + 12b - 432 = 0
⇒b2 + 6b - 216 = 0
⇒b2 + 18b - 12b - 216 = 0
⇒b(b + 18) - 12(b + 18) = 0
⇒(b + 18)(b - 12) = 0
⇒b + 18 = 0 or b - 12 = 0
⇒b = -18 = 0 or b = 12
Side cannot be negative and hence b = 12 m.
Therefore, a = b + 6 = 12 + 6 = 18 m.
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