Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square .
Answers
✎Δnsɯer࿐
Let the length of the side of the smaller square be x .
Perimeter of smaller square = 4x
Let the length of the side of the larger square be y m.
Perimeter of larger square = 4y
Now, 4y−4x=24
or, y−x=6 (Dividing both sides by 4)
or, y=x+6
Now, area of larger square = y 2=(x+6)2
= x2 +12x+36
Area of smaller square = x2
Now, x2+12x+36+x2=468 (Sum of areas)
or, 2 x2+12x+36=468
or, x2 +6 x +18=234 (Dividing both sides by 2)
or, x2 +6x+18−234=0
or, x2 +6x−216=0
or, (x+18)(x−12)=0 (Factoring the equation)
or, x=−18,12
Since x cannot be negative,therefore,
x=12m
Therefore, x + 6 = 18m
So the sides are 18m and 12m respectively.
Hope that helps.
Let the two squares be A and B. A has sides of a, while B has sides of b,
Now for the two equations.
a^2 + b^2 = 468 …(1) [Sum of the areas]
4a - 4b = 24 …(2) (perimeter of the two squares), which can be written as
a - b = 6, or
a = b+6. Put that value in (1) to get
(b+6)^2 + b^2 = 468, or
b^2 + 12b + 36 + b^2 = 468, or
2b^2 + 12b + 36 = 468, or
b^2 + 6b + 18 = 234, or
b^2 + 6b + 18 - 234 = 0, or
b^2 + 6b -216 = 0, or
b^2 + 18b - 12b - 216 = 0, or
b(b + 18) - 12(b + 18) = 0, or
(b-12)(b+18) = 0
b = 12 m or -18m (inadmissible)
Then a = b+6 = 12 + 6 = 18m
So A is a square of 18 m and B is a square of 12 m
Check: Area of A = 18^2 = 324 sq m. area of B = 12^2 = 144 sq m and their sum is 324 + 144 = 468 sq m. Correct.
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=468a2+12a+36+a2
=4682a2+12a+36
=468a2+6a+18
=234a2+6a−216
=0a2+18a−12a−216
=0a(a+18)−12(a+18)
=0(a−12)(a+18)=0
=12,−18
So, the side of the first square is 12 m. and the side of the second square is 18 m.