Sum of the areas if two squares is 468m sq. if the difference of their perimeter is 24m find the sides of the two squares
Answers
Area= sidexside = (x)^2
area1 = a, let and area2 = b
a^2 + b^2 = 468
perimeter of square = 4a
perimeter of square 1 = 4a and square 2 = 4b
4a - 4b = 24
a - b = 6
a = (6+b) -----------(1)
putting 1 in equation a^2 + b^2 = 468 from above
(6+b)^2 + b^2 = 468
36 + b^2 + 12b + b^2 = 468
2b^2 + 12b + 36 = 468
b^2 + 6b + 18 = 234
b^2 + 6b - 216 = 0
b^2 + 18b - 12b - 216 = 0
b (b + 18) - 12 ( b + 18) = 0
b = -18 and 12
ignoring negative value
we get b = 12 putting in (1)
a = (6+b) -----------(1)
a = (6+12)
a = 18
Answer:
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.