Question

Sum of the areas of two square is 468 square meter.If the difference of theirs perimeters is 24 m,find the sides of the squares.

Answer 1

Answer:

18m and 12m

Step-by-step explanation:

Answer 2

Let side of First square be 'S' and side of second square be 's'

Area of square = side × side = side^2

ATQ -

S^2 + s^2 = 468

Perimeter of squares = 4×side

So, 4(S) - 4(s) = 24m

So, the system of equations we have here is -

S^2 + s^2 = 468 ---(1)

4S - 4s = 24 ----(2)

In equation (2),

4S = 24 + 4s

S = 24+4s ÷ 4

S = 6+s

Put, S = 6+s in equation (1)

(6+s)^2 + s^2 = 468

6^2 + 2(6)(s) + s^2 + s^2 = 468

36 + 12x + 2s^2 = 468

12x + 2s^2 = 468-36

2s^2 + 12s = 432

2s^2 + 12s - 432 = 0

Divide both sides by 2, we get

s^2 + 6s - 216 = 0

s^2 + (18 - 12)s - 216 = 0

s^2 + 18s - 12s - 216 = 0

s(s + 18) - 12(s + 18) = 0

(s - 12) (s + 18) = 0

Either, s-12 = 0 or s + 18 = 0

s = 12 and s = -18

Side cannot be negative, so s = 12m

Put s = 12 in eq. (2)

4S - 4×12 = 24

4S = 24 + 48

4S = 72

S = 72÷4

S = 18m

So, Side of First square, S = 18m and second square, s = 12m