Question

The sum of the areas of two square fields is 468 m². If the difference of their perimeter is 24 m, find the sides of the two squares.

Answer 1

Answer:

12 m

18 m

Step-by-step explanation:

Hi,

Let A₁ be the area of the first square,

Let s₁ be the side of the first square,

Let A₂ be the area of the second square,

Let s₂ be the side of the second square,

Given A₁ + A₂ = 468 m² ,

4(s₁ - s₂) = 24 m

s₁ - s₂ = 6

We know that A₁ = s₁²

A₂ = s₂²

⇒ s₁² + s₂² = 468

(s₂ + 6)² + s₂² = 468

2s₂² + 12s₂ + 36 - 468 = 0

2s₂² + 12s₂ - 432 = 0

s₂² + 6s₂ - 216 = 0

s₂² + 18s₂ - 12s₂ - 216 = 0

s₂(s₂ + 18) - 12(s₂ + 18) = 0

(s₂ - 12)(s₂ + 18) = 0

Since s₂ > 0,

s₂ = 12 m and

s₁ = 12 + 6 = 18 m

Hence the sides are 12 m and 18 m.

Hope, it helps !

Answer 2

Answer:

12 m and 18 m

Step-by-step explanation:

Define x and y:

Let the side of one of the square be x and the other be y.

The sum of the squares is 468 m²

x² + y² = 486

The difference inn their perimeter is 24 m

4x - 4y = 24

x - y = 6

x = y + 6

Solve x and y:

x² + y² = 468  ---------- [ 1 ]

x = y + 6 ---------- [ 2 ]

Sub [ 2 ] into [ 1 ]:

(y + 6)²  + y²  = 468

y²  + 12y + 36 + y² = 468

2y²  + 12y - 432 = 0

y²  + 6y - 216 = 0

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

y = 12 or y = - 18 (rejected, y cannot be negative)

Sub y = 12 into  [ 2 ]:

x = 12 + 6

x = 18

Find the sides:

Side of one of the square = x = 18 m

Side of the other square = y = 12 m

Answer: The sides are 12 m and 18 m