Question

sum of the areas of two squares is 468m2 if the difference of their perimeter is 24m find the sides of the two squares​

Answer 1

Answer:-

Let the side of first square be a m and side of second square be b m.

Given:-

Sum of the areas of two squares = 468 m²

We know that,

Area of a square = (side)²

So,

⟹ a² + b² = 468 m² -- equation (1)

Also given that,

Difference between their perimeters = 24 m.

We know,

Perimeter of a square = 4 × side

So,

⟹ 4a - 4b = 24

⟹ 4(a - b) = 24

⟹ a - b = 24/4

⟹ a - b = 6

⟹ a = 6 + b

Putting the value of a in equation (1) we get,

⟹ (6 + b)² + b² = 468

using (a + b)² = a² + b² + 2ab in LHS we get,

⟹ (6)² + b² + 12b + b² - 468 = 0

⟹ 2b² + 12b - 432 = 0

⟹ 2(b² + 6b - 216) = 0

⟹ b² + 6b - 216 = 0

⟹ b² + 18b - 12b - 216 = 0

⟹ b(b + 18) - 12(b + 18) = 0

⟹ (b - 12)(b + 18) = 0

★ b - 12 = 0

⟹ b = 12 m

★ b + 18 = 0

⟹ b = - 18 m

Side of a square can't be negative. so, - 18 is neglected.

Substitute b = 12 in equation (1).

⟹ a² + (12)² = 468

⟹ a² = 468 - 144

⟹ a² = 324

⟹ a = 18 m

∴

• Side of first square (a) = 18 m
• Side of second square (b) = 12 m

Answer 2

Answer:

Given :-

• Sum of area of two square = 468 m²
• Perimeter = 24 m

To Find :-

Sides

SoluTion :-

According to the question

Let the two square be x and y.

x² + y² = 468 (1)

4x - 4y = 24

4(x - y) = 24

x - y = 24/4

x - y = 6

x = 6 + y (2)

Now,

Applying Identity (a + b)² = a² + 2ab + b²

Now,

(6)² + y² + 12y + y² - 468 = 0

2y² + 12y - 432 = 0

2(y² + 6y - 216) = 0

y² + 6y - 216 = 0

• Spilting middle term

y² + (18y - 12y) - 216 = 0

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

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

Now,

Length of square can't be negative

y = 12

Now,

x² + 12² = 468

x² + 144 = 468

x² = 324

x = √324

x = 18