Question

sum of the areas of two squares is 468 m2.if the difference of their perimetres is 24 m ,find the slides of the two squares by quadratic formula by quadratic formula​

Answer 1

Step-by-step explanation:

ANSWER

Let the side of the first square be 'a'm and that of the second be

′

A

′

m.

Area of the first square =a

2

sq m.

Area of the second square =A

2

sq m.

Their perimeters would be 4a and 4A respectively.

Given 4A−4a=24

A−a=6 --(1)

A

2

+a

2

=468 --(2)

From (1), A=a+6

Substituting for A in (2), we get

(a+6)

2

+a

2

=468

a

2

+12a+36+a

2

=468

2a

2

+12a+36=468

a

2

+6a+18=234

a

2

+6a−216=0

a

2

+18a−12a−216=0

a(a+18)−12(a+18)=0

(a−12)(a+18)=0

a=12,−18

So, the side of the first square is 12 m. and the side of the second square is 18 m.

Answer 2

Step-by-step explanation:

Sum of the areas of two squares = 468 m2

Let a and b be the sides of the two squares.

⇒a2 + b2 = 468…(1)

Also given that,

the difference of their perimeters = 24m

⇒4a - 4b = 24

⇒a - b = 6

⇒a = b + 6…(2)

We need to find the sides of the two squares.

Substituting the value of a from equation (2) in equation (1), we have,

(b + 6)2 + b2 = 468

⇒b2 + 62 + 2 × b × 6 + b2 = 468

⇒2b2 + 36 + 12b = 468

⇒2b2 + 36 + 12b - 468 = 0

⇒2b2 + 12b - 432 = 0

⇒b2 + 6b - 216 = 0

⇒b2 + 18b - 12b - 216 = 0

⇒b(b + 18) - 12(b + 18) = 0

⇒(b + 18)(b - 12) = 0

⇒b + 18 = 0 or b - 12 = 0

⇒b = -18 = 0 or b = 12

Side cannot be negative and hence b = 12 m.

Therefore, a = b + 6 = 12 + 6 = 18 m.