Math, asked by taherkanch, 4 days ago

# sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 metres find the sides of the two squares

27
HEYA !

Let the side of two squares be 'a' and 'A'

Now
According to the question ,

Sum of Area of two square = 468 m²

a² + A² = 468 m² ----------(1)

[ ∵ Area of square = (side)² ]

Also,

Difference of their perimeter = 24 m

4A – 4a = 24 m

[ ∵ Perimeter of square = 4 × side ]

→ 4 ( A – a ) = 24

→ A – a = 6

→ A = 6 + a ---------(2)

Putting Value of A from (2) in (1) , We get

a² + (6 + a)² = 468

→ a² + 36 + a² + 12a = 468

→ 2a² + 12a + 36 – 468 = 0

→ 2a² + 12a – 432 = 0

→ a² + 6a – 216 = 0

→ a² + 18a – 12a – 216 = 0

→ a ( a + 18 ) – 12 ( a + 18 ) = 0

→ ( a – 12 ) ( a + 18 ) = 0

→ a – 12 = 0 ...or... a + 18 = 0

→ a = 12 ...or... a = – 18

Since , a = – 18 is not possible as side cannot be negative

Therefore ,
→ a = 12

Putting a = 12 in (2) , We get

A = 6 + 12 = 18

→ A = 18

Hence,

9
Heya !!!

Let the sides of first and second square be X and Y .

Area of first square = (X)²

And,

Area of second square = (Y)²

According to question,

(X)² + (Y)² = 468 m² ------------(1).

Perimeter of first square = 4 × X

and,

Perimeter of second square = 4 × Y

According to question,

4X - 4Y = 24 -----------(2)

From equation (2) we get,

4X - 4Y = 24

4(X-Y) = 24

X - Y = 24/4

X - Y = 6

X = 6+Y ---------(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 468

(6+Y)² + (Y)² = 468

(6)² + (Y)² + 2 × 6 × Y + (Y)² = 468

36 + Y² + 12Y + Y² = 468

2Y² + 12Y - 468 +36 = 0

2Y² + 12Y -432 = 0

2( Y² + 6Y - 216) = 0

Y² + 6Y - 216 = 0

Y² + 18Y - 12Y -216 = 0

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

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

(Y+18) = 0 Or (Y-12) = 0

Y = -18 OR Y = 12

Putting Y = 12 in EQUATION (3)

X = 6+Y = 6+12 = 18

Side of first square = X = 18 m

and,

Side of second square = Y = 12 m.