Question

Q. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.​

Answer 1

Answer:

Let the sides of the first and second square be X and Y . Area of the first square = (X)²

Area of the 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

Side of second square = Y = 12 m

Answer 2

Step-by-step explanation:

⭐Question:-

Q. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

⭐Answer:-

Solution:-

Sum of the areas of two squares is 468 m²

∵ x² + y² = 468 ………..(1) [ ∵ area of square = side²] → The difference of their perimeters is 24 m.

∵ 4x – 4y = 24 [ ∵ Perimeter of square = 4 × side] ⇒ 4( x – y ) = 24

⇒ x – y = 24/4

⇒ x – y = 6

∴ y = x – 6 ……….(2)

From equation (1) and (2),

∵ x² + ( x – 6 )² = 468

⇒ x² + x² – 12x + 36 = 468

⇒ 2x² – 12x + 36 – 468 = 0

⇒ 2x² – 12x – 432 = 0

⇒ 2( x² – 6x – 216 ) = 0

⇒ x² – 6x – 216 = 0

⇒ x² – 18x + 12x – 216 = 0

⇒ x( x – 18 ) + 12( x – 18 ) = 0

⇒ ( x + 12 ) ( x – 18 ) = 0

⇒ x + 12 = 0 and x – 18 = 0

⇒ x = – 12m [ rejected ] and x = 18m

∴ x = 18m

Put the value of ‘x’ in equation (2),

∵ y = x – 6

⇒ y = 18 – 6

∴ y = 12m

[Hence, sides of two squares are 18m and 12m respectively.]