Question

The sum of the areas of two squares is 400 sq.m. If the difference
between their perimeters is 16 m, find the sides of two squares.​

Answer 1

Answer:answer is 12m and 16m.

Step-by-step explanation:

Answer 2

Given data :

• The sum of the areas of two squares is 400 m².
• The difference between their perimeter is 16 m.

Solution:

Let, x be the side of first of square. & y be the side of second square.

We know formulae of area & perimeter of square.

→ Area of square = (side)²

→ Area of first square = ( x )² __[1]

→ Area of second square = ( y )² __[2]

→ Perimeter of square = 4(side)

→ Perimeter of first square = 4x ___[3]

→ Perimeter of second square = 4y ___[4]

{Accodring to given}

For area of square

{from eq. [1] & eq. [2]}

→ ( x )² + ( y )² = 400 ___[5]

{Accodring to given}

For perimeter of square

{from eq. [3] & eq. [4]}

→ 4x - 4y = 16

→ 4(x - y) = 16

→ (x - y) = 16/4

→ x - y = 4

→ x = 4 + y __[6]

Put value of x in eq. [ 5 ]

→ ( x )² + ( y )² = 400

→ ( 4 + y )² + ( y )² = 400

{By algebric identity}

→ (4² + [2 × 4 × y] + y² ) + y² = 400

→ 16 + 8y + y² + y² = 400

→ 2y² + 8y + 16 = 400

→ y² + 4y + 8 = 200

→ y² + 8y + 8 - 200 = 0

→ y² + 8y - 192 = 0

→ y² + 16y - 12y - 192 = 0

→ y (y + 16) - 12 (y + 16 ) = 0

→ (y + 16) - (y - 12) = 0

→ y + 16 = 0 or y - 12 = 0

→ y = - 16 or y = 12

We know that side of square is can not negative hence,

→ y = 12 & y ≠ - 16

Hence,side of first square is 12 m.

Put value of y in eq. ( 6 )

→ x = 4 + y

→ x = 4 + 12

→ x = 16 m

Hence,side of second square is 16 m.