Question

Sum of the areas of two squares is 260 m². If the difference of their perimeters is 24 m then find the sides of the two squares.​

Answer 1

➫ Given :-

⟶ Sum of the areas of two squares is 260 m².

⟶ the difference of their perimeters is 24 m

➫ To Find :-

⟶ the sides of the 2 squares

➫ Solution :-

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

Then, their areas are (a²) m² and (b²) m² respectively.

And, their perimeters are (4a) m and (4b) m respectively.

$⟹$4a - 4b = 24

$⟹$ 4(a - b) = 24

$⟹$a-b= 6

$⟹$ b = (a-6). ......(i)

Sum of their areas = 260 m².

∴ a² + b² = 260

$⟹$ a² + (a - 6)² = 260 (using (i)]

$⟹$ 2a² - 12a - 224 = 0

$⟹$ a² - 6a - 112 = 0

$⟹$ a² - 14a + 8a -112 = 0

$⟹$ a(a - 14) +8(a - 14) = 0

$⟹$ (a - 14)(a + 8) = 0

$⟹$ a - 14 = 0 or a +8= 0

$⟹$ a = 14 or a = -8

$⟹$ a = 14

[∵ side of a square cannot be negative]

∴ a = 14 and b = (14-6) = 8.

Hence, the sides of the square are 14 m and 8 m

Answer 2

Step-by-step explanation:

sides of the square are

14 cm

8cm

for explanation refer attachment