Question

The sum of the areas of two squares is 260m*m. if the difference of their perimeters is 24m, then what is the side of the smaller square?

Answer 1

let side of first square = x
side of second square = y
sum of area of two square = 260 sq m
i.e     x^2 + y^2 = 260    (i)
now the difference of their perimeter = 24 m
i.e     4x - 4y = 24
=> x - y = 6     (ii)   (taken 4 common)
squarring equation (ii)
(x - y)^2 = 6^2
x^2 + y^2 - 2 xy = 36
260 - 2 xy =36           (value from equation (i)
224 - 2 (6+y) y =0       (value from equation (ii)
2y^2 + 12 y - 224 = 0    (multiplying both side with  -1)
y^2 +6y - 112 =0
y^2 + 14 y - 8y - 112 = 0
y (y+14) - 8 (y+14) =0
y= 8 or -14 m
here we take positive number 
so y= 8
therefore  x-y = 6
=> x= 6+8 = 14 m
thus , the side of small square is 8m.

Answer 2

Answer:

Step-by-step explanation:

Solution :-

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

Their areas are (a²) m² and (b²) m².

Perimeter are 4a m and 4b m.

According to the Question,

⇒ 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  [From (i)]

⇒ 2a² - 12a - 224 = 0

⇒ a² - 14a + 8a - 112 = 0

⇒ a² - 6a - 112 = 0

⇒ a(a - 14) + 8(a - 14) = 0

⇒ (a - 14) (a + 8) = 0

⇒ a - 14 = 0 or a + 8 = 0

⇒ a = 14, - 8 (As x can't be negative)

⇒ a = 14

Putting a's value in Eq (i), we get

⇒ b = a - 6

⇒ b = 14 - 6

⇒ b = 8

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