Question

Question 11 Sum of the areas of two squares is 468 m^2 . If the difference of their perimeters is 24 m, find the sides of the two squares.

Class 10 - Math - Quadratic Equations Page 88

Answer 1

let the sides of sq.=x and y
area of sq.=x^ and y^
a.t.q=4x-4y=24
4 (x-y)=24
x-y =6.
x=6+y. eq. 1
x^ + y^=468
put value of x. (6+y)+y^ =468
36 +y^ +12y +y^=468
2y^ +12y +y^ =468-36
2y^ +12y=432
2(y^ +6y)=432
y^ +6y=432/2
y^ +6y=216
y^ +6y-216=0.
y^ +18y-12y-216 =0. (factorisation method)
y(y+18)-12(y+18)
(y+18)(y-12)
consider, y+18=0. y-12=0
y= -18. y =12
therefore
side of square
12,18

Answer 2

Let the sides of the two squares be x m and ym. Therefore, their perimeter will be 4x and 4yrespectively and their areas will be x2 and y2 respectively.
It is given that
4x - 4y = 24
x - y = 6
x = y + 6
Also, x2 + y2 = 468
⇒ (6 + y2) + y2 = 468
⇒ 36 + y2 + 12y + y2 = 468
⇒ 2y2 + 12y + 432 = 0
⇒ y2 + 6y - 216 = 0
⇒ y2 + 18y - 12y - 216 = 0
⇒ y(y +18) -12(y + 18) = 0
⇒ (y + 18)(y - 12) = 0
⇒ y = -18, 12
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.