CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
Solution

Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively 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+y)2+y2=468
36+y2+12y+y2=468
2y2+12y+432=0
y2+6y216=0
y2+18y12y216=0
y(y+18)12(y+18)=0
(y+18)(y12)=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.

flag
Suggest Corrections
thumbs-up
12
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Algebraic Solution
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon