MENU
Question -

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



Answer -

Let the sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y respectively
And area of the squares will be x2 and y2 respectively.
Given,
4x тАУ 4y = 24
x тАУ y = 6
x = y + 6

Also,┬аx2┬а+┬аy2┬а=468

тЗТ (6┬а+┬аy2)┬а+┬аy2┬а=468

тЗТ 36┬а+┬аy2┬а+ 12y┬а+┬аy2┬а=468

тЗТ 2y2┬а+ 12y┬а+ 432 = 0

тЗТ┬аy2┬а+ 6y тАУ 216 = 0

тЗТ┬аy2┬а+ 18y┬атАУ 12y┬атАУ216 = 0

тЗТ┬аy(y┬а+18) -12(y┬а+ 18) = 0

тЗТ (y┬а+ 18)(y┬атАУ 12) = 0

тЗТ┬аy┬а= -18, 12

As we know, the side of a square cannot benegative.

Hence, the sides of the squares are 12 m and(12 + 6) m = 18 m.


Comment(S)

Show all Coment

Leave a Comment

Free - Previous Years Question Papers
Any questions? Ask us!
×