Question

Heya!!!

Answer the question don't send pics

Only for class 10th CBSE

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

Don't answer rubbish
Explain it clearly

I will mark brainliest to the required answer

Answer 1

Let the sides of the two squares be x and y.

Given that sum of the areas of two squares is 468m^2.

x^2 + y^2 = 468    ----------- (1)


We know that perimeter of a square = 4 * side

Given that difference of their perimeter is 24cm.

4x - 4y = 24

x - y = 6

x = y + 6      ---------- (2)


Substitute (2) in (1), we get

(y + 6)^2 + y^2 = 468

y^2 + 36 + 12y + y^2 = 468

2y^2 + 12y - 432 = 0

y^2 + 6y - 216 = 0

y^2 - 12y + 18y - 216 = 0

y(y - 12) + 18(y - 12) = 0

(y - 12)(y + 18) = 0

y = 12 (or) y = -18

y cannot be negative, so y = 12.

Substitute y = 12 in (1), we get

x^2 + y^2 = 468

x^2 + (12)^2 = 468

x^2 + 144 = 468

x^2 = 468 - 144

x^2 = 324

x = 18


Therefore the sides of the two squares = 18m and 12m.


Hope this helps!

Answer 2

Hey mate !!!


let the side of first square be x


therefore , area will be =>. x²

similarly ,

let the side be y


area will be y²


now , according to question ,

x² + y ² = 468 cm ² -------------(1)




now the difference between their perimeter

will be


4x - 4y = 24


4( x-y ) = 24


x-y = 6 --------------(2)




x = y+6 ------- (3)



now , putting the Value of equation (3) in first we will get



( y+ 6) ² + y² = 468



on factorize we get =>

Y² + 6y = 216

=> y² + 6y - 216 = 0




y² + 18 y - 12y -216 =0


=>y (. y + 18 ) - 12 ( y + 18 ) =0





=> (y+18 ) - (y-12) = 0


y= 12 , -18



hence ,


side of 1st square = 12


and second


x = 12 + 6


x = 18



hence , sides is 12 cm and 18 cm




hope it helps :D



thanks