Question

Sum of two sqaure is 468 if the differene their perimeter is 24 km find tve sicle of 2 sqaure

Answer 1

Answer:

Let the sides of squares be a m & b m long such that a > b.

By 1st condition,  

a$^{2}$ + b$^{2}$  = 468     ... (i)

By 2nd condition,

4a-4b = 24

a - b = 6.

a = b + 6.

Putting this value of a in equation (i),

(b+6)$^{2}$  + b$^{2}$  = 468

b$^{2}$  + 12b + 36 + b$^{2}$  = 468

b$^{2}$  + 6b + 18 = 234

b$^{2}$  + 6b - 216 = 0.

b$^{2}$  + 18b - 12b - 216 = 0.

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

(b+18)(b-12) = 0.

b = -18 or b = 12.

But b = -18 is unacceptable so  

b = 12.

a = 18.

The side of two square = 12 and 18

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@GauravSaxena01