Question

The sum of areas of two squares is 400 sq.m .If the difference between their perimeter is 16 m , find the sides of squares​

Answer 1

Step-by-step explanation:

let first = x

second = y

• difference in perimeter = 16m

so...

4x - 4y = 16

4(x-y) = 16

x-y = 16/4 = 4

so.... x - y = 4 (where x and y are sides)

x = 4+y

• now.. sum of their area =400

x²+y² = 400

putting value of x in this eq..

(4+y)² + y² = 400

( 4²+y²+2*4*y) +y² = 400 (using Identity (a+b)²)

16 +y² +8y +y² = 400

16 +2y² +8y -400 = 0

taking 2 common

2(8+y²+4y-200) = 0

y²+4y - 192 = 0/2

• By middle term spillting

y² +16y-12y -192 = 0

y(y+16)- 12(y+16)

(y+16)(y-12) =0

• taking first factor = y = -16
• taking second factor = y =12

• so y = 12 (because in first it is -16 and the side never be negative)

• now.. we know that

x-y = 4 (proved above)

x - 12 = 4

x = 4+12 = 16

$\huge{\bold{\green{\underline{\mathcal{x=16cm}}}}}$

$\huge{\bold{\green{\underline{\mathcal{y=12cm}}}}}$

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