Question

sum of the area of two square is 260 m square if the difference of their perimeter is 24 M then find the sides of the two squares

Answer 1

Hope it will help you

Answer 2

$\red\bigstar$ SOLUTION:-

Let the sides of the two squares be a metres and b metres.

Then ,Their area are (a²)m² and (b²)m² respectively.

And their perimeters are (4a)m and (4b) m respectively.

∴4a-4b=24

⇛4(a-b)=24

⇛a-b=6

⇛b=(a-6) .....(1)

Sum of their areas =260m²

∴ a²+b²=260

⇛a²+(a-6)²=260. [using (1)]

⇛2a²-12a-224=0

⇛a²-6a-112=0

⇛a²-14a +8a -112=0

⇛a(a-14)+8(a-14)=0

⇛(a-14)(a+8)=0

⇛a-14=0 or a+8=0

⇛a=14 or a=-8

But the sides of the square cannot be in negative

∴a=14 and b=(14-6)=8

Hence, the sides of the square are 14 m and 8 m respectively.

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Another method

Let the sides of square be x and y

Then their peimeters are 4x and 4y and areas be x² and y²

a/q

x²+y²=260...........(1) 

Also,

4x-4y=24

4(x-y)=24

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

Solving above equation  

(x+y)² +(x-y)² =2(x²+y²)

(x+y)²+(6)² =2(260) [from 2]

(x+y)²+ 36=520

(x+y)²=520-36=484

x+y=22.............(3)

Now adding eqn. 2 and 3...........

x-y+x+y=6+22

2x=28

x=14m

y=22-x=22-14=8m

Hence, the sides of the squares are 14 m and 8m respectively

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