Question

7.
Sum of the areas of two squares is 544 m2. If the difference of their
perimeters is 32 m, find the sides of the two squares.​

Answer 1

The side of squares are 20 and 12

Step-by-step explanation:

let the sides of the two squares are x and y

then according to question

x²+y²=544..(1)

4x-4y=32

4(x-y) = 32

x-y = 8...(2)

x = 8+y..(3)

put the value of x in equation 1

x²+y²=544

(8+y)²+y²=544

8²+y²+16y+y²=544

64+y²+16y+y²=544

2y²+16y = 544-64

2y²+16y = 480

y²+8y-240=0

y²+20y-12y-240=0

y(y+20) -12(y+20)

(y-12)(y+20)

y-12=0

y = 12

y+20=0

y =-20 (neglected )

then put the value of y in equation 3

x = 8+y

x = 8+12 = 20

hence , The side of squares are 20 and 12

#Learn more:

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.

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