Question

please answers this question plz and fast

Answer 1

let side of first square = x

side of second square = y

sum of area of two square = 260 sq m

i.e     x^2 + y^2 = 260    (i)

now the difference of their perimeter = 24 m

i.e     4x - 4y = 24

=> x - y = 6     (ii)   (taken 4 common)

squarring equation (ii)

(x - y)^2 = 6^2

x^2 + y^2 - 2 xy = 36

260 - 2 xy =36           (value from equation (i)

224 - 2 (6+y) y =0       (value from equation (ii)

2y^2 + 12 y - 224 = 0    (multiplying both side with  -1)

y^2 +6y - 112 =0

y^2 + 14 y - 8y - 112 = 0

y (y+14) - 8 (y+14) =0

y= 8 or -14 m

here we take positive number 

so y= 8

therefore  x-y = 6

=> x= 6+8 = 14 m

hope it helps

Answer 2

Answer:

Step-by-step explanation:

let side of first square = x

side of second square = y

sum of area of two square = 260 sq m

i.e     x^2 + y^2 = 260    (i)

now the difference of their perimeter = 24 m

i.e     4x - 4y = 24

=> x - y = 6     (ii)   (taken 4 common)

squarring equation (ii)

(x - y)^2 = 6^2

x^2 + y^2 - 2 xy = 36

260 - 2 xy =36           (value from equation (i)

224 - 2 (6+y) y =0       (value from equation (ii)

2y^2 + 12 y - 224 = 0    (multiplying both side with  -1)

y^2 +6y - 112 =0

y^2 + 14 y - 8y - 112 = 0

y (y+14) - 8 (y+14) =0

y= 8 or -14 m

here we take positive number 

so y= 8

therefore  x-y = 6

=> x= 6+8 = 14 m

please mark brainliest