Question

27. The sum of areas of two squares is 468

and difference of their perimeters is 24 m, find the length

of sides of each square.​

Answer 1

Step-by-step explanation:

$Let \: the \: sides \: of \: the \: two \: squares \: be \: of \: length \: a \: m \: and \: b \: m \: respectively \: with \: a > b \\ \\ according \: to \: the \: question \: \\ \\ {a}^{2} + {b}^{2} = 468 \: \: \: \: \: \: \: \: say \: eq \: 1 \\ \\ 4(a - b) = 24 \\ (a - b) = 6 \: \: \: \: say \: eq \: 2 \\ \\ using \: equation \: 2 \: in \: 1 \: we \: get \: \\ (6 + b) {}^{2} + {b}^{2} = 468 \\ {2b}^{2} + 12b + 36 = 468 \\ {2b}^{2} + 12b - 432 = 0 \\ {b}^{2} + 6b - 216 = 0 \\ (b + 18) \: (b - 12) = 0 \\ \\ b \: is \: always \: postive \: \\ b = 12 \\ \\ then \: \\ from \: equation \: 2 \: we \: get \: \\ a - 12 = 6 \\ a = 18 \\ \\ So \: the \: sides \: are \: of \: length \: 18m \: and \: 12m \\ \\ \\ \\ \\ \\ hope \: its \: help \: uh \\ be \: brainly \:$

Answer 2

Answer:

1 st squar side = 12m

2nd squar side = 18m

Step-by-step explanation:

the perimeter of big squar=

4s=24-4s

=4s+4s=24

8s=24

s=24/8

=3

the 1st squar side is = 4×3=12m

and perimeter is = 4×s

4×12=48m

and the perimeter of 2nd squar is =

48 + 24=72m

then the side is 4s=72

s=72/4

=18m