Question

The perimeters of a square and rectangle are equal. One side of the rectangle is 11 cm and the area of the square is 4 cm sq. more than the area of the rectangle. Find the side of the square.

Answer 1

length of rect=11
let the other side =x
let the side of sq be y
2(11+x)=4y
11+x=2y-eq1
y2+4=11x-eq2
solve both the eqs and u get x&y

Answer 2

The correct answer is 9 and 13.

Given: The perimeters of a square and rectangle are equal.

One side of the rectangle is 11 cm.

To Find: The side of square.

Solution:

Perimeter of square = Perimeter of rectangle

4a = 2(l+b)

2a = l+b

One side of rectangle = 11cm.

2a = 11 + b

b = 2a - 11

Area of square = 4 + area of rectangle

$a^{2} =4+lb$

$a^{2} =4+(11)b$

Put b = 2a - 11 in tis equation.

$a^{2} =4+11(2a-11)$

$a^{2} =4+22a-121$

$a^{2} -22a+117=0$

$a^{2} -13a-9a+117=0\\ a(a-13)-9(a-13)$

(a-13)(a-9)

a = 9, 13.

Hence, there can be two values of side of square 9 and 13.

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