Question

A square and a rectangle each have a perimeter of 40 m. The difference between areas of the two figures is 9
m2. What are the possible dimensions of the rectangle?
(A) 13 m, 7 m
(B) 14 m, 6 m
(C) 108 m, 1 m
(D) both (A) and (C)​

Answer 1

Step-by-step explanation:

hope this is helpful for you...

Answer 2

Answer:

(a) 13 m, 7 m

Step-by-step explanation:

Given :

A square and a rectangle have equal perimeter, and is 40 m.

| Area of Rectangle - Area of Square | = 9m²

Formulae :

Perimeter of a square = 4s units

Perimeter of a rectangle = 2(l + b) units

Area of a rectangle = (l × b) units²

Area of a square = s² units²

Procedure :

4s = 2(l + b) = 40 m

Dividing the whole equation by 2,

⇒ 2s = l + b = 20 m

⇔ 2s = 20 m and l + b = 20 m

∴ s = 10 m.

We found out that the side of the square is 10 m.

| Area of Rectangle - Area of Square | = 9m²

⇒ | (l × b) - (10 m)² | = 9m²

⇒ | lb - 100 m² | = 9 m²

Hence lb = (100 + 9) m² or (100 - 9) m²

∴ lb = 109 m² or 91 m²

As l + b = 20 m

From both the equations,

l + b = 20 m

lb = 109 m² or 91 m²

⇒ lb = 91 m² [109 does not have any factors, ∵ It is a prime number]

l + b = 20 m

lb = 91 m²

Lets take and find b :

b² - 20b + 91 = 0

[Converted into the form of x² - (α + β)x + (αβ) = 0]

-13, -7 are the factors of 91 which add up to -20.

b² - 13b - 7b + 91 = 0

⇒ b(b - 13) - 7(b - 13)

⇒ (b - 13)(b - 7) = 0

Hence b = 13 m and l = 7 m (OR) b = 7 m and l = 13 m.

b = 13 m and l = 7 m is incorrect, ∵ Length is always greater than breadth.

Hence option(a) ⇔ 13 m, 7 m, is correct.

Thanks !