Question

If the area of rectangle is increased by 13% and its
breadth is increased by 5%, then what is the per-
centage increase in its length? (Approximately)​

Answer 1

Answer:

Percentage increase in perimeter is 17.14 %

Step-by-step explanation:

Let the rectangle have sides: L and B.

Area = LB and the perimeter is 2(L+B).

The area of the rectangle is increased by 32%, by increasing the breadth by 10%.

That is for 1.32LB =L(n)*1.1B or

L(n) = 1.32LB/1.1B = 1.2L. It means the length is increased by 20% and the breadth by 10%.

The new rectangle is 1.2Lx1.1B, and the perimeter is 2(1.2L+1.1B).

However the increase in perimeter is not consistent and cannot be predicted without actual dimensions of the rectangle.

Case 1: Let L = 20 and B = 10. Area = 20*10 = 200. P = 2(20+10) = 60

New L = 1.2*20 = 24 and B = 1.1*10 = 11. Area = 264. P =2(24+11) = 70.

Percentage increase in area = (264–200)*100/200 = 32%.

Percentage increase in perimeter = (70–60)*100/60 = 16.6666%.

Case 2: Let L = 40 and B = 30. Area = 40*30 = 1200. P = 2(40+30) = 140

New L = 1.2*40 = 48 and B = 1.1*30 = 33. Area = 1584. P =2(48+33) = 162.

Percentage increase in area = (1584–1200)*100/1200 = 32%.

Percentage increase in perimeter = (162–140)*100/140 = 15.714%.

Case 3: Let L = 50 and B = 20. Area = 50*20 = 1000. P = 2(50+20) = 140

New L = 1.2*50 = 60 and B = 1.1*20 = 22. Area = 1320. P =2(60+22) = 164.

Percentage increase in area = (1320–1000)*100/1000 = 32%.

Percentage increase in perimeter = (164–140)*100/140 = 17.14%.