Question

The length a rectangle is increased by 20% and its breadth is increased by 10 % . Which of the following is a possible value of the percentage increase in its perimeter?
A. 13%
B. 14%
C. 12%
D. 16%​

Answer 1

Answer:

Option D

Step-by-step explanation:

Given :-

The length a rectangle is increased by 20% and its breadth is increased by 10 % .

To find :-

Which of the following is a possible value of the percentage increase in its perimeter?

A. 13%

B. 14%

C. 12%

D. 16%

Solution :-

Let the length of the rectangle be l units

Let the breadth of the rectangle be b units

Then, Perimeter of the rectangle = 2(l+b) units

P = 2(l+b) units --------(1)

If the length of the rectangle is Increased by 20% then

The new length = l+20% of l

=> l + 20% × l

=> l +(20/100)×l

=> l + (1/5)×l

=>l + (l/5)

=> (5l+l)/5

=> 6l/5 units

and

If the breadth of the rectangle is increased by 10% then the new breadth

=> b+10% of b

=> b+10% × b

=> b+(10/100)×b

=> b+(1/10)×b

=> b+(b/10)

=> (10b+b)/10

=> 11b/10

We have

New length = 6l/5 units

New breadth = 11b/10 units

Then

The new perimeter

=> 2[(6l/5)+(11b/10)]

=> 2[(12l+11b]/10

=> (12l+11b)/5 units

Increasing in perimeter

= New perimeter - Original perimeter

= (12l+11b)/5 - 2(l+b)

=> [(12l+11b)-(10(l+b))]/5

=> (12l+11b-10l-10b)/5

=> (2l+b)/5 units

Now

Percentage of increasing in perimeter

=> (increased perimeter/Original perimeter)×100

=> [[(2l+b)/5]/[2(l+b)]]×100

=> [(2l+b)/10(l+b)]×100

=> [(2l+b)/(l+b)]×10

=> (20l+10b)/(l+b) %

The increased percentage in the perimeter is (20l+10b)/(l+b) %

Now,on applying for any lengths and breadths of any rectangle we get increasing percentage in its perimeter as given below:

Let l = 10 units and b = 5 units then

=> (20×10+10×5)/(10+5)

=> (200+50)/15

=> 250/15

=> 50/3

=> 16.666...

=> 16.67% ~ 16%

and on taking l = 2 units and b = 1 units then

=> (20×2+10×1)/(2+1)

=> (40+10)/3

=> 50/3

=> 16.67%~16%

Answer:-

The possible value of the percentage increase in its perimeter is 16%

Used formulae:-

• Perimeter of the rectangle = 2(l+b) units

Where , l = length of the rectangle

and , b = breadth of the rectangle

Answer 2

Answer:

16%

Step-by-step explanation:

Let the length of the rectangle be l units

Let the breadth of the rectangle be b units

Then, Perimeter of the rectangle = 2(l+b) units

P = 2(l+b) units --------(1)

If the length of the rectangle is Increased by 20% then

The new length = l+20% of l

=> l + 20% × l

=> l +(20/100)×l

=> l + (1/5)×l

=>l + (l/5)

=> (5l+l)/5

=> 6l/5 units

and

If the breadth of the rectangle is increased by 10% then the new breadth

=> b+10% of b

=> b+10% × b

=> b+(10/100)×b

=> b+(1/10)×b

=> b+(b/10)

=> (10b+b)/10

=> 11b/10

We have

New length = 6l/5 units

New breadth = 11b/10 units

Then

The new perimeter

=> 2[(6l/5)+(11b/10)]

=> 2[(12l+11b]/10

=> (12l+11b)/5 units

Increasing in perimeter

= New perimeter - Original perimeter

= (12l+11b)/5 - 2(l+b)

=> [(12l+11b)-(10(l+b))]/5

=> (12l+11b-10l-10b)/5

=> (2l+b)/5 units

Now

Percentage of increasing in perimeter

=> (increased perimeter/Original perimeter)×100

=> [[(2l+b)/5]/[2(l+b)]]×100

=> [(2l+b)/10(l+b)]×100

=> [(2l+b)/(l+b)]×10

=> (20l+10b)/(l+b) %

The increased percentage in the perimeter is (20l+10b)/(l+b) %

Now,on applying for any lengths and breadths of any rectangle we get increasing percentage in its perimeter as given below:

Let l = 10 units and b = 5 units then

=> (20×10+10×5)/(10+5)

=> (200+50)/15

=> 250/15

=> 50/3

=> 16.666...

=> 16.67% ~ 16%

and on taking l = 2 units and b = 1 units then

=> (20×2+10×1)/(2+1)

=> (40+10)/3

=> 50/3

=> 16.67%~16%