We have a rectangle whose length and breadth is l m and b m respectively. (i) 2l × 2b (ii) l × b (iii) l × b (iv) None.

Question

We have a rectangle whose length and breadth is l m and b m respectively. The length of a rectangle is increased by 20% and breadth is decreases by 20%. 

Now, answer the following question that follows: 

(a) The original area of rectangle is: (in m²)

(i) 2l × 2b 

(ii) l × b

(iii) l × b

(iv) None.

(b) The new length is:

(i) \(\frac{6l}{5}\)m

(ii) \(\frac{3l}{5}\)m

(iii) \(\frac{5l}{3}\)m

(iv) None.

(c) The new breadth is:

(i) \(\frac{4b}{5}\)m

(ii) \(\frac{3b}{5}\)m

(iii) \(\frac{5b}{3}\)

(iv) None.

(d) The new area of rectangle is: (in m²)

(i) \(\frac{lb}{25}\)

(ii) \(\frac{lb}{24}\)

(iii) \(\frac{24lb}{25}\)

(iv) None.

(e) Percentage change in area will be:

(i) 2% 

(ii) 3% 

(iii) 4% 

(iv) None.

Answer 1

Given that length and breadth of rectangle is l m and b m respectively.

(a) Therefore, the area of rectangle is l × b m².

Hence, option (ii) is correct.

(b) Given that length of rectangle is increased by 20 %.

∴ Length of modified rectangle is l + \(\frac{20}{100} \)l = l + \(\frac{1}{5}\)l = \(\frac{6}{5}\)l m.

Hence, the new length is \(\frac{6}{5}\)l m.

Hence, option (i) is correct.

(c) Given that breadth of rectangle is decreased by 20%.

∴ Breadth of modified rectangle = b \(-\frac{20}{100}\)b = b \(-\frac{1}{5}\)b = \(\frac{4b}{5}\)m.

Hence, the new breadth is \(\frac{4b}{5}\)m.

Hence, option (i) is correct.

(d) Since, length and breadth of modified rectangle are \(\frac{6l}{5}\)m and \(\frac{4b}{5}\)m respectively.

∴ The area of modified rectangle is \(\frac{6l}{5}\times\frac{4b}{5}=\frac{24lb}{25}\) m².

Hence, the new area of rectangle is \(\frac{24lb}{25}\) m².

Hence, option (iii) is correct.

(e) The original area of rectangle is lb m2.

(e) The new area of rectangle after modification is \(\frac{24lb}{25}\) m².

Hence, the change in area = lb – \(\frac{24lb}{25}=\frac{lb}{25}.\)

Hence, option (iii) is correct.