Question

The length of the rectangle increase by 15 and breadth is increased by 10%, what is the perentage of increase in area of the rectangle

Answer 1

Hey there!
Here's your answer :

Let the initial length, breadth be l, b.

Initial area = lb.

Length is increased by 15% ,
New length = l + 3/20l = 23/20l .

Breadth is increased by 10% ,
New breadth = b + 10/100b = b + 1/10b = 11/10b .

New Area = 23/20 * 11/10 lb = 253 / 200 lb .

Increase in Area = 253/200lb - lb = 53/200 lb.

Percentage increase in Area = 53/200 lb / lb * 100 = 53/200 * 100
= 53/2
= 26.5%

There is an increase of 26.5%

Answer 2

Hey there! ☺️
Thanks for your Brainliest Question.

Let the length and breadth of the original rectangle be x & y respectively. So its area will be xy

Now, the length of the rectangle is increased by 15% i.e. 
$x {}^{2} = x + \frac{15}{100} x = x + \frac{1}{2} x = \frac{3}{2} x$
And the breadth of the rectangle is increased by 10% i.e.
$y {}^{2} = y + \frac{10}{100} y= y + \frac{1}{10} y = \frac{11}{10} y$
Thus the area of the new rectangle will be 
$x {}^{2} y {}^{2} = \frac{3}{2} x \times \frac{11}{10} y = \frac{33}{20} xy$
Thus, the area of the new rectangle will be 33/20 times the area of the original rectangle.

Now let's calculate the relative percentage increase in the area of the rectangle

$\frac{A 2 - A1}{ A1 } \times 100 = \frac{ \frac{33}{20} xy - xy}{xy} \times 100 \\ = \frac{ \frac{13}{20}xy }{xy} \times 100 \\ = \frac{13}{20} \times 100 = 13 \times 5 = 65$
Thus, there is a total increase of 65% in the area of the original rectangle.