Question

there is a rectangle abcd.the length of the rectangle is increased by 10%.the width of the rectangle is increased by 5%,what is the percentage increase in the recatangle.

Answer 1

Answer:

Therefore the area of rectangle is increased by 15.5%.

Step-by-step explanation:

Assume the length and width of the rectangle be x and y respectively.

The area of the rectangle is = length×width

                                               =xy.

Now the length of the rectangle is increased by 10%.

Then the length of the rectangle increased = $x\times \frac{10}{100}$.

New length of the rectangle is $=x+\frac{10x}{100}$

                                                    $=\frac{100x+10x}{100}$

                                                    $=\frac{110x}{100}$

The width of the rectangle is increased by 5%.

Then the width of the rectangle increased = $y\times \frac{5}{100}$.

New length of the rectangle is $=y+\frac{5y}{100}$

                                                    $=\frac{100y+5y}{100}$

                                                    $=\frac{105y}{100}$

New area of the rectangle is = $\frac{110x}{100}\times\frac{105y}{100}$

                                               = 1.155 xy.

The percentage of area increase is

$=\frac{\textrm{New area - Original area}}{\textrm{Original area}}\times 100$

$=\frac{1.155xy-xy}{xy}\times 100$

$=\frac{0.155xy}{xy}\times 100$

=15.5.

Therefore the area of rectangle is increased by 15.5%.

Answer 2

Answer:

Step-by-step explanation: