Question

The length of the rectangle should be increased by 37 1/2% and the width should be reduced by 30%. Find the change in area?​

Answer 1

ANSWER:

Given:

• Length of the rectangle is increased by 37 1/2%(⁷⁵/₂%).

• Breadth of the rectangle is decreased by 30%.

To Find:

• Change in area of the rectangle

Solution:

Let the initial length and breadth of the rectangle be L and B respectively.

So,

Initial Area of the rectangle = (L*B)sq. units ————(1)

We are given that, the length of the rectangle is increased by ⁷⁵/₂%.

So,

⇒ Increased Length = L + ⁷⁵/₂%of L

⇒ Increased Length = L + ⁷⁵/₂ * ¹/₁₀₀ *L

⇒ Increased Length = L + ⁷⁵/₂₀₀ L

⇒ Increased Length = (200 L + 75 L)/200

⇒ Increased Length = 275L/200 = ¹¹/₈ L ————(2)

We are given that, the breadth of the rectangle is decreased by 30%.

So,

⇒ Decreased Breadth = B - 30%of B

⇒ Decreased Breadth = B - 30 * ¹/₁₀₀ * B

⇒ Decreased Breadth = B - ³⁰/₁₀₀ B

⇒ Decreased Breadth = (100 B - 30 B)/100

⇒ Decreased Breadth = 70B/100 = ⁷/₁₀ B ————(3)

⇒ New Area of the rectangle = Changed Length * Changed Breadth

from (2) & (3)

⇒ New Area of the rectangle = ¹¹/₈ L * ⁷/₁₀ B

⇒ New Area of the rectangle =(⁷⁷/₈₀ L*B) sq. units ————(4)

Change in area = New area - Initial area

from (1) & (4)

⇒ Change in area = ⁷⁷/₈₀ L*B - L*B

⇒ Change in area = (77 LB - 80LB)/80

⇒ Change in area = -3LB/80

As the answer is in negative, there is decrease in the area.

%. change = (change in area)/(initial area) * 100

%. change = (3LB/80) / (LB) * 100

%. change = (3LB)/(80LB) *100

%. change = 0.0375 * 100

%. change = 3.75%

Hence, the area of the rectangle is decreased by (3/80) of the initial area.

OR

The area of the rectangle is decreased by 3.75%.

Answer 2

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\bf \:Let - \begin{cases} &\sf{Length_{(rectangle)} = x \: units} \\ &\sf{Breadth_{(rectangle)} = y \: units} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf \:Area_{(rectangle)} = (A_1) - \begin{cases} &\sf{xy \: square \: units}  \end{cases}\end{gathered}\end{gathered}$

Now,

• New dimensions of rectangle,

According to statement,

$\rm :\longmapsto\:Length_{(rectangle)} \: is \: increased \: by \: 37 \: \dfrac{1}{2} \%$

and

$\rm :\longmapsto\:Breadth_{(rectangle)} \: is \: reduced \: by \: 30\%$

So,

$\rm :\longmapsto\:Length_{(rectangle)} = x + \dfrac{75}{200}x = x + \dfrac{3x}{8} = \dfrac{11x}{8}$

$\rm :\longmapsto\:Breadth_{(rectangle)} = y - \dfrac{30}{100}y = y - \dfrac{3y}{10} = \dfrac{7y}{10}$

$\rm :\longmapsto\:Area_{(rectangle)} = A_2 = \dfrac{11x}{8} \times \dfrac{7y}{10} = \dfrac{77xy}{80}$

$\rm :\longmapsto\:\%age \: change = \dfrac{A_2 - A_1}{A_1} \times 100\%$

$\rm :\longmapsto\:\%age \: change =\dfrac{\dfrac{77xy}{80} -xy}{xy} \times 100\%$

$\rm :\longmapsto\:\%age \: change ={\dfrac{77xy - 80xy}{80xy}} \: \times 100\%$

$\rm :\longmapsto\:\%age \: change =\dfrac{ - 3xy}{80xy} \: \times 100 \: \%$

$\rm :\longmapsto\:\%age \: change = - \dfrac{30}{8}\% = - 3.75\%$

$\rm :\implies\:Area_{(rectangle)} \: get \: reduced \: by \: 3.75\%$

$</p><p>\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$