Question

a door of length 2.2 M and breadth 0.9 M is fitted in a wall the dimension of the wall are 5 litre and 3.5 M find the cost of whitewashing the wall at 35 per square metre
plz answer with a figure ​

Answer 1

Appropriate Question :-

A door of length 2.2 m and breadth 0.9 n is fitted in a wall and the dimension of the wall are 5 m and 3.5 m. Find the cost of whitewashing the wall at 35 per square metre.

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

Given that, a door of length 2.2 m and breadth 0.9 m is fitted in a wall and the dimension of the wall are 5 m and 3.5 m

Dimensions of wall

Length of wall = 5 m

Breadth of wall = 3.5 m

So,

$\rm \: Area_{(wall)} = Length \times Breadth$

$\rm \: Area_{(wall)} = 5 \times 3.5$

$\rm\implies \:Area_{(wall)} = 17.5 \: {m}^{2}$

Dimensions of door

Length of door = 2.2 m

Breadth of door = 0.9 m

So,

$\rm \: Area_{(door)} = Length \times Breadth$

$\rm \: Area_{(door)} = 2.2 \times 0.9$

$\rm\implies \:Area_{(door)} = 1.98 \: {m}^{2}$

So,

$\rm \: Area_{(wall \: to \: be \: white \: washed)}$

$\rm \: = \: Area_{(wall)} - Area_{(door)}$

$\rm \: = \: 17.5 - 1.98$

$\rm \: = \: 15.52 \: {m}^{2}$

Now, Given that

$\rm \: Cost\:of\:whitewashing\:1 \: {m}^{2} = Rs \: 35$

Thus,

$\rm \: Cost\:of\:whitewashing\:15.52 \: {m}^{2} =15.52 \times 35 = Rs \: 543.20$

$\rule{190pt}{2pt}$

Additional Information :-

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