Question

A room is 15 m long and 9.5 m wide. A square carpet of side 11 m is laid on the floor. How much area is left uncarpeted?​

Answer 1

Answer:

Diagram :

Diagram of rectangular room :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large{15\ m}}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large{9.5\ m}}\put(-0.5,-0.4){\bf}\put(-0.5,3.2){\bf}\put(5.3,-0.4){\bf}\put(5.3,3.2){\bf}\end{picture}$

$\rule{300}{1.5}$

Diagram of square carpet :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large{11\ m}}\put(4.4,2){\bf\large{11\ m}}\end{picture}$

Given :

• Lenght of rectangular room = 15 m.
• Breadth of rectangular room = 9.5 m.
• Side of square carpet = 11 m.

$\begin{gathered}\end{gathered}$

To Find :

• Area of left uncarpeted?.

$\begin{gathered}\end{gathered}$

Using Formulas :

${\longrightarrow{\small{\underline {\boxed{\sf{Area \: of \: rectangle = Length \times Breadth}}}}}}$

${\longrightarrow{\small{\underline {\boxed{\sf{Area \: of \: square = (side)^2}}}}}}$

${\longrightarrow{\small{\underline {\boxed{\sf{Uncarpeted \: area = Area_{(Rectangle \: room)} - Area_{(Square \: carpet)}}}}}}}$

$\begin{gathered}\end{gathered}$

Solution :

Finding the area of rectangular room ;

${\longrightarrow \: \: {\sf{Area \: of \: rectangle = Length \times Breadth}}}$

${\longrightarrow \: \: {\sf{Area \: of \: rectangle = 15 \times 9.5}}}$

${\longrightarrow \: \: {\sf{Area \: of \: rectangle = 142.5 \: {m}^{2}}}}$

${\bigstar \:{\underline{\boxed{\sf{\red{Area \: of \: rectangular \: room = 142.5 \: {m}^{2}}}}}}}$

∴ The area of rectangular room is 142.5 m².

$\rule{300}{1.5}$

Finding the area of square carpet :

${\longrightarrow \: \: {\sf{Area \: of \: square = (side)^2}}}$

${\longrightarrow \: \: {\sf{Area \: of \: square = (11)^2}}}$

${\longrightarrow \: \: {\sf{Area \: of \: square = (11 \times 11)}}}$

${\longrightarrow \: \: {\sf{Area \: of \: square = 121 \: {m}^{2}}}}$

${\bigstar \: {\underline{\boxed{\sf{\red{Area \: of \: square \: carpet= 121 \: {m}^{2}}}}}}}$

∴ The area of square carpet is 121 m².

$\rule{300}{1.5}$

Now, finding the area is left uncarpeted :

${\longrightarrow \: \: {\sf{Uncarpeted \: area = Area_{(Rectangle \: room)} - Area_{(Square \: carpet)}}}}$

${\longrightarrow \: \: {\sf{Uncarpeted \: area = 142.5 - 121}}}$

${\longrightarrow \: \: {\sf{Uncarpeted \: area = 142.5 - 121.0}}}$

${\longrightarrow \: \: {\sf{Uncarpeted \: area = 21.5 \: {m}^{2}}}}$

${\bigstar \: {\underline{\boxed {\sf{ \red{Uncarpeted \: area = 21.5 \: {m}^{2}}}}}}}$

∴ The uncarpeted area is 21.5 m².

$\begin{gathered}\end{gathered}$

Learn More :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

$\rule{220pt}{3pt}$