Question

guys find the answer​

Answer 1

Answer:

Q12 ans 9cm×3cm

27m^2

Q13 ans 4cm×4cm

16cm^2

Answer 2

Answer:

$\large \star \:\underline{ \sf \pmb{Question.1}}$

Find the area of Rectangle with lenght = 9 cm and Breadth = 3 cm.

$\large \star \: \underline{\sf {\pmb{Given}}}$

• ➥ Lenght of Rectangle = 9 cm

• ➥ Breadth of Rectangle = 3 cm

$\large \star \: \underline{\sf \pmb{To \: Find }}$

• ➥ Area of Rectangle

$\large \star \: \underline{\sf \pmb{Using \: Formula }}$

$\underline{ \boxed{\sf{Rectangle_{(Area)} = Length \times Breadth}}}$

$\large \star \: \underline{\sf \pmb{Solution}}$

${\implies \: \sf{Rectangle_{(Area)} = Length \times Breadth}}$

• Substituting the values

${ \implies{\sf{Rectangle_{(Area)} =9 \: cm \times 3 \: cm}}}$

${ \implies{\sf{Rectangle_{(Area)} = 27 \: {cm}^{2} }}}$

$\: \: \underline{\boxed {\mathsf \purple{Area \: of \: Rectangle = 27 \: {cm}^{2}}}}$

$\large \star \: \underline{\sf \pmb{Diagram }}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 9 cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 3 cm}\put(-0.5,-0.4){\bf }\put(-0.5,3.2){\bf }\put(5.3,-0.4){\bf }\put(5.3,3.2){\bf }\end{picture}$

$\large \star \: \underline{\sf {\pmb{Therefore}}}$

• ➠ The Area of Rectangle is 27cm²

━━━━━━━━━━━━━━━

$\large \star \: \underline{\sf \pmb{Question.2}}$

Find the Area of Square with side 4 cm.

$\large \star \: \underline{\sf {\pmb{Given}}}$

• ➥ Side of Square = 4 cm

$\large \star \: \underline{\sf \pmb{To \: Find }}$

• ➥ Area of Square

$\large \star \: \underline{\sf \pmb{Using \: Formula }}$

${\underline{\boxed{\sf{Square_{(Area)} = {a}^{2}}}}}$

Where

• Side of Square

$\large \star \: \underline{\sf \pmb{Solution }}$

$\implies{\sf{Square_{(Area)} = {a}^{2}}}$

• Substituting the values

$\implies{\sf{Square_{(Area)} = {(4 \: cm)}^{2}}}$

$\implies{\sf{Square_{(Area)} =4 \times 4 }}$

$\implies{\sf{Square_{(Area)} =16 \: {cm}^{2} }}$

$\: \: \underline{\boxed{\sf \purple{Square_{(Area)} = 16 \: {cm}^{2} }}}$

$\large \star \: \underline{\sf \pmb{Diagram }}$

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

$\large \star \: \underline{\sf \pmb{Therefore}}$

• ➠ The Area of Square is 16 cm².

━━━━━━━━━━━━━━━

$\large \star \: \underline{\sf \pmb{Additional \: Information}}$

$\begin{gathered}\begin{gathered}\small\begin{gathered}\begin{gathered} \bigstar \: \bf \underline{More \: Useful \: formulae} \: \bigstar\\ \begin{gathered} \\ {\boxed{\begin{array}{cccc} \\ {\displaystyle \star\sf{Perimeter \: of \: Triangle }} = \sf{Length +Breadth + Height} \\ \\ \star\sf{Area \: of \: Rectangle = Lenght×Breadth} \\ \\ {\star\small\sf{Perimeter \: of\: Rectangle = 2(Length + Breadth)}} \\ \\ {\star\small \sf{Diagonal \: of \: Rectangle = \sqrt{ {Length }^{2} + { Breadth}^{2} }}} \\ \\ \star\small\sf{Perimeter \: of \: Square = 4a } \\ \\ \star\small\sf{Diagonal \: of \: Square = a \sqrt{2} }\end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$