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This is the second of our many quizzes on Net area. Click on this to access more such competitive quizzes on Mensuration.

Net area Quiz 2

mathematics

 A square of side $5cm$ is inside of a $8cm\times 10cm$ rectangle. Find the area of the remaining region. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6240e823-9f61-413c-92d6-aae24e181e223237812466659278010.png"/>

 Subtract the area of square EFGH  from the area of Rectangle ABCD to find the area of the remaining region.

  $\text{Area}\,\text{of}\,\text{remaining}\,\text{part}\,\text{=}\,\text{Area of rectangle - Area of square} \\ =10\left( 8 \right)-{{\left( 5 \right)}^{2}} \\ =80-25 \\  =55\,c{{m}^{2}}$

 Find the area bounded by the curve $y=-{{x}^{3}}+4{{x}^{2}}-3x$ , where $x\ge 0$ above the X-axis.

 First draw the curve of  $y=-{{x}^{3}}+4{{x}^{2}}-3x$ and select the region where $x\ge 0$ and then integrate it to find the required area.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/56f354de-31c6-4bfe-b220-531951d656ca6995208493427899456.png"/> $ \\ \text{When  }y=0\text{ , }x=0,1,3 \\ \text{From the graph area above X-axis is the shaded area} \\ \therefore \text{The area is} \\ =\int\limits_{1}^{3}{\left( -{{x}^{3}}+4{{x}^{2}}-3x \right)\,dx} \\ =\left[ -\dfrac{{{x}^{4}}}{4}+\dfrac{4{{x}^{3}}}{3}-\dfrac{3{{x}^{2}}}{2} \right]_{1}^{3} \\  =\left[ -\dfrac{{{3}^{4}}}{4}+\dfrac{4{{\left( 3 \right)}^{3}}}{3}-\dfrac{3{{\left( 3 \right)}^{2}}}{2} \right]-\left[ -\dfrac{{{1}^{4}}}{4}+\dfrac{4{{\left( 1 \right)}^{3}}}{3}-\dfrac{3{{\left( 1 \right)}^{2}}}{2} \right] \\ =\dfrac{8}{3}\,sq.units $

 Find the area bounded by the lines $y=2x-3$ and $y=-3$, $y=3$ and Y-axis.  

 First draw the graph of $y=2x-3$ ,  $y=-3$ ,$y=3$  and Y-axis , and then find the area bounded by these lines.

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ebe77843-8017-40d7-be53-8bf4d496d51d2110668748655772403.png"/> $ \\ \text{From the graph}, \\  \text{We see that it is a Right-angled triangle with }b=3;\,h=6 \\ \therefore \text{The area is} \\ \Rightarrow \dfrac{1}{2}\times b\times h \\  =\dfrac{1}{2}\times 3\times 6 \\  =9\,sq.units $

 Find the area bounded by the curve $y=9-{{x}^{2}}$ and the lines $y=x-1;\,x=-2$ and $x=1$.

 First draw the graph of $y=9-{{x}^{2}}$ and then select the region bounded by the lines $y=x-1;\,x=-2$ and $x=1$ and then integrate it to find the area.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/27d97330-6022-4cf6-bca9-ed0d7c25700e9068328805788167866.png"/> $ \\ \text{The area of the shaded region is} \\ \int\limits_{-2}^{1}{\left( \left( 9-{{x}^{2}} \right)-\left( x-1 \right) \right)\,dx} \\  =\int\limits_{-2}^{1}{\left( -{{x}^{2}}-x+10 \right)\,dx} \\  =\left[ \dfrac{-{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+10x \right]_{-2}^{1} \\ =\left[ \dfrac{-{{1}^{3}}}{3}-\dfrac{{{1}^{2}}}{2}+10\left( 1 \right) \right]-\left[ \dfrac{-{{\left( -2 \right)}^{3}}}{3}-\dfrac{{{\left( -2 \right)}^{2}}}{2}+10\left( -2 \right) \right] \\ =28.5\,sq.units $

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3447372d-4d6c-4f22-bef8-75ca5995f9ef7969900276002573906.png"/> $ \\ \text{From the graph,} \\ \text{We can see that they are all right-angled triangles} \\ \therefore \text{ The area is} \\ =\left( \dfrac{1}{2}\times 2\times 2 \right)+\left( \dfrac{1}{2}\times 1\times 1 \right)+\left( \dfrac{1}{2}\times 1\times 1 \right)+\left( \dfrac{1}{2}\times 2\times 2 \right) \\ =2+\dfrac{1}{2}+\dfrac{1}{2}+2 \\ =5\,sq.units $