8th Class Mathematics Mensuration Question Bank Mensuration

  • question_answer A room of the hall is such that the ratio of the height of the room to its semi perimeter is 6 : 10. the cost of paper to cover the wall of the room is Rs. 1700, when the width of the paper is 100 cm and the rate to cover the wall is Rs. 10 per meter. The door and window whose area is 40 m2 are not to be covered. The height of the room is

    A)  \[2\sqrt{6}\,m\]            

    B)  \[6\sqrt{0.475}\,m\] 

    C)  \[4\sqrt{6}\,m\]                        

    D)  \[5\sqrt{6}\,m\]

    Correct Answer: B

    Solution :

    (b): semi perimeter \[=\left( 1+b \right)\] \[\therefore \frac{h}{1+b}=\frac{6}{10}\]            \[\begin{align}   & h=6 \\  & 1+b=10x \\ \end{align}\] Effective area of wall \[=perimeter\times h=40\] \[=2\left( 1+b \right)h-40\] \[=2\times 10x\times 6x-40\] \[=120{{x}^{2}}-40.\] Cost to cover = area (in \[{{m}^{2}}\]) \[\times \] cost per \[{{m}^{2}}\] For 100cm (0.1 m width) and 1m length i.e. for 0.1 \[{{m}^{2}}\], cost = Rs. 10. For 1 \[{{m}^{2}}\], cost = Rs. 100. Cost to cover \[=1700=\left( 120{{x}^{2}}-40 \right)\times 100\] \[\Rightarrow 120{{x}^{2}}=57\] \[\Rightarrow x=\sqrt{\frac{57}{120}}=\sqrt{\frac{19}{40}}=\sqrt{0.475}\] \[\therefore 6x=6\sqrt{0.475}.\]

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