Ex.11.2 Q7 Mensuration Solution - NCERT Maths Class 8

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Question

The floor of a building consists of \(3000\) tiles which are rhombus shaped and each of its diagonals is \(45 \,\rm{cm}\) and \(30\,\rm{ cm}\) in length. Find the total cost of polishing the floor, if the cost per \(\rm{m}^2\) is \(4.\)

Text Solution

What is Known?

The tiles used are rhombus shaped. Dimensions of single tile used and cost per \(\,{m^2}\) of polishing the floor.

What is unknown?

Total cost of polishing the floor.

Reasoning:

By using the method of splitting into triangles (triangulation method) we can find area of the rhombus visually diagonals \(AD\) and \(BC\) are perpendicular bisectors of each other hence area of rhombus \(ABCD\) will be the sum of area of triangle \(ACB\) and area of the triangle \(DBC.\)

Steps:

Area of rhombus

\[\begin{align}& = \! {\text{Area of }}\Delta ABC \! + \! {\text{Area of }}\Delta DCB\\& = \! \frac{1}{2}  \! \times \!  (BC  \! \times \!  AO) \! + \! \frac{1}{2} \! \times \! (BC \! \times \! OD)\\& = \! \frac{1}{2} \! \times  \! BC \! \times \! (AO \! + \! OD)\\&= \! \frac{1}{2} \! \times \! BC \! \times \! AD\\&= \! \frac{1}{2} \! \times \! 45\,{\rm{cm}} \! \times \! 30\,{\rm{cm}}\\& = \! 675\,{\rm{c}}{{\rm{m}}^2}\\\end{align}\]

Area of rhombus \(ABCD\)

\[\begin{align} &= \! \text{Area of }\Delta ABC \! + \! {\text{Area of }}\Delta DCB\\ &= \! \frac{1}{2} \! \times \! (BC \! \times \! AO) \! + \! \frac{1}{2} \! \times \! (BC \! \times \! OD)\\&= \! \frac{1}{2} \! \times \! BC \! \times \! (AO \! + \! OD)\\& = \! \frac{1}{2} \! \times \! BC \! \times \! AD\\&= \! \frac{1}{2} \! \times \! 45\,{\rm{cm}} \! \times \! 30\,{\rm{cm}}\\& = \! 675\,{\rm{c}}{{\rm{m}}^2}\end{align}\]

Area of each tile \( = 675\,\rm{m^2}\)

Area covered by \(3000\) tiles

\[\begin{align}& = (675 \times 3000){\rm{c}}{{\rm{m}}^2}\\& = 2025000\,{\rm{c}}{{\rm{m}}^2}\\ &= 202.5\,{{\rm{m}}^2}\end{align}\]

The cost of polishing is \(\rm{Rs.}\, 4\) per \({m^2}\).

\(\therefore \) Cost of polishing for \(202.5\,\rm{m^2}\)area \( = \rm{Rs.}(4 \times 202.5) = \rm{Rs.}810.0\)

Thus, the cost of polishing the floor is \(\rm{Rs }\,810.\)

  
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