Question

The floor of a building consists of 1000 tiles which are triangle shaped. The base and height of each triangle are 15 cm and 10 cm respectively. The cost of polishing the floor at the rate of Rs. 4 per sq.cm is?​

Answer 1

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The floor of a building consists of 1000 tiles which are triangle shaped. The base and height of each triangle are 15 cm and 10 cm respectively. The cost of polishing the floor at the rate of Rs. 4 per cm²

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• Cost of polishing a floor

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Here as we can see first we have to find area of triangle because the cost of polishing the floor is given per cm² which is measurement unit of area . Then after finding area of a tile we have to multiply it with no. of tiles i.e. 1000 because we have to find cost of polishing floor and floor consists of 1000 similar tiles. Then finally we have multiply it with Rs. 4.

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• Total no. of tiles = 1000
• height of a triangular tile = 10 cm
• base of a triangular tile = 15 cm
• Cost of polishing floor per cm²= Rs 4

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• $\sf{ Are a \: of \: triangle = \dfrac{1}{2} \times height\times base}$
• $\sf{Total \: Surface \: area = area \: of \: 1 \: tile \times total \: no. \: of \: tiles}$
• $\sf{total \: cost = total \: surface \: area \times \: Rs \: 4}$

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To find area of a triangular tile:

$:\implies \sf{ Are a \: of \: triangle = \dfrac{1}{2} \times height\times base}$

$\sf{:\implies Area \: of \: triangle = \dfrac{1}{2} \times 10 \times 15}$

$:\implies \sf{ Area \: of \: triangle = 5\times 15}$

$:\implies \sf{ Are \: of \: triangle =75 \: {cm}^{2} }$

To find surface area:

$:\implies \sf{Total \: Surface \: area = area \: of \: 1 \: tile \times total \: no. \: of \: tiles}$

$:\implies \sf{Total \: Surface \: area = 75\times 1000}$

$:\implies \sf{Total \: Surface \: area = 75000}$

To find total cost of polishing the floor:

$:\implies \sf{total \: cost = total \: surface \: area \times \: Rs \: 4}$

$:\implies \sf{total \: cost = 75000 \times \: Rs \: 4}$

$:\implies\large\star \boxed {{ \sf{total \: cost = \: Rs \: 300000}}}\star$

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• $\sf{Area \: of \: e quilateral \: triangle = {side}^{2} \times \dfrac{ \sqrt{3} }{4} }$

• $\sf{Area \: of \: square = {side}^{2} }$

• $\sf{Area \: of \: rectangle=length \times breadth}$

• $\sf{Area \: of \: cone =\pi \: radius(radius + \sqrt{ {height}^{2} + {radius}^{2} } })$

• $\sf{Area \: of \: parallogram=base \times height}$

• $\sf{Area \: of \: rhombus = \dfrac{digonal1 \times digonal2}{2} }$

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$\sf\orange{And \:all \:we\: are\: done!}$