Question

13. A floor of a building consists of 2,500 tiles. All the tiles are rhombus shaped with diagonal 40 cm and 25 cm. Find the total cost of polishing the floor at the rate of ₹8 per square meter.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• A floor of a building consists of 2,500 tiles.

• All the tiles are rhombus shaped with diagonal 40 cm and 25 cm.

• The total cost of polishing the floor is ₹8 per square meter.

Now, Area of 1 tile = Area of rhombus having diagonals 40 cm and 25 cm.

So,

$\rm \: Area_{(1\:tile)} = \dfrac{1}{2} \times product \: of \: two \: diagonals$

$\rm \: Area_{(1\:tile)} = \dfrac{1}{2} \times 25 \times 40$

$\rm \: Area_{(1\:tile)} = 25 \times 20$

$\rm\implies \:Area_{(1\:tile)} = 500 \: {cm}^{2}$

Since, floor consisting of 2500 tiles.

$\rm\implies \:Area_{(2500\:tile)} = 500 \: \times 2500 = 1250000 \: {cm}^{2}$

$\rm\implies \:Area_{(2500\:tile)} = = 125 \: {m}^{2} \: \: \: \{ \because \: 1 \: m = 100 \: cm \}$

Now, further given that

Cost of polishing 1 m² of the floor = ₹ 8

So,

Cost of polishing 125 m² of the floor = ₹ 8 × 125 = ₹ 1000

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Answer:

Given :-

• A floor of a building consists of 2500 tiles.
• All the tiles are rhombus shaped with diagonal 40 cm and 25 cm.

To Find :-

• What is the total cost of polishing the floor at the rate of ₹8 per square metre.

Formula Used :-

$\clubsuit$ Area Of Rhombus Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{Area_{(Rhombus)} =\: \dfrac{d_1 \times d_2}{2}}}}\: \: \: \bigstar$

where,

• d₁ = First Diagonal of Rhombus
• d₂ = Other Diagonal of Rhombus

Solution :-

First, we have to find the area of one tiles :

So,

$\longrightarrow \sf\bold{\green{Area\: of\: one\: tile =\: Area_{(Rhombus)}}}$

Given :

• First Diagonal (d₁) = 40 cm
• Other Diagonal (d₂) = 25 cm

According to the question by using the formula we get,

$\implies \bf Area_{(Rhombus)} =\: \dfrac{d_1 \times d_2}{2}$

$\implies \sf Area_{(Rhombus)} =\: \dfrac{40\: cm \times 25\: cm}{2}$

$\implies \sf Area_{(Rhombus)} =\: \dfrac{\cancel{1000}\: cm^2}{\cancel{2}}$

$\implies \sf Area_{(Rhombus)} =\: \dfrac{500\: cm^2}{1}$

$\implies \sf\bold{\blue{Area_{(Rhombus)} =\: 500\: cm^2}}$

Hence, area of one tiles is 500 cm² .

Now, we have to find the area of 2500 tiles :

Given :

• Area of one tiles = 500 cm²

So,

$\leadsto \sf Area\: of\: 2500\: tiles =\: 2500 \times 500$

$\leadsto \sf Area\: of\: 2500\: tiles =\: 1250000$

$\leadsto \sf\bold{\orange{Area\: of\: 2500\: tiles =\: 1250000\: cm^2}}$

Now, we have to convert the area of 2500 tiles :

$\mapsto \sf Area\: of\: 2500\: tiles =\: 1250000\: cm^2$

$\mapsto \sf Area\: of\: 2500\: tiles =\: \dfrac{125\cancel{0000}}{1\cancel{0000}}\: m^2$

$\mapsto \sf\bold{\purple{Area\: of\: 2500\: tiles =\: 125\: m^2}}$

Now, we have to find the total cost of polishing the floor :

Given :

• Area of floor = 125 m²
• Cost of polishing the floor = ₹8 per m²

So,

$\small \dashrightarrow \sf Total\: Cost\: of\: polishing\: the\: floor =\: 8 \times 125$

$\small \dashrightarrow \sf\bold{\red{Total\: Cost\: of\: polishing\: the\: floor =\: ₹1000}}$

$\therefore$ The total cost of polishing the floor at the rate of ₹8 per m² is ₹1000.

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