Question

The floor of a building consists of 2000 tiles that are rhombus shaped and each of its diagonals are 1.6m and 1.2 m in length. The total cost of polishing the floor if the cost per sq.m is ₹10 is ₹_ _ _ _ _

Answer 1

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

Given that,

The floor of a building consists of 2000 tiles that are rhombus shaped and each of its diagonals are 1.6m and 1.2 m in length.

Diagonal of rhombus are 1.6 m and 1.2 m long.

Let assume that

$\rm \: d_1 = 1.6 \: m$

$\rm \: d_2 = 1.2 \: m$

We know,

$\boxed{\tt{ \: \: Area_{(rhombus)} \: = \: \dfrac{1}{2} \: \times \: d_1 \: \times \: d_2 \: }}$

So, on substituting the values, we get

$\rm \: Area_{(1 \: rhombus \: shaped \: tile)} \: = \: \dfrac{1}{2} \times 1.2 \times 1.6$

$\rm\implies \: Area_{(1 \: rhombus \: shaped \: tile)} \: = \:0.96 \: {m}^{2}$

Since, floor consists of 2000 tiles.

So,

$\rm\implies \: Area_{(2000 \: rhombus \: shaped \: tile)} \: =2000 \times \:0.96 \: = 1920 {m}^{2}$

Now, further it is given that,

Cost of polishing 1 sq. m of the floor = ₹ 10

So,

Cost of polishing 1920 sq. m of the floor = ₹ 19200

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