The floor of a building consists of 3000 tiles which are rhombus shape

1.	The floor of a building consists of 3000 tiles which are rhombus shaped and eachof its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing thefloor, if the cost per m2 is 4
Answer» $\large\underline{\sf{Given- }}$ The floor of a building consist of 3000 rhombus shaped tiles. Length of diagonals of rhombus are 45 cm and 30 cm. Cost of polishing 1 square meter = Rs 4 $\large\underline{\sf{To\:Find - }}$ Total cost of polishing the floor. $\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}} \end{gathered}$ $\boxed{ \sf \: Area_{(rhombus)} \: = \dfrac{1}{2} \times product \: of \: diagonals \: }$ $\large\underline{\sf{Solution-}}$ Dimensions of rhombus shaped tile Length of first diagonal = 45 cm = 0.45 m Length of other diagonal = 30 cm = 0.3 m So, Area of ONE tile, $\rm :\longmapsto\:Area_{(1 \: tile)} \: = \dfrac{1}{2} \times product \: of \: diagonals$ $\rm :\longmapsto\:Area_{(1 \: tile)} = \dfrac{1}{2} \times 0.45 \times 0.3$ $\rm :\implies\:Area_{(1 \: tile)} = 0.0675 \: {m}^{2}$ As number of tiles are 3000, So, $\rm :\longmapsto\:Area_{(3000 \: tile)} = 3000 \times 0.0675$ $\bf :\implies\:Area_{(3000 \: tile)} = 202.5 \: {m}^{2}$ Now, $\sf \: Cost \: of \: polishing \: 1 \: {m}^{2} = Rs \: 4$ $\sf \: Cost \: of \: polishing \: 202.5 \: {m}^{2} = \: 4 \times 202.5 = Rs \: 810$ ─━─━─━─━─━─━─━─━─━─━─━─━─ Additional INFORMATION :- $\boxed{ \sf \: Area_{(rhombus)} = base \times height}$ $\boxed{ \sf \: Area_{(rectangle)} =length \times breadth}$ $\boxed{ \sf \: Area_{(square)} = {(side)}^{2}}$ $\boxed{ \sf \: Area_{(right \: angle \triangle)} =\dfrac{1}{2} \times base \times height}$ $\boxed{ \sf \: Area_{(circle)} =\pi \: {r}^{2}}$