Question

. To tile the floor of a room of dimensions 8m x 4.8m, rhombus-shaped tiles of diagonals 16 cmand 12cm costing rupess 24 and square shaped tiles of diagonal 16cm costing rupees 28 are available.Both the tiles are of same quality. Which type of tiles should be chosen to minimise the cost??How much can be saved by choosing it?​

Answer 1

Square shaped tile should be chosen to minimize cost and

saved money by choosing it=Rs 24000

Step-by-step explanation:

Dimension of floor=8m by 4.8 m

Area of  floor=$length\times breadth$

Area of floor=$8\times 4.8=38.4 m^2$

Diagonal of rhombus shaped tile

$d_1=16 cm,d_2=12 cm$

Area of rhombus shaped tile=$\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times 16\times 12=96cm^2$

$1cm^2=10^{-4}m^2$

Area of rhombus shaped tile=$96\times 10^{-4}m^2$

Number of tiles=$\frac{38.4}{96\times 10^{-4}}=4000$

Cost of 1 tile=Rs 24

Cost of 4000 rhombus shaped tile=$4000\times 24=$Rs96000

Diagonal of square=16cm

Let side of square x

$x^2+x^2=(16)^2$

Because diagonal of square bisect perpendicularly

Using Pythagoras theorem

$(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2$

$2x^2=256$

$x^2=\frac{256}{2}=128$

$x=\sqrt{128}=8\sqrt 2$cm

Side of square shaped tile=$8\sqrt 2$cm

Area of square shaped tile=$(side)^2=(8\sqrt 2)^2=128cm^2$

Area of square shaped tile=$128\times 10^{-4} m^2$

Number of square shaped tile=$\frac{38.4}{128\times 10^{-4}}$=3000

Cost of 3000 tiles =$24\times 3000=$Rs 72000

Square shaped tile should be chosen because cost of square shaped tiles is minimum.

Difference=96000-72000=Rs 24000

Hence, saved money by choosing it=Rs 24000

#Learns more:

https://brainly.in/question/5398843

Answer 2

Answer: ₹2400

Step by Stpe Explaination :

The abv guy is correct