To tile the floor of a room of dimensions
8 m x 4.8 m, rhombus-shaped tiles of
diagonals 16 cm and 12 cm costing 24
and square-shaped tiles of diagonal
16 cm costing 28 are available. Which
type of tiles should be chosen to minimize
the cost? How much can be saved by
choosing the right type over the other?
Answers
Answer:
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 breadthlength×breadth
Area of floor=8\times 4.8=38.4 m^28×4.8=38.4m
2
Diagonal of rhombus shaped tile
d_1=16 cm,d_2=12 cmd
1
=16cm,d
2
=12cm
Area of rhombus shaped tile=\frac{1}{2}\times d_1\times d_2=\frac{1}{2}\times 16\times 12=96cm^2
2
1
×d
1
×d
2
=
2
1
×16×12=96cm
2
1cm^2=10^{-4}m^21cm
2
=10
−4
m
2
Area of rhombus shaped tile=96\times 10^{-4}m^296×10
−4
m
2
Number of tiles=\frac{38.4}{96\times 10^{-4}}=4000
96×10
−4
38.4
=4000
Cost of 1 tile=Rs 24
Cost of 4000 rhombus shaped tile=4000\times 24=4000×24= Rs96000
Diagonal of square=16cm
Let side of square x
x^2+x^2=(16)^2x
2
+x
2
=(16)
2
Because diagonal of square bisect perpendicularly
Using Pythagoras theorem
(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2(Hypotenuse)
2
=(Base)
2
+(Perpendicularside)
2
2x^2=2562x
2
=256
x^2=\frac{256}{2}=128x
2
=
2
256
=128
x=\sqrt{128}=8\sqrt 2x=
128
=8
2
cm
Side of square shaped tile=8\sqrt 28
2
cm
Area of square shaped tile=(side)^2=(8\sqrt 2)^2=128cm^2(side)
2
=(8
2
)
2
=128cm
2
Area of square shaped tile=128\times 10^{-4} m^2128×10
−4
m
2
Number of square shaped tile=\frac{38.4}{128\times 10^{-4}}
128×10
−4
38.4
=3000
Cost of 3000 tiles =24\times 3000=24×3000= 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
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