Question

- Find the cost of tiling the floor of a trapezoidal hall whose parallel sides are 25 m and 20 m and the distance
between them is 16 m at the rate of 40 per m2​

Answer 1

Step-by-step explanation:

parallel sides= 25m and 20m

distance between them =16m

area= 1/2 ×(20+25)m×16m

= 45m×8m

=360m^2

cost per m^2= 40

cost of 360m^2 floor = 360×40

=14400

therefore the cost for tilling the floor is 14400

Answer 2

The cost of tiling the floor would be Rs. 14400.

Step-by-step explanation:

Since we have given that

Dimensions of parallel sides = a= 25 m and b= 20 m

Distance between them h= 16 m

So, Area of trapezium would be

$\dfrac{1}{2}\times (a+b)\times h\\\\=\dfrac{1}{2}\times (25+20)\times 16\\\\=\dfrac{1}{2}\times 45\times 16\\\\=45\times 8\\\\=360\ m^2$

Cost of per sq. m = Rs. 40

So, Cost of 360 sq. m would be

$360\times 40\\\\=Rs.\ 14400$

Hence, The cost of tiling the floor would be Rs. 14400.

# learn more:

Find the cost of watering a trapezoidal field whose parallel sides are 10 m and 25 m respectively the perpendicular distance between them is 15cm and rate of watering is rs4 per m square

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