Question

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A rectangular lawn, 75m by 60m, has two roads, each road 4m wide, running through the middle of the lawn, one parallel to length and the other parallel to breadth, as shown in the figure. Find the cost of gravelling the roads at ₹50 per m².​​

Answer 1

GIVEN :-

Length of the rectangular lawn = $\sf 75 \: m$

Breadth of the lawn =$\sf 60 \: cm$

Width of the road = $\sf 4 \: m$

SOLUTION :-

Area of the road (ABCD) parallel to the length of the lawn

$\sf = 75 \: m \times 4 \: m$

$\sf = 300 {m}^{2}$

Area of the road (EFGH) parallel to the breadth of the lawn

$\sf = 60 \: m \times 4 \: m = {240 \: m}^{2}$

As we can see in attachment ABCD and EFGH have common area.

Area of the common area in both ABCD and EFGH

$\sf = 4 \: m \times 4 \: m = {16 \: m}^{2}$

In question it is given to gravel the roads only so we have to find area of both roads only.

Total area of both the rectangular roads

$\sf = 300 + 240 - 16 = {524m}^{2}$

Cost of gravelling the roads per m² = ₹50

Cost of gravelling the roads for $\sf {540m}^{2}$

$\sf = 50 \times 524$

$\fbox{\sf = 26200}$

Cost of gravelling the road is ₹26200.

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Answer 2

Answer:

Cost of gravelling roads at ₹50/$m^2$ is Rs. 2358.

Explanation:

$\large\mathtt{\star Given:}$

•Length of the rectangular lawn=75m

•Breadth of the lawn=60m

•Width of the road=4m

$\large\mathtt{\star Solution:}$

➾Area of the road(ABCD) parallel to the length of the lawn = 75m×4m

=300$m^2$

➾Area of the road(EFGH) parallel to the breadth of the lawn = 60m×4m

=240$m^2$

➾Area of the common area in both ABCD and EFGH = 4m×4m

=16$m^2$

➾Total area of both the rectangular roads

= 300+240−16

=524$m^2$

➾Cost of gravelling the roads per $m^2$= Rs.4.50

➾Cost of gravelling the roads for 524$m^2$

=4.50×524

=2358

➾Cost of gravelling the road is Rs.2358.

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