Question

There are two rectangular paths in the middle of a park as shown in the adjoining figure. Find the area of the path, it is give that
AB = CD = 100m
AD = BC = 70m
HE = SR = 5 m​

Answer 1

Answer:

width =5 m

length of first parallel road 70 m

area of first road 70×5=350 sq.m

length of second road 45 m

area of second road =45×5=225 sq.m

area of the common part of cross road with 5 m width that lies at the

center of the park 5×5 = 225 sq.m

area of roads = area of first road +area of second cross road−common area

=(350+225)−25

=575−25

=550sqm

cost of the construction per sq.m =Rs.105

total cost =550×105=Rs 57750

Answer 2

The area of the path is $825\;m^{2}$.

(The diagram is attached at the end)

The lengths and breadths of the entire park is already given along with the widths of the paths in between the park which is shaped like a cross as shown in the figure.

The paths are shaped like a rectangle and hence to find the area of these two paths the area of the rectangle formula must be applied.

$Area\; of\; a\; rectangle=Length\times Breadth$

In order to find the area of the paths the area of the rectangles named SRPQ and HEGF respectively (shaded portions in the figure) need to be computed.

To find the area of the first rectangle path SRPQ we need its corresponding lengths and breadths.

The width SR is already given to be 5 m.

From the figure, we can see the length SP or RQ is parallel to the breadth of the park (AD or BC) which means that it is of the same length as that of the park breadth which is given as 70 m.

Thus, we have:

$l=70\;m\\b=5\;m$

$\implies Area\; of\; 1^{st} path\;(SRPQ) =70\times 5$

$\implies Area=350\;m^{2}$  ---------(1)

Similarly, the length EF and HG are parallel to length of the park (AB or CD) given to be 100 m. The breadth HE is already given to be 5m.

Thus, we have:

$l=100\;m\\b=5\;m$

$\implies Area\; of\; 2^{nd} path\;(HEGF) =100\times 5$

$\implies Area =500\;m^{2}$   ---------(2)

To find the area covered by both these paths we need to add their respective areas but however from the figure it must be observed that since the paths are crossing each other there is an overlap in their paths.

This extra area must be discarded or deducted from the total area of these two paths (This is the dark grey shaded portion in the figure)

The extra area will be the area of the square LMNO. The side LM is parallel to the side HE and hence it will be equal to it and similarly the side LO is parallel to SR so it is equal.

Both these sides are of length 5 m.

$Area\; of\; LMNO= side\times side$

$\implies Area =5\times5$

$Area\; of\; LMNO= 25\;m^{2}$

We have:

Area of the 2 paths=Area of SRPQ + Area of HEGF -Area of LMNO

By substituting the corresponding areas from equations (1), (2) and (3) we have:

$Total\;area\; of\; the\; 2\; paths=(350+500)-25$

$\implies Total\;area=850-25$

$\implies Total\;area=825\;m^{2}$

Thus, this is the area of the two rectangular paths of the park.

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