Course / Category - 8th

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Maths / Mensuration / Area of Path and Verandah
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## Area of Path and Verandah

### Area of Path and Verandah:

It is observed that in square or rectangular gardens or parks. Some space in the form of path is left inside or outside or in between as cross paths.

Illustration: A footpath of uniform width 5 m runs round the inside of a rectangular park 38 m long and 32 m wide. Find the area of the path.

Solution: Let ABCD be the rectangular park and PQRS be the internal boundaries of the path.

We have, Length AB= 38 m and Breadth BC= 32 m:

Thus, Area of rectangle ABCD = AB $\dpi{120} \fn_jvn \large \times$ BC = 38 m $\dpi{120} \fn_jvn \large \times$ 32 m = 1216 $\dpi{120} \fn_jvn \large ^{m^{2}}$

Now, Length PQ  = 38 m - 5 m - 5 m = 28 m and Breadth QR = 32 m - 5 m - 5 m = 22m

Area of rectangle PQRS = PQ $\dpi{120} \fn_jvn \large \times$ RS = 28 m $\dpi{120} \fn_jvn \large \times$ 22 m = 616 $\dpi{120} \fn_jvn \large ^{m^{2}}$

Area of footpath = Area of rectangle ABCD - Area of rectangle PQRS = 1216 - 616=600 $\dpi{120} \fn_jvn \large ^{m^{2}}$

Thus, Area of footpath =600 $\dpi{120} \fn_jvn \large ^{m^{2}}$

The area of the crossroads = Area of the two colored rectangles - Area of the red colored rectangle at the centre.

We are subtracting the area of the rectangle because we are taking it into consideration two times.

Illustration: A rectangular lawn is 30 m by 20 m .It has two roads each 2 m wide running in the middle of it, one parallel to the length and the other parallel to the breadth . Find the area of the roads.

Solution: Let ABCD and PQRS be the cross roads,

we have AB = 2 m and BC = 30 m, then

Area of path ABCD =  AB $\dpi{120} \fn_jvn \large \times$ BC = 2 m  $\dpi{120} \fn_jvn \large \times$  30 m  = 60 $\dpi{120} \fn_jvn \large ^{m^{2}}$

Now, for the other crossroad, PQ = 2 m and QR = 20 m,

then Area of path  PQRS = PQ $\dpi{120} \fn_jvn \large \times$  QR = 2 m  $\dpi{120} \fn_jvn \large \times$  20 m = 40 $\dpi{120} \fn_jvn \large ^{m^{2}}$

Area EFGH is common to the both the paths: EF = 2 m and FG = 2 m, then Area of EFGH  =  EF $\dpi{120} \fn_jvn \large \times$ FG = 2 m  $\dpi{120} \fn_jvn \large \times$ 2 m = 4 $\dpi{120} \fn_jvn \large ^{m^{2}}$

Total area covered  as roads = Area of road ABCD + Area of road PQRS - Area of  EFGH = 60 + 40 - 4 =  100 - 4  = 96 $\dpi{120} \fn_jvn \large ^{m^{2}}$

Thus, total area of the roads = 96 $\dpi{120} \fn_jvn \large ^{m^{2}}$

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