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**There is a rectangular field of size 94 m Ã— 32 m. Three roads each of 2 m width pass through the field such that two roads are parallel to the breadth of the field and the third is parallel to the length. Calculate: (i) area of the field covered by the three roads (ii) area of the field not covered by the roads.**

Let* ABCD *be the rectangular field.

Here,

Two roads which are parallel to the breadth of the field *KLMN* and *EFGH* with width 2 m each.

One road which is parallel to the length of the field *PQRS* with width 2 m.

Length of the rectangular field* AB* = 94 m and breadth of the rectangular field* BC* = 32 m

âˆ´ Area of the rectangular field = Length x Breadth = 94 m x 32 m = 3008 m^{2}

Area of the road* KLMN* = 32 m x 2 m = 64 m^{2}

Area of the road* EFGH *= 32 m x 2 m = 64 m^{2}

Area of the road *PQRS *= 94 m x 2 m = 188 m^{2}

Clearly area of *TUVI *and *WXYZ* is common to these three roads.

Thus,

Area of* **TUVI *= 2 m x 2 m = 4 m^{2}

Area of *WXYZ* = 2 m x 2 m = 4 m^{2}

Hence,**(i)** Area of the field covered by the three roads:

= Area (*KLMN*) + Area (*EFGH)* + Area (*PQRS*) âˆ’ {Area (*TUVI *) + Area (*WXYZ*)}

= [ 64+ 64 + 188 âˆ’ (4 + 4 ^{ })] m^{2}

= 316 m^{2} âˆ’ 8 m^{2}

= 308 m^{2}**(ii) **Area of the field not covered by the roads:

= Area of the rectangular field* ABCD* âˆ’ Area of the field covered by the three roads

= 3008 m^{2} âˆ’ 308 m^{2}

= 2700 m^{2}

**A school has a hall which is 22 m long and 15.5 m broad. A carpet is laid inside the hall leaving all around a margin of 75 cm from the walls. Find the area of the carpet and the area of the strip left uncoverd. If the width of the carpet is 82 cm, find the cost at the rate of Rs 18 per metre.**

We have,

Length of the hall *PQ *= 22 m and breadth of the hall *QR* = 15.5 m

âˆ´ Area of the school hall *PQRS* = 22 m x 15.5 m = 341 m^{2}

Length of the carpet *AB* = 22 m âˆ’ ( 0.75 m + 0.75 m) = 20.5 m [ Since 100 cm = 1 m]

Breadth of the carpet *BC* = 15.5 m âˆ’ ( 0.75 m + 0.75 m) = 14 m

âˆ´ Area of the carpet *ABCD *= 20.5 m x 14 m = 287 m^{2}

Area of the strip = Area of the school hall *PQRS* âˆ’ Area of the carpet *ABCD*

= 341 m^{2} âˆ’ 287 m^{2}

= 54 m^{2}

Again,

Area of the 1 m length of carpet = 1 m x 0.82 m = 0.82 m^{2}

Thus,

Length of the carpet^{ }whose area is 287 m^{2}^{ }= 287 m^{2}^{ }Ã· 0.82 m^{2}^{ }= 350 m

Cost of the 350 m long carpet = Rs. 18 x 350 = Rs. 6300

**Two cross roads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of Rs 105 per m ^{2}.**

Let* ABCD *be the rectangular park then *EFGH* and* IJKL* the two rectangular roads with width 5 m.

Length of the rectangular park *AD* = 70 cm

Breadth of the rectangular park *CD* = 45 m

âˆ´ Area of the rectangular park = Length x Breadth = 70 m x 45 m = 3150 m^{2}

Area of the road *EFGH* = 70 m x 5 m = 350 m^{2}

Area of the road* JKIL *= 45 m x 5 m = 225 m^{2}

Clearly area of *MNOP* is common to the two roads.

Thus, Area of* MNOP* = 5 m x 5 m = 25 m^{2}

Hence,

Area of the roads = Area (*EFGH*) + Area (*JKIL*) âˆ’ Area (*MNOP*)

= (350 + 225 ) m^{2}âˆ’ 25 m^{2} = 550 m^{2}

Again, it is given that the cost of constructing the roads = Rs. 105 per m^{2}

Therefore,

Cost of constructing 550 m^{2} area of the roads = Rs. (105 Ã— 550)

= Rs. 57750.

**The length and breadth of a rectangular park are in the ratio 5 : 2. A 2.5 m wide path running all around the outside the park has an area 305 m ^{2}. Find the dimensiions of the park.**

We have,

Area of the path = 305 m^{2}

Let the length of the park be 5*x** *m and the breadth of the park be 2*x** *m

Thus,

Area of the rectangular park = 5*x* x 2*x* = 10*x ^{2}* m

Width of the path = 2.5 m

Outer length

Outer breadth

Area of

âˆ´ Area of the path = [(10

â‡’ 305 = 35

â‡’ 305 âˆ’ 25 = 35

â‡’ 280 = 35

â‡’

Therefore,

Length of the park = 5

Breadth of the park = 2

**A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m ^{2}, find the area of the lawn.**

Let the side of the lawn be *x *m*.*

Given that width of the path = 2.5 m

Side of the lawn including the path = (*x* + 2.5 + 2.5) m = (*x* + 5 ) m

So, area of lawn = (Area of the lawn including the path) âˆ’ (Area of the path)

We know that the area of a square = (Side)^{2}

âˆ´ Area of lawn (*x*^{2}^{ }) = (*x* + 5)^{2} âˆ’ 165

â‡’ *x*^{2}^{ }= (*x*^{2} + 10*x** *+ 25) âˆ’ 165

â‡’ 165^{ }= 10*x** *+ 25

â‡’ 165 âˆ’ 25 ^{ }= 10*x** *

â‡’ 140 = 10*x** *

Therefore *x* = 140 Ã· 10 = 14

Thus the side of the lawn = 14 m

Hence,

The area of the lawn = (14 m)^{2} = 196 m^{2}