Question

A rectangular garden is 185 m long and 220 m wide. It has two roads in its centre of uniform width of 4 m, one parallel to its length and the other parallel to its breadth. Find the cost of levelling the roads at 1.25 per sq.m. (3) = 4,010 (2) = 3,010 (1) 5 3,900 (4) 32,005​

Answer 1

Answer:

Rs. 2005

Step-by-step explanation:

Breadth= 220m

Length= 185m

The width of the crossroads are: 4m

Road 1

The length of the crossroad parallel to the breadth of the

garden should be: 220m

Road 2

The length of the crossroad parallel to the length of the

garden should be: 185m

Now area of a rectangle= breadth x length

                         Road 1=            220x4m = 880sqm

                          Road 2=.          185x4 = 740sqm

Now ATQ the roads intersect each other at the centre.

The intersection would be 4m each side as the width of the roads.

The area = SxS = 4x4m

=16sqm

Now the area of the path to be leveller = 880+ 740-16 sqm

= 1604

Cost of level long per square meter= rs.1.25

so, 1604x1.25

=2005 rs.

Answer 2

Step-by-step explanation:

Given :-

A rectangular garden is 185 m long and 220 m wide. It has two roads in its centre of uniform width of 4 m, one parallel to its length and the other parallel to its breadth.

To find :-

The cost of levelling the roads at 1.25 per sq.m.

Solution :-

Given that

Length of the rectangular garden = 185 m

Breadth of the rectangular garden = 220 m

Consider ABCD rectangle

EFGH is the road parallel to the length.

IJKL is the road parallel to the width.

Width of the road = 4 m

We know that

Area of a rectangle is "lb" sq.units

Area of the rectangle EFGH = EF×EH

=> Area of the rectangle = 185×4 m²

Area of the rectangle = 740 m²

Area of the rectangle IJKL = IL×IJ

=> Area of the rectangle IJKL = 220×4 m²

Area of the rectangle IJKL = 880 m²

In two areas, area of the square MNOP is calculated twice.

We know that

Area of a square whose side is 'a' units is a² sq.units

Area of the square MNOP = MN×MN

=> Area of the square = 4×4 m²

Area of the square MNOP = 16 m²

Area of the roads

= Area of EFGH + Area of IJKL - Area of MNOP

= 740+880-16

= 1620-16

= 1604 m²

Area of the roads = 1604 m²

Given that

The cost of levelling the roads per 1 m²

= Rs. 1.25

Total cost of levelling the roads of

1604 m² = 1604×1.25

= Rs. 2005

Total cost is Rs. 2005

Alternative Method :-

If two perpendicular paths of uniform width (w) run inside a rectangle one is parallel to the length (l) and another is parallel to the breadth (b) then area of the path = w(l+b-w) sq.units

We have,

length (l) = 185 m

Breadth (b) = 220 m

Width (w) = 4 m

Area of the roads

=> A = 4(185+220-4) m²

=> A = 4(405-4) m²

=> A = 4(401)

=> A = 1604 m²

Given that

The cost of levelling the roads per 1 m² = Rs. 1.25

Total cost of levelling the roads of 1604 m²

= 1604×1.25

= Rs. 2005

Answer:-

The total cost of levelling the roads is Rs. 2005

Used formulae:-

→ Area of a rectangle = lb sq.units

• l = length
• b = breadth

→ Area of a square = a² sq.units

• a = side

→ If two perpendicular paths of uniform width (w) run inside a rectangle one is parallel to the length (l) and another is parallel to the breadth (b) then area of the path = w(l+b-w) sq.units