S Chand ICSE Mathematics Class 9 Solution Sixteenth Chapter Area of Plane Figures Exercise 16D (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)

S Chand ICSE Mathematics Class 9 Solution Sixteenth Chapter Area of Plane Figures Exercise 16D (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)

(1) Find the area of the trapezium if the:

(i) Parallel sides are 3 cm and 6 cm and perp. distance between them is 10 cm.

(ii) Parallel sides are 25 m and 33 m and perp. distance between them is 20 m.

Solution: (i) Area of the trapezium = ½ x (3 + 6) x 10 cm2

= 9 x 10 / 2 cm2

= 45 cm2

(ii) Area of the trapezium = ½ x (25 + 33) x 20 m2

= 580 m2

(2) The area of a trapezium is 240 m2 and the sum of the parallel sides is 48 m. Find the height.

Let, ‘h’ be the height of the trapezium

1/2 x 48 x h = 240

Or, h = 10 m

(3) The parallel sides of a trapezium are 4.36 cm and 3.18 cm and area is 18.85 cm2. Find the distance between the parallel sides.

Solution: 

Let ‘h’ be the distance between the parallel side,

1/2 x (4.36 + 3.18) x h = 18.85

Or, 7.54 x h = 37.70

Or, h = 37.7 / 7.54

= 5 cm

(4) The area of a trapezium is 475 cm2 and the height is 19 cm. Find its two parallel sides if one side is 4 cm greater than the other.

Solution: 

Let, one side is = x cm

another side is = (x + 4) cm

∴ 1/2 x 19 x (2x + 4) = 475

2x + 4 = 50

2x = 46

x = 23

So, two parallel sides are 23 cm, 27 cm cm.

(5) The parallel sides of a trapezium are in the ratio 2:5 and the distance between the parallel sides is 10 cm. if the area of the trapezium is 350 cm2, find the lengths of its parallel sides.

Solution: Let, the common ratio be x

So, sides of the trapezium are 2x and 5x

∴ ½ x 10 x 7x = 350

Or, x = 10

So, lengths of parallel sides are = 20 cm and 50 cm.

(6) In the given figure, AD = BC = 5 cm, AB = 7 cm. The parallel sides AB, DC are 4 cm apart DC =  cm. Find x and the area of the trapezium ABCD.

Solution: 

(7) The parallel sides of a trapezium are 7.5 cm, 3.9 cm, and the other sides are each 2.6 cm. Find its area.

Solution: 

Area = 1/2 x (7.5 + 3.9) x 1.87 cm2

= 10.7 cm2

(9) In the figure find:

(i) AB

(ii) area of the trapezium ABCD.

Solution: (i) DC2 = DE2 + EC2

(10)2 = 62 + EC2

EC2 = 100 – 36 = 64

EC = 8

∴ AB = 8 cm

(ii) Area = ½ x (8 + 2) x 8

= 4 x 10

= 40 cm2

(10) The cross-section of a tunnel perpendicular to its length is a trapezium ABCD as shown in the figure. AM = BN; AB = 4.4 m; CD = 3 m. The height of the tunnel is 2.4 m. The tunnel is 50 m long.

Calculate:

(i) the cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. 5 per m2

(ii) the cost of paving the floor at the rate of Rs. 18 per m2

Solution

(12) The cross section of a canal is in the shape of a trapezium. If the canal is 12 m wide at the top and 8 m wide at the bottom and the area of its cross section is 84 m2, determine its depth.

Solution: 

Let, the depth be x

Thus, 1/2 x (12 + 80 x x = 84

Or, 10x = 84

Or, x = 8.4 m


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  1. Please provide chapter test of this lesson

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