**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 cm^{2}

= 9 x 10 / 2 cm^{2}

= 45 cm^{2}

(ii) Area of the trapezium = ½ x (25 + 33) x 20 m^{2}

= 580 m^{2}

**(2) The area of a trapezium is 240 m ^{2} 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 cm ^{2}. 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 cm ^{2} 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 cm ^{2}, 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 cm^{2}

= 10.7 cm^{2}

**(9) In the figure find:**

**(i) AB**

**(ii) area of the trapezium ABCD.**

**Solution:** (i) DC^{2} = DE^{2} + EC^{2}

(10)^{2} = 6^{2} + EC^{2}

EC^{2} = 100 – 36 = 64

EC = 8

∴ AB = 8 cm

(ii) Area = ½ x (8 + 2) x 8

= 4 x 10

= 40 cm^{2}

**(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 m ^{2}**

**(ii) the cost of paving the floor at the rate of Rs. 18 per m ^{2}**

**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 m ^{2, 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

Please provide chapter test of this lesson