Parents can use S Chand Class 9 Maths Solutions ICSE Chapter 16 Area of Plane Figures Ex 16(D) to provide additional support to their children.

## S Chand Class 9 ICSE Maths Solutions Chapter 16 Area of Plane Figures Ex 16(D)

Question 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) Length of parallel sides are

3 cm and 6 cm

and perpendicular distance (h) between them = 10 cm

∴ Area = \(\frac { 1 }{ 2 }\) (Sum of parallel sides) × height

= \(\frac { 1 }{ 2 }\) (6 + 3) × 10 = 9 × 5 = 45 cm^{2}

(ii) Parallel sides are 25 m and 33 m

and perpendicular distance (h) = 20 m

∴ Area = \(\frac { 1 }{ 2 }\) (sum of parallel sides × height)

= \(\frac { 1 }{ 2 }\) (25 + 33) × 20 = 58 × 10 = 582 m^{2}

Question 2.

The area of a trapezium is 240 m^{2} and the sum of the parallel sides is 48 m. Find the height.

Solution:

Area of trapezium = 240 m^{2}

Sum of parallel sides = 48 m

Question 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:

Parallel sides of trapezium are 4.36 cm and 3.18 cm

Area = 18.85 cm^{2}

∴ Distance between the parallel sides (h)

Question 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:

Area of trapezium = 475 cm^{2}

Height (h) = 19 cm

∴ Sum of two parallel sides = \(\frac{\text { Area } \times 2}{\text { Height }}\)

= \(\frac{475 \times 2}{19}\) = 50 cm

Let one of the two parallel sides = x

Then second side = x + 4

∴ x + x + 4 = 50 ⇒ 2x = 50 – 4 = 46

x = \(\frac { 46 }{ 2 }\) = 23

∴ First side = 23 cm

and second side = 23 + 4 = 27 cm

Question 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:

Ratio in parallel sides = 2 : 5

Distance between them (h) = 10 cm

and area = 350 cm^{2}

∴ Sum of parallel sides = \(\frac{\text { Area } \times 2}{\text { Height }}\)

= \(\frac{350 \times 2}{10}\) = 70 cm

Sum of ratio = 2 + 5 = 7

∴ First side = \(\frac{70 \times 2}{7}\) = 20 cm

and second side = \(\frac{70 \times 5}{7}\) = 50 cm

Question 6.

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

Solution:

In the figure, ABCD is a trapezium in which

AD = BC = 5 cm

AB = 7 cm, DC = x cm

and distance between the parallel sides = 4 cm

Draw AL and BM ⊥s on DC

AL = BM = 4 cm

AB = LM = 4 cm

∵ AD = BC = 7 cm

∴ ABCD is an isosceles trapezium

Now in right △ALD

AD^{2} = AL^{2} + LD^{2} (Pythagoras Theorem)

(5)^{2} = (4)^{2} + LD^{2}

⇒ 25 = 16 + LD^{2}

LD^{2} = 25 – 16 = 9 = (3)^{2}

∴ x = LD = 3 cm

∴ DC = AB + 2 LD

= 7 + 2 × 3 = 7 + 6 = 13 cm

Now area of trapezium = \(\frac { 1 }{ 2 }\) (sum of parallel sides) × height

= \(\frac { 1 }{ 2 }\) (7 + 13) × 4

= \(\frac { 1 }{ 2 }\) × 20 × 4 = 40 cm^{2}

∴ x = 13 and area = 40 cm^{2}

Question 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:

In trapezium ABCD

AB = 7.5 cm, CD = 3.9 cm

AD = BC = 2.6 cm

Draw CE || DA, CL ⊥ AB

∴ CE = 2.6 cm and AE = DC = 3.9 cm

∴ EB = AB – AE = 7.5 – 3.9 = 3.6 cm

In an isosceles △CEB, CL ⊥ EB

Which bisects EB at L

∴ EL = \(\frac { 1 }{ 2 }\) EB = \(\frac { 1 }{ 2 }\) × 3.6 = 1.8 cm

Now in right △CEL

CE^{2} = CL^{2} + EL^{2} (Pythagoras Theorem)

⇒ (2.6)^{2} = CL^{2} + (1.8)^{2}

⇒ 6.76 = CL^{2} + 3.24

⇒ CL^{2} = 6.76 – 3.24 = 3.52

∴ CL = \(\sqrt{3.52}\) = 1.876

Now area of trapezium = \(\frac { 1 }{ 2 }\) (sum of parallel sides) × height

= \(\frac { 1 }{ 2 }\) (7.5+3.9) × 1.876

= \(\frac { 1 }{ 2 }\) × 11.4 × 1.186 = 5.7 × 1.876 cm^{2}

=10.6932 cm^{2}

= 10.7 cm^{2}

Question 8.

The given figure shows that cross section of a concrete structure with the measurments, as given. Calculate the area of the cross section.

Solution:

In the figure, produce DE to meet AB at G Now we have one trapezium GBCD and one rectangle AGEF

Now GB = AB – AG = AB – FE = 1.8 – 0.3 = 1.5 cm

and DG = DE + EG = DE + FA = 1.2 + 2.4 = 3.6 cm

Now area of rectangle AGEF = l × b = 2.4 × 0.3 = 0.72 cm^{2}

and area of trapezium GBCD = \(\frac { 1 }{ 2 }\) (sum of parallel sides) × height

= \(\frac { 1 }{ 2 }\) (GB + DC) × DG

= \(\frac { 1 }{ 2 }\) (1.5 + 0.6) × 3.6 cm^{2}

= \(\frac { 1 }{ 2 }\) × 2.1 × 3.6 = 2.1 × 1.8 cm^{2}

= 3.78 cm^{2}

Area of the whole figure = 0.72 + 3.78

= 4.50 cm^{2} = 4.5 cm^{2}

Question 9.

In the figure, find

(i) AB

(ii) Area of the trapezium ABCD.

Solution:

(i) In the figure, draw CE || AB meeting AD in E, then

AE = BC = 2 and DE = AD – EA

=8 – 2 = 6 cm

In right △ECD

CD^{2} = D^{2} + EC^{2} (Pythagoras Theorem)

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

100 = 36 + E^{2}

⇒ EC^{2} = 100 – 36 = 64 = (8)^{2}

∴ EC = 8 cm or AB = EC = 8 cm

(ii) Now area of trapezium ABCD

= \(\frac { 1 }{ 2 }\) (sum of parallel sides) × height

= \(\frac { 1 }{ 2 }\) (AD + BC) × AB

= \(\frac { 1 }{ 2 }\) (8 + 2) × 8 = \(\frac { 1 }{ 2 }\) × 10 × 8 cm^{2} = 40 cm)^{2}

Question 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:

In the figure, ABCD is a trapezium in which AM = BN, AB = 4.4 m, CD = 3 m and height of the tunnel = 2.4 m

Length of tunnel = 50 m

∵ DM and CN are perpendiculars on AB and

AM = NB = \(\frac { 1 }{ 2 }\) (AB – DC)

= \(\frac { 1 }{ 2 }\) (4.4 – 3.0) m

= \(\frac { 1 }{ 2 }\) (1.4) = 0.7 m

In right △ADM,

AD^{2} = DM^{2} + AM^{2} = (2.4)^{2} + (0.7)^{2}

= 5.76 + 0.49 = 6.25 = (2.5)^{2}

∴ AD = CB = 2.5 m

(i) Now area of the interval surface of the tunnel (excluding floor) = Perimeter of the face × Length

= (3 + 2.5 + 2.5) × 50 m^{2}

= 8 × 50 = 400 m^{2}

Rate of painting = Rs. 5 per m^{2}

Total cost = Rs. 400 × 5 = Rs. 2000

(ii) Area of floor = 4.4 × 50 = 220 m^{2}

Rate of paving = Rs. 18 per m^{2}

∴ Total cost of paving the floor = Rs. 18 × 220 = Rs. 3960

Question 11.

A trapezium with its parallel sides in the ratio 11 : 3 is cut off from a rectangle whose sides measure 98 cm and 12 cm. The area of the trapezium is \(\frac { 3 }{ 7 }\) of the area of the rectangle. Find the lengths of the parallel sides of the trapezium, when its height is equal to the smaller side of the rectangle.

Solution:

Sides of rectangle are 98 cm and 12 cm

∴ Area = l × b = 98 × 12 cm^{2} = 1176 cm^{2}

∴ Area of trapezium = \(\frac { 3 }{ 7 }\) of 1176 = 504 cm^{2}

Height of trapezium (h) = smaller side of rectangle = 12 cm

∴ Sum of parallel sides = \(\frac{\text { Area } \times 2}{\text { Height }}\) = \(\frac{504 \times 2}{12}\) cm

=42 × 2 = 84 cm

Ratio of the parallel sides = 11 : 3

Sum of ratios = 11 + 3 = 14

∴ First side = \(\frac{84 \times 11}{14}\) = 66 cm

and second side = \(\frac{84 \times 3}{14}\) = 18 cm

Question 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:

Area of cross-section = 84 m^{2}

Sides are 12 m and 8 m

Let h be its depth, then

h = \(\frac{\text { Area } \times 2}{\text { Sum of parallel sides }}\)

= \(\frac{84 \times 2}{(12+8)}\) = \(\frac{84 \times 2}{20}\) = \(\frac { 168 }{ 20 }\) m = 84 m

∴ Depth of the canal = 8.4 m