OP Malhotra Class 9 Maths Solutions Chapter 16 Area of Plane Figures Ex 16(D)

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 cm²

(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 m²

Question 2.
The area of a trapezium is 240 m² and the sum of the parallel sides is 48 m. Find the height.
Solution:
Area of trapezium = 240 m²
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 cm². Find the distance between the parallel sides.
Solution:
Parallel sides of trapezium are 4.36 cm and 3.18 cm
Area = 18.85 cm²
∴ Distance between the parallel sides (h)

Question 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:
Area of trapezium = 475 cm²
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 cm², find the lengths of its parallel sides.
Solution:
Ratio in parallel sides = 2 : 5
Distance between them (h) = 10 cm
and area = 350 cm²
∴ 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
AD² = AL² + LD² (Pythagoras Theorem)
(5)² = (4)² + LD²
⇒ 25 = 16 + LD²
LD² = 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 cm²
∴ x = 13 and area = 40 cm²

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
CE² = CL² + EL² (Pythagoras Theorem)
⇒ (2.6)² = CL² + (1.8)²
⇒ 6.76 = CL² + 3.24
⇒ CL² = 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 cm²
=10.6932 cm²
= 10.7 cm²

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 cm²
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 cm²
= \(\frac { 1 }{ 2 }\) × 2.1 × 3.6 = 2.1 × 1.8 cm²
= 3.78 cm²
Area of the whole figure = 0.72 + 3.78
= 4.50 cm² = 4.5 cm²

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
CD² = D² + EC² (Pythagoras Theorem)
(10)² = (6)² + E²
100 = 36 + E²
⇒ EC² = 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 cm² = 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 m².
(ii) the cost of paving the floor at the rate of Rs. 18 per m².
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,
AD² = DM² + AM² = (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 m²
= 8 × 50 = 400 m²
Rate of painting = Rs. 5 per m²
Total cost = Rs. 400 × 5 = Rs. 2000

(ii) Area of floor = 4.4 × 50 = 220 m²
Rate of paving = Rs. 18 per m²
∴ 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 cm² = 1176 cm²
∴ Area of trapezium = \(\frac { 3 }{ 7 }\) of 1176 = 504 cm²
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 m², determine its depth.
Solution:
Area of cross-section = 84 m²
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