NCERT Solutions for Class 7 Maths Chapter 11 Perimeter And Area

Q.1 Find the area of each of the following parallelograms:

Answer :

(a) Area of the parallelogram
= base × altitude = 7 cm × 4 cm
= 28 \( cm^2\)
(b) Area of the parallelogram
= base × altitude = 5 cm × 3 cm
= 15\( cm^2\)
(c) Area of the parallelogram
= base × altitude = 2.5 cm × 3.5 cm
= 8.75\( cm^2 \)
(d) Area of the parallelogram
= base × altitude = 5 cm × 4.8 cm
= 24.0 \( cm^2 \)
(e) Area of the parallelogram
= base × altitude = 2 cm × 4.4 cm
= 8.8 \( cm^2\)

Q.2 Find the area of each of the following triangles:

Answer :

Area of the triangle =\(\frac{1}{2} \) × b × h
= \(\frac{1}{2} \) × 4 cm × 3 cm
=\( 6m^2\)
(b) Area of the triangle = \(\frac{1}{2} \) × b × h
= \(\frac{1}{2} \) × 5 cm × 3.2 cm
= 8.0\( cm^2\)
(c) Area of the triangle = \(\frac{1}{2} \) × b × l
= \(\frac{1}{2} \) × 3 cm × 4 cm
= 6\( cm^2 \)
(d) Area of the triangle = \(\frac{1}{2} \) × b × h
= \(\frac{1}{2} \) × 3 cm × 2 cm
= 3 \(cm^2\)

Q.3 Find the missing values:

Answer :

(a) Area of the parallelogram =b × h
Or, b×h =\( 246 cm^2 \)
Or, 20 cm × h =\( 246 cm^2\)
Or, h = \( 246cm^2\) /20 cm = 12.3 cm
(b) Area of the parallelogram =b × h
Or, b×h =\( 246 cm^2 \)
Or, b × 15 cm =\( 154.5 cm^2\)
Or, b = \( 154.5 cm^2\) /15 cm = 10.3 cm
(c) Area of the parallelogram =b × h
Or, b×h =\( 48.72 cm^2 \)
Or, 8.4 cm × h =\( 48.72 cm^2\)
Or, h = \( 48.72cm^2\) /8.4 cm = 5.8 cm
(d) Area of the parallelogram =b × h
Or, b×h =\( 16.38 cm^2 \)
Or, b × 15.6 cm =\( 16.38 cm^2\)
Or, b = \( 16.38 cm^2\) /15.6 cm = 1.05 cm

Q.4 Find the missing values:

Answer :

(i) Area of the triangle = \(\frac{1}{2} \)× b × h
\(87cm^2 \) = \(\frac{1}{2} \) × 15 × h
Or, h = \(\frac{87×2}{15} \) = \(\frac{174}{15} \) = 11.6cm
(ii) Area of the triangle = \(\frac{1}{2} \)× b × h
\(1256cm^2 \) = \(\frac{1}{2} \) × b × 31.4
Or, b = \(\frac{1256×2}{31.4} \) = 40×2 = 80 mm
(iii) Area of the triangle = \(\frac{1}{2} \)× b × h
\(170.5cm^2 \) = \(\frac{1}{2} \) × 22 × h
Or, h = \(\frac{170.5×2}{22} \) = 15.5 cm

Q.5 PQRS is a parallelogram. QM is the height of Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm

Answer :

(a) Area of the parallelogram PQRS = Base × Height
= SR × QM
= 12 cm × 7.6 cm
= 91.2 \( cm^2 \)
(b) Area of the parallelogram PQRS = Base × Height
Or, 91.2 = PS×QN
Or, 91.2 = 8× QN
Or, QN = 91.2 / 8 = 11.4 cm

Q.6 DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470\( cm^2 \), AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Answer :

Area of the parallelogram ABCD = base × height
= AB × DL= 1470 \(cm^ 2 \)
Or, 35 cm × DL = 1470 \(cm^ 2 \)
Or, DL= 1470/ 35
\(\therefore \) DL = 42 cm
Area of the parallelogram ABCD = AD × BM
1470 \( cm^2 \) = 49 cm × BM
Or, 1470/ 49 = BM
\(\therefore \) BM = 30 cm
Hence, BM = 30 cm and DL = 42 cm

Q.7 \(\triangle \) ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of\(\triangle \)ABC. Also find the length of AD.

Answer :

Area of triangle = \(\frac{1}{2} \) × base ×height
\(\therefore \) \(\frac{1}{2} \) ×AB×AC = \(\frac{1}{2} \)×BC×AD
Or, \(\frac{1}{2} \)× 5 ×12 = \(\frac{1}{2} \)×13×AD
Or, AD = \(\frac{60}{13} \) = 4.62 cm

Q.8 \(\triangle \)ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ?ABC. What will be the height from C to AB i.e., CE?

Answer :

Area of \(\triangle \)ABC = \(\frac{1}{2} \) × base × height
\(\therefore \) \(\frac{1}{2} \)×AB×CE= \(\frac{1}{2} \)×BC×AD
Or, \(\frac{1}{2} \)× 7.5× CE = \(\frac{1}{2} \)× 9×6
Or, CE= \(\frac{54}{7.5} \)= 7.2 cm

Q.1 Find the circumference of the circles with the following radius. (Take \(\pi =\frac{22}{7} \))
(a) 14 cm
(b) 28 mm
(c) 21 cm

Answer :

(a) Given: Radius (r) = 14 cm
Circumference =\( 2\pi r = 2 × \frac{22}{7} × 14 = 88 cm \)
(b) Given: Radius (r) = 28 mm
Circumference =\( 2\pi r = 2 × \frac{22}{7} × 28 = 176 mm \)
(c) Given: Radius (r) = 21 cm
Circumference = \( 2\pi r = 2 × \frac{22}{7} × 21 = 132 cm \)

Q.2 Find the area of the following circles, given that (Take \(\pi =\frac{22}{7} \))
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm

Answer :

(a) Given, r = 14 mm \(\therefore \) Area of the circle =\( \pi r^2 \)
= \(\pi × 14 × 14 = \frac{22}{7} × 14 × 14 = 616mm^2\)
(b) Given, r = 49 m \(\therefore \)Area of the circle =\(\pi r^2 \)
= \(\pi × 49 × 49 = \frac{22}{7} × 49 × 49= 1886.5 m^2 \)
(c) Given, r = 5 cm \(\therefore \) Area of the circle =\( \pi r^2 \)
= \(\pi × 5 × 5 = \frac{22}{7} × 5 × 5= 78.57 cm^2 \)

Q.3 If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take \(\pi =\frac{22}{7} \))

Answer :

Given: Circumference = 154 m
\(\therefore 2 \pi r = 154 \)
Or, \( 2 × \frac{22}{7} × r = 154 \)
Or, r = \(\frac{154×7}{2×22} = \frac{49}{2} m \)
\(\therefore \) area of circle =\( \pi r^2 = \frac{22}{7} × \frac{49}{2}×\frac{49}{2} \)
= \(\frac{3773}{2} m^2 = 1886.5 m^2 \)

Q.4 A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs Rs. 4 per meter. (Take \(\pi = \frac{22}{7} \))

Answer :

Given: Diameter of the circular garden = 21 m
\(\therefore Radius = \frac{21}{2} \) m
\(\therefore \)Circumference =\( 2\pi r = 2×\frac{22}{7}×\frac{21}{2} = 66 m \)
Length of rope needed for 2 rounds = 2 × 66 m = 132 m
Cost of the rope = Rs. 4 × 132 = Rs. 528

Q.5 From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take \(\pi \)= 3.14)

Answer :

Given: Radius of the circular sheet = 4 cm
Radius of the circle to be removed = 3 cm
Area of the remaining sheet = area of ring = \( \pi (R^2-r^2)\)
\( = 3.14 (4^2-3^2) = 3.14 × (16-9) = 3.14× 7 = 21.98 cm^2 \)
Hence, the required area = 21.98 \(cm^2.\)

Q.6 Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one metre of the lace costs Rs.15. (Take \(\pi \)= 3.14)

Answer :

Diameter of the table cover = 1.5 m
\(\therefore \)Radius\( =\frac{1.5}{2}= 0.75 m \)
\(\therefore \)Length of the lace \(= 2\pi r = 2 × 3.14 × 0.75 = 4.71 m \)
Cost of the lace = Rs. 15 × 4.71 = Rs. 70.65

Q.7 Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Answer :

Given: Diameter = 10 cm
Then r = 5 cm
Perimeter of the adjoining figure = \( \frac{2 \pi r}{2} + 2r = \pi r + 2r \)
\(= \frac{22}{7} × 5 + 2×5 = \frac{110}{7} + 10 = \frac{180}{7} cm \) = 25.71 cm

Q.8 Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs.15\( m^2. (Take \pi = 3.14) \)

Answer :

Given: Diameter = 1.6 m
\(\therefore Radius = \frac{1.6}{2} = 0.8 m \)
Area of the table-top =\( \pi r^2\)
= 3.14 × 0.8 × 0.8 \(m^2\)
= 2.0096 \(m^2\)
\(\therefore \) Cost of polishing = Rs. 15 × 2.0096
= Rs. 30.144 (approx.)

Q.9 Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take \(\pi =\frac{22}{7} \))

Answer :

Length of the wire to be bent into a circle = 44 cm
2\(\pi \)r = 44 cm
Or, 2 ×\(\frac{22}{7} \) × r = 44 cm
Or, r = \(\frac{44×7}{22×2} \)
Area of circle \(\pi r^2 = \frac{22}{7} × 7×7 = 154 cm^2 \)
Now, the length of the wire is bent into a square.
Here perimeter of square= 44cm
Length of each side of the square
= Perimeter/ 4=44/4=11cm
Area of the square = \((Side)^2 = (11)^2 = 121 cm^2 \)
Since, \( 154 cm^2 >121 cm^2 \)
Thus, the circle encloses more area.

Q.10 From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure below). Find the area of the remaining sheet. (Take \(\pi = \frac{22}{7} \) )

Answer :

Given: Radius of the circular sheet = 14 cm
\(\therefore Area = \pi r^2 =\frac{22}{7} × 14 × 14 cm^2 = 616 cm^2 \)
Area of 2 small circles \( = 2 × \pi × r^2 \)
\( = 2 × \frac{22}{7} × 3.5 × 3.5 cm^2 \) = 77.0 \( cm^2 \)
Area of the rectangle = l × b
=\( 3 × 1 cm^2 = 3 cm^2 \)
Area of the remaining sheet after removing the 2 circles and 1 rectangle
= \( 616 cm^2 – (77 + 3) cm^2 \)
= \( 616 cm^2 – 80 cm^2 = 536 cm^2 \)

Q.11 A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take \(\pi = 3.14)

Answer :

Side of the square sheet = 6 m
\(\therefore Area of the sheet = (Side)^2 = (6)^2 = 36 cm^2 \)
Radius of the circle = 2 cm
\(\therefore area of the circle to be cut out = ?r^2 \)
=\( \frac{22}{7}×2×2=\frac{88}{7}cm^2 \)
Area of the left over sheet = \( 36cm^2 - \frac{88}{7} cm^2 \)
\(\frac{252-88}{7} cm^2 = \frac{164}{7} cm^2 = 23.44 cm^2

Q.12 The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take \(\pi \) = 3.14)

Answer :

Given: Circumference of the circle = 31.4 cm
2\(\pi\)r = 31.4
\(\therefore r= \frac{31.4}{2×3.14} = 5cm \)
Area of the circle =\( \pi r^2 = 3.14 × 5 × 5 = 78.5 cm^2 \) Hence, the required radius = 5 cm and area =\( 78.5 cm^2.\)

Q.13 A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (Take \(\pi \) = 3.14)

Answer :

Given: Diameter of the flower bed = 66 m .
\(\therefore Radius (r) = 66/2 = 33 m \)
Width of the path = 4 m
Radius of the flower bed included path
(R )= 33 m + 4 m = 37m
Area of the circular path = \(\pi(R^2?r^2) \)
\( = 3.14 (37^2– 33^2) = 3.14 × (37 + 33) (37 – 33) \)
\( = 3.14 × 70 × 4 = 879.20 m^2 \)
\( Hence, the required area = 879.20 m^2 \)

Q.14 A circular flower garden has an area of\( 314 m^2. \) A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler can water the entire garden? [Take\(\pi \) = 3.14]

Answer :

Given: Area of the flower garden =\( 314 m^2 \) Radius of the circular portion covered by the sprinkler = 12 m
\(\therefore area = \pi r^2 = 3.14 × 12 × 12 = 3.14 × 144 m^2 = 452.16 m^2 \)
Since\( 452.16 m^2 > 314 m^2 \)
Yes, the sprinkler will water the entire garden.

Q.15 Find the circumference of the inner and the outer circles, shown in the given figure. (Take \(\pi \)= 3.14)

Answer :

Given: Radius of the outer circle = 19 m
\(\therefore Circumference of the outer circle = 2\pi r \)
= 2 × 3.14 × 19 = 3.14 × 38 m = 119.32 m
Radius of the inner circle = 19m – 10m = 9m
\(\therefore Circumference = 2 \pi r = 2 × 3.14 × 9 = 56.52 m \)
Here the required circumferences are 56.52 m and 119.32 m.

Q.16 How many times a wheel of radius 28 cm must rotate to go 352 m? (Take\( \pi =\frac{22}{7} \))

Answer :

Given: Radius of the wheel = 28 cm
\(\therefore Circumference = 2\pi r = 2 × \frac{22}{7} × 28 = 176 cm \)
Number of rotations made by the wheel in going 352 m or 35200 cm
\( =\frac{35200}{176} \)=200
Hence, the required number of rotation = 200.

Q.17 The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take \(\pi \) = 3.14)

Answer :

Length of minute hand = 15 cm
\(\therefore \)Radius = 15 cm
Circumference = 2\(\pi \) r = 2 × 3.14 × 15 cm = 94.2 cm
Since the minute hand covers the distance in 1 hour equal to the circumference of the circle. Here the required distance covered by the minute hand = 94.2 cm.

Q.1 A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Answer :

Given: Length = 90 m
Breadth = 75 m
Area of the garden = l × b
= 90 m × 75 m = 6750\( m^2 \)
Length of the garden including path
= 90m + 5m + 5m = 100 m
Breadth of the garden including path
= 75m + 5m + 5m = 85m
Area of the garden including path = l × b
= 100 m × 85 m = 8500 \( m^2 \)
Area of the path = 8500 \( m^2 – 6750 m^2 = 1750 m^2 \)
Hence, required area of path = \(1750 m^2\) and area of the garden =\( 6750 m^2 = 0.675 hactare \)

Q.2 . A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.

Answer :

Length of the park = 125 m
Breadth of the park = 65 m
Area of the park = l × b
= 125 m × 65 m = \( 8125 m^2 \)
Length of the park including path = 125 m + 3m + 3m = 131 m
Breadth of the park including path = 65m + 3m + 3m = 71m
Area of the park including path = 131 m × 71 m = \( 9301 m^2 \)
\(\therefore \)Area of the path\(= 9301 m^2 – 8125 m^2 = 1176 m^2 \)
Hence, the required area =\( 1176 m^2. \)

Q.3 A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1. 5 cm along each of its sides. Find the total area of the margin.

Answer :

Length = 8 cm, breadth = 5 cm
Area of the cardboard = l × b \( = 8 cm × 5 cm = 40 cm^2\)
Width of the margin = 1.5 cm
Length of the inner cardboard= 8 cm – 1.5 × 2 cm
= 8 cm – 3 cm = 5 cm
Breadth of the inner cardboard= 5 cm – 1.5 × 2 cm
= 5 cm – 3 cm = 2 cm
Area of the inner rectangle = l × b
= 5 cm × 2 cm = 10 \(cm^2 \)
Area of the margin\(= 40 cm^2 – 10 cm^2 = 30 cm^2 \) Hence, the required area = 30\( cm^2\)

Q.4 A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate of Rs. 200 per \( m^2. \)

Answer :

Length of the room = 5.5 m
Breadth of the room = 4 m
\(\therefore \) Area of the room = l × b = 5.5 m × 4 m = 22 \( m^2 \)
Width of the verandah = 2.25 m
Length of the room including verandah
= 5.5 m + 2 × 2.25 m = 10 m
Breadth of the room including verandah
= 4 m + 2 × 2.25 m = 8.50 m
Area of the room including verandah = l × b
= 10 m × 8.50 m =\( 85 m^2 \)

Q.5 A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
(i) the area of the path
(ii) the cost of planting grass in the remaining portion of the garden at the rate of Rs. 40 per \( m^2.\)

Answer :

Area of the square garden =\( (Side)^2 \)
= 30 m × 30 m =\( 900 m^2 \)
Length of the garden excluding the path = 30 m – 2 × 1 m = 28 m
\(\therefore \)Area of the garden excluding the path = 28 m × 28 m = 784\( m^2 \)
(i) Area of the path =\( 900 m^2 – 784 m^2 = 116 m^2 \)
(ii) Cost of the planting the remaining portion
at the rate of Rs. 40 per\( m^2 \) = Rs. 40 × 784 = Rs. 31,360

Q.6 Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.

Answer :

Length of the road parallel to the length of the park = 700 m
Width of the road = 10 m
\(\therefore \) Area of the road = l × b = 700 m × 10 m = 7000\( m^2 \)
Length of the road parallel to the breadth of the park = 300 m
Width of the road = 10 m Area of this road = l × b = 300 m × 10 =\( 3000 m^2 \)
Area of the both roads = \( 7000 m^2 + 3000 m^2 \)– Area of the common portion
=\( 10,000 m^2 – 10 m × 10 m = 10,000 m^2 – 100 m^2 \)
\(= 9900 m^2 = 0.99 hactare \)
Area of the park = l × b = 700 m × 300 m = \(210000 m^2 \)
Area of the park without the roads
=\( 210000 m^2 – 9900 m^2 \) =\( 200100 m^2 = 20.01 hactare \)

Q.7 Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
(i) the area covered by the roads.
(ii) the cost of constructing the roads at of the rate of Rs. 110 per\( m^2. \)

Answer :

Length of the road along the length of the field = 90 m Breadth = 3 m
\(\therefore Area of this road = l × b
= 90 m × 3 m =\( 270 m^2 \)
Similarly, the area of the road parallel to the breadth of the field = l × b
= 60 m × 3 m =\( 180 m^2\)
Area of the common portion =\( 3m × 3m = 9m^2\)
(i) Area of the two roads = \(270 m^2 + 180 m^2 – 9 m^2 \) =\( 450 m^2 – 9 m^2 = 441 m^2 \)
(ii) Cost of constructing the roads
= Rs. 110 × 441 = Rs. 48,510

Q.8 Pragya wrapped a card around a circular pipe of radius 4 cm and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (\(\pi \)= 3.14)

Answer :

Length of the cord = Circumference of the circular pipe
= 2\(\pi \)r = 2 × 3.14 × 4 = 25.12 cm
Perimeter of the square box
= 4 × side = 4 × 4 cm = 16 cm
Length of the cord left
= 25.12 cm – 16 cm = 9.12 cm
Yes, 9.12 cm cord is left.

Q.9 The given figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land.
(ii) the area of the flower bed.
(iii) the area of the lawn excluding the area of the flower bed.
(iv) the circumference of the flower bed.

Answer :

(i) Length of the lawn = 10 m
Breadth of the lawn = 5 m
Area of the lawn = l × b
=\( 10 m × 5 m = 50 m^2\)
(ii) Area of the circular flower bed = \(\pi r^2 \)
=\(\frac{22}{7} ×2×2=\frac{88}{7}m^2=12.57m^2 \)
(iii) Area of the lawn excluding the area of the flower bed
= \( 50m^2 - \frac{88}{7} m^2 = \frac{350-88}{7} m^2 = \frac{262}{7} \) = \( 37. 43m^2 \)
(iv) circumference of the flower bed = 2\(\pi r \)
= \( 2 \frac{22}{7} ×2 m^2 = \frac{88}{7} m^2 = 12.57 m^2 \)

Q.10 In the following figures, find the area of the shaded portion.

Answer :

(i) Area of the rectangle = l × b
= 18 cm × (6 cm + 4 cm) = 18 cm × 10 cm =\( 180 cm^2\) Area of right triangle \(\triangle \)AEF
=\(\frac{1}{2} ×b×h= \frac{1}{2} ×6×10=30 cm^2 \)
Area of right \(\triangle BCE = \frac{1}{2} × b × h = \frac{1}{2} × 8 × 10 =40 cm^2 \)
Area of the two right triangles =\( 30 cm^2 + 40 cm^2 = 70 cm^2 \)
Area of the shaded portion
= \( 180 cm^2 – 70 cm^2 = 110 cm^2 \)
(ii) Area of the square PQRS =\( (Side)^2 = (20)^2 = 400 cm^2 \)
Area of right triangle \(\triangle \)TSU
=\(\frac{1}{2} ×b×h= \frac{1}{2} ×10×10=50 cm^2 \)
Area of right triangle \(\triangle \)QRU
=\(\frac{1}{2} ×b×h= \frac{1}{2} ×20×10=100 cm^2 \)
Area of right triangle \(\triangle \)PQT
=\(\frac{1}{2} ×b×h= \frac{1}{2} ×20×10=100 cm^2 \)
Area of the three triangles
\( = 50 cm^2 + 100 cm^2 + 100 cm^2 = 250 cm^2 \)
Area of the shaded portion
=\( 400 cm^2 – 250 cm^2 = 150 cm^2 \)

Q.11 Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM \(\perp \)AC, DN \(\perp \)AC.

Answer :

Area of \(\triangle ABC = \frac{1}{2} × b×h = \frac{1}{2}×22×3 = 33^2 \)
Area of \(\triangle ADC = \frac{1}{2} × b×h = \frac{1}{2}×22×3 = 33^2 \)
Area of the quadrilateral ABCD
= Area of \(\triangle \)ABC + Area of \(\triangle \)ADC \( = 33 cm^2 + 33 cm^2 = 66 cm^2 \)
Hence, the required area = \(66 cm^2. \)