NCERT Solutions for Class 7 Maths Chapter 9 Perimeter and Area

NCERT
SOLUTIONS
CLASS - 7TH

aglase .co

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Class : 7th
Subject : Maths
Chapter : 11
Chapter Name : Perimeter and Area

Exercise 11.1

Q1 The length and the breadth of a rectangular piece of land are 500 m and 300 m
respectively. Find (i) its area (ii) the cost of the land, if 1 m2 of the land costs Rs 10,000.

Answer. (i) Area = Length x Breadth
= 500 × 300
= 150000m2
(ii) Cost of 1m2 land = Rs10000
Cost of 150000m2 land = 10000 × 150000 = Rs1500000000

Page : 208 , Block Name : Exercise 11.1

Q2 Find the area of a square park whose perimeter is 320 m.

Answer. Perimeter = 320 m
4 x Length of the side of park = 320
320
Length of the side park = 4 = 80m
Area = ( Length of the side of park )2 = (80)2 = 6400m2

Page : 208 , Block Name : Exercise 11.1

Q3 Find the breadth of a rectangular plot of land, if its area is 440m2 and the length is 22
m. Also �nd its perimeter.

Answer. Area = Length × Breadth = 440m2
22 × Breadth = 440
440
Breadth = 22 = 20m
Perimeter = 2 (Length + Breadth )
= 2(22 + 20) = 2(42) = 84m2

Page : 208 , Block Name : Exercise 11.1

Page 1 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q4 The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, �nd its
breadth. Also �nd the area.

Answer. Perimeter = 2 ( Length + Breadth ) = 100 cm

Page : 208 , Block Name : Exercise 11.1

Q5 The area of a square park is the same as of a rectangular park. If the side of the square
park is 60 m and the length of the rectangular park is 90 m, �nd the breadth of the
rectangular park.

2
Answer. Area of square park = (One of its sides) = (60)2 = 3600m2
Area of rectangular park = Length × Breadth = 3600
90 × Breadth = 3600
Breadth = 40m

Page : 208 , Block Name : Exercise 11.1

Q6 A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the
same wire is rebent in the shape of a square, what will be the measure of each side. Also
�nd which shape encloses more area?

Answer. Perimeter of rectangle = Perimeter of square
2 (Length + Breadth) = 4 x side
2(40 + 22) = 4× Side
2 × 62 = 4× Side
Side = 124
4
= 31cm
Area of rectangle = 40 × 22 = 880cm2
Area of square = ( Side )2 = 31 × 31 = 961cm2
Therefore, the square-shaped wire encloses more area.

Page : 208 , Block Name : Exercise 11.1

Q7 The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, �nd
its length. Also �nd the area of the rectangle.

Answer. Perimeter = 2( Length + Breadth ) = 130
2( Length + 30) = 130
Length +30 = 65
Length = 65 − 30 = 35cm
Area = Length x Breadth = 35 x 30 = 1050 cm2

Page 2 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Page : 208 , Block Name : Exercise 11.1

Q8 A door of length 2 m and breadth 1m is �tted in a wall. The length of the wall is 4.5 m
and the breadth is 3.6 m (Figure). Find the cost of white washing the wall, if the rate of
white washing the wall is Rs 20 per m2 .

Answer. Area of wall = 4.5 × 3.6 = 16.2m2
Area of door = 2 × 1 = 2m2
Area to be white-washed = 16.2 − 2 = 14.2m2
Cost of white-washing 1m2 area = Rs20
∴ cost of white-washing 14.2m2 area = 14.2 × 20 = Rs284

Page : 208 , Block Name : Exercise 11.1

Exercise 11.2

Q1 Find the area of each of the following parallelograms:

Answer. Area of parallelogram = Base x Height

Page 3 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

(a) Height = 4 cm
Base = 7 cm
Area of parallelogram = 7 x 4 = 28cm2
(b) Height = 3 cm
Base = 5 cm
Area of parallelogram = 5 × 3 = 15cm2
(c) Height = 3.5 cm
Base = 2.5 cm
Area of parallelogram = 2.5 × 3.5 = 8.75cm2
(d) Height = 4.8 cm
Base = 5 cm
Area of parallelogram = 5 × 4.8 = 24cm2
(e) Height = 4.4 cm
Base = 2 cm
Area of parallelogram = 2 × 4.4 = 8.8cm2

Page : 216 , Block Name : Exercise 11.2

Q2 Find the area of each of the following triangles:

Answer. Area of triangle = 12 × Base ×
(a) Base = 4 cm, height = 3 cm
Area = 12 × 4 × 3 = 6cm2
(b) Base = 5 cm, height = 3.2 cm
Area = 12 × 5 × 3.2 = 8cm2
(c) Base = 4 cm, height = 3 cm
Area = 12 × 4 × 3 = 6cm2
(d) Base = 3 cm, height = 2 cm
Area = 12 × 2 × 3 = 3cm2

Page : 216 , Block Name : Exercise 11.2

Q3 Find the missing values:

Page 4 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer. Area of parallelogram = Base x Height
(a) b = 20 cm
h=?
Area = 246cm2
20 × h = 246
h = 246
20
= 12.3cm
Therefore, the height of such parallelogram is 12.3 cm.
(b) b = ?
h = 15cm
Area = 154.5cm2
b × 15 = 154.5
b = 10.3cm
Therefore, the base of such parallelogram is 10.3 cm.
(c) b = ?
h = 8.4 cm
Area = 48.72cm2
b × 8.4 = 48.72
b = 48.72
8.4
= 5.8cm
Therefore, the base of such parallelogram is 5.8 cm.
(d) b = 15.6 cm
h=?
Area = 16.38cm2
15.6 × h = 16.38
h = 16.38
15.6
= 1.05cm
Therefore, the height of such parallelogram is 1.05 cm.

Page : 216 , Block Name : Exercise 11.2

Q4 Find the missing values:

Page 5 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer. Area of triangle = 12 × Base ×
(a) b = 15 cm
h=?
Area = 12 × b × h = 87cm2
1
2
× 15 × h = 87cm2
h = 87×2
15
= 11.6cm
Therefore, the height of such triangle is 11.6 cm.
(b) b = ?
h = 31.4 mm
Area = 12 × b × h = 1256mm2
1
2
× b × 31.4 = 1256
b = 1256×2
31.4
= 80mm
Therefore, the base of such triangle is 80 mm.
(c) b = 22 cm
h=?
Area = 12 × b × h = 170.5cm2
1
2
× 22 × h = 170.5cm2
h = 170.5×2
22
= 15.5cm
Therefore, the height of such triangle is 15.5 cm.

Page : 217 , Block Name : Exercise 11.2

Q5 PQRS is a parallelogram (Figure). QM is the height from 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 parallegram PQRS
(b) QN, if PS = 8 cm

Page 6 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer. (a) Area of parallelogram = Base x Height = SR X QM
= 7.6 × 12 = 91.2cm2

Area of parallelogram = Base x Height = AB X DL
1470 = 35 × DL
DL = 1470
35
= 42cm
Also, 1470 = AD × BM
1470 = 49 × BM
BM = 147049
= 30cm

Page : 217 , Block Name : Exercise 11.2

Q7 ∆ABC is right angled at A (Figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13
cm and AC = 12 cm, Find the area of ∆ABC. Also �nd the length of AD.

Answer. Area = 12 × Base × Height = 12 × 5 × 12
Also,area of triangle = 12 × AD × BC
30 = 12 × AD × 13
30×2
13
= AD
AD = 4.6cm

Page : 217 , Block Name : Exercise 11.2

Q8 ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Figure). 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 ΔABC = 12 × Base × Height = 12 × BC × AD

Page 7 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

= 12 × 9 × 6 = 27cm2
Area of ΔABC = 12 × Base × Height = 12 × AB × CE
27 = 12 × 7.5 × CE
CE = 7.2cm

Page : 217 , Block Name : Exercise 11.2

Exercise 11.3

Q1 Find the circumference of the circles with the following radius: ( Take π = 22
7
)
(a) 14 cm (b) 28 mm (c) 21 cm

Answer. (a) r = 14 cm
Circumference = 2πr = 2 × 22
7
× 14 = 88 cm
(b) r = 21 cm
Circumference = 2πr = 2 × 22
7
× 28 = 176 mm
(c) r = 21 cm
Circumference = 2πr = 2 × 22
7
× 21 = 132 cm

Page : 223 , Block Name : Exercise 11.3

Q2 Find the area of the following circles, given that:
(a) radius = 14 mm ( Take π = 227
)
(b) diameter = 49 m
(c) radius = 5 cm

Answer. (a) r = 14 mm
Area = πr2 = 22 7
× 14 × 14 = 616mm2
(b) d = 49 m
r = 49
2
m
Area = πr2 = 22
7
× 49
2
× 49
2
= 1886.5m2
(c) r = 5 cm
Area = πr2 = 22
7
× 5 × 5 = 550
7
= 78.57cm2

Page : 223 , Block Name : Exercise 11.3

Page 8 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q3 If the circumference of a circular sheet is 154 m, �nd its radius. Also �nd the area of
the sheet. ( Take π = 22
7
)

Answer. Circumference = 2πr = 154m
2 × 22
7
× r = 154
r = 154 × 447
= 49
2
= 24.5m
Area = πr2 = 22
7
× r2
= 22
7
× 49
2
× 49
2
= 1886.5m2

Page : 223 , Block Name : Exercise 11.3

Q4 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 �nd the cost of the rope, if
it costs Rs 4 per meter. ( Take π = 22
7
)

Answer. d = 21 m
r = 21/2 m
Circumference = 2πr = 2 × 22
7
× 21
2
= 66m
Length of rope required for fencing = 2 × 66m = 132m
Cost of 1 m rope = Rs 4
Cost of 132 m rope = 4 x 132 = Rs 528

Page : 223 , Block Name : Exercise 11.3

Q5 From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area
of the remaining sheet. (Take π = 3.14)

Answer. Outer radius of circular sheet = 4 cm
Inner radius of circular sheet = 3 cm
Remaining area = 3.14 × 4 × 4 − 3.14 × 3 × 3
= 50.24 − 28.26
= 21.98cm2

Page : 223 , Block Name : Exercise 11.3

Page 9 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q6 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 �nd its cost if one meter of the lace costs Rs 15.
(Take π = 3.14)

Answer. Circumference = 2πr
= 2 × 3.14 × d2
= 2 × 3.14 × 1.5
2
= 4.71m
Cost of 1 m lace = Rs 4.71 m
Cost of 4.71 m lace = 4.71 × 15 = Rs 70.65

Page : 223 , Block Name : Exercise 11.3

Q7 Find the perimeter of the adjoining �gure, which is a semicircle including its
diameter.

Answer. Radius = 5 cm
Length of curved part = πr
= 22
7
×5
= 15.71cm
Total perimeter = Length of curved part + Length of diameter
= 15.71 + 10 = 25.71cm

Page : 223 , Block Name : Exercise 11.3

Q8 Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of
polishing is Rs 15/m2 . (Take π = 3.14)

Answer. Diameter = 1.6 m
Radius = 1.6/2 = 0.08 m
Area = 3.14 × 0.8 × 0.8
= 2.0096m2
Cost of polishing 1m2 area = Rs15
Cost of polishing 2.0096m2 area = 15 × 2.0096 = 30.14
Therefore, it will cost Rs 30.14 for polishing such circular table.

Page 10 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Page : 223 , Block Name : Exercise 11.3

Q9 Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the
radius of that circle. Also �nd 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 �gure encloses more area, the circle or
the square? ( Take π = 22 7
)

Answer. Circumference = 2πr = 44cm
2 × 22
7
× r = 44
r = 7cm
Area = πr2 = 22
7
× 7 × 7 = 154cm2
If the wire is bent into a square, then the length of each side would be = 44
4
= 11cm
Area of square = (11)2 = 121cm2
Therefore, circle encloses more area.

Page : 223 , Block Name : Exercise 11.3

Q10 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 1cm are removed. (as shown in the adjoining
�gure). Find the area of the remaining sheet.( Take π = 227
)

Answer. Area of bigger circle
Area of 2 small circles
Area of rectangle = Length x Breadth
Remaining area of sheet = 616 − 77 − 3 = 536cm2

Page : 223 , Block Name : Exercise 11.3

Q11 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 π = 3.14)

Answer. Area of square-shaped sheet = ( Side )2 = (6)2 = 36cm2
Area of circle = 3.14 × 2 × 2 = 12.56cm2
Remaining area of sheet = 36 − 12.56 = 23.44cm2

Page 11 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Page : 224 , Block Name : Exercise 11.3

Q12 The circumference of a circle is 31.4 cm. Find the radius and the area of the circle?
(Take π = 3.14)

Answer. Circumference = 2πr = 3.14 cm
2 × 3.14 × r = 31.4
r = 5cm
Area = 3.14 × 5 × 5 = 78.50cm2

Page : 224 , Block Name : Exercise 11.3

Q13 A circular �ower bed is surrounded by a path 4 m wide. The diameter of the �ower
bed is 66 m. What is the area of this path? (π = 3.14)

66
Answer. Radius of �ower bed = 2
= 33m
Radius of �ower bed and path together = 33 + 4 = 37m
Area of �ower bed and path together = 3.14 × 37 × 37 = 4298.66m2
Area of �ower bed = 3.14 × 33 × 33 = 3419.46m2
Area of path = Area of �ower bed and path together - Area of �ower bed
= 4298.66 − 3419.46 = 879.20m2

Page : 224 , Block Name : Exercise 11.3

Q14 A circular �ower garden has an area of 314m2 . A sprinkler at the centre of the
garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire
garden? (Take π = 3.14)

Answer. Area = πr2 = 314m2
3.14 × r2 = 314
r2 = 100
r = 10m
Yes, the sprinkler will water the whole garden.

Page : 224 , Block Name : Exercise 11.3

Page 12 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q15 Find the circumference of the inner and the outer circles, shown in the adjoining
�gure? (Take π = 3.14)

Answer. Radius of outer circle = 19 m
Circumference = 2πr = 2 × 3.14 × 19 = 119.32m
Radius of inner circle = 19 − 10 = 9m
Circumference = 2πr = 2 × 3.14 × 9 = 56.52m

Page : 224 , Block Name : Exercise 11.3

Q16 How many times a wheel of radius 28 cm must rotate to go 352 m? ( Take π = 22
7
).

Answer. r = 28cm
Circumference = 2πr = 2 × 22
7
× 28 = 176 cm
Total distance to becovered 352m
Number of rotations = Circumference of wheel
= 176cm = 35200
176
= 200
Therefore, it will rotate 200 times.

Page : 224 , Block Name : Exercise 11.3

Q17 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 π = 3.14)

Answer. Distance travelled by the tip of minute hand Circumference of the clock
= 2πr = 2 × 3.14 × 15
= 94.2cm

Page : 224 , Block Name : Exercise 11.3

Exercise 11.4

Page 13 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q1 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 �nd the area of the garden in hectare.

Answer.

Length (l) of garden = 90 m
Breadth ( b) of garden = 75 m
Area of garden = l x b = 90 x 75 = 6750 m2
From the �gure, it can be observed that the new length and breadth of the garden, when
path is also included, are 100 m and 85 m respectively.
Area of path = Area of the garden including the path - Area of garden
= 8500 − 6750 = 1750m2
1 hectare = 10000m2
6750
Therefore, area of garden in hectare = 10000 = 0.675

Page : 226 , Block Name : Exercise 11.4

Q2 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 (l) of park = 125m
Breadth (b) of park = 65m
Area of park = l × b = 125 × 65 = 8125m2
From the �gure, it can be observed that the new length and breadth of the park, when
path is also included, are 131 m and 71 m respectively.
Area of the park including the path = 131 × 71 = 9301m2
Area of path = Area of the park including the path - Area of park
= 9301 − 8125 = 1176m2

Page 14 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Page : 226 , Block Name : Exercise 11.4

Q3 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 (l) of cardboard = 8cm
Breadth (b) of cardboard = 5 \mathrm { cm }
Area of cardboard including margin = l × b = 8 × 5 = 40cm2
From the �gure, it can be observed that the new length and breadth of the cardboard,
when margin is not included, are 5 cm and 2 cm respectively.
Area of the cardboard not including the margin = 5 × 2 = 10cm2
Area of the margin = Area of cardboard including the margin - Area of cardboard not
including the margin
= 40 − 10 = 30cm2

Page : 226 , Block Name : Exercise 11.4

Q4 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 �oor of the verandah at the rate of Rs 200 per m2

Answer.

(i) Length (l) of room = 5.5m
Breadth (b) of room = 4m
Area of room = l × b = 5.5 × 4 = 22m2

Page 15 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

From the �gure, it can be observed that the new length and breadth of the room, when
verandah is also included, are 10 m and 8.5 m respectively.
Area of the room including the verandah = 10 × 8.5 = 85m2
Area of verandah = Area of the room including the verandah - Area of room
= 85 − 22 = 63m2
(ii) Cost of cementing 1m2 area of the �oor of the verandah = Rs 200
Cost of cementing 63m2 area of the �oor of the verandah = 200 x 63
= Rs 12600

Page : 226 , Block Name : Exercise 11.4

Q5 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 m2 .

Answer.

(i) Side (a) of square garden 30 m
Area of square garden including path = a2 = (30)2 = 900m2
From the �gure, it can be observed that the side of the square garden, when path is not
included, is 28 m.
Area of the square garden not including the path = (28)2 = 784m2
Area of the path = Area of the square garden including the path - Area of square garden
not including the path
= 900 − 784 = 116m2
(ii) Cost of planting grass in 1m2 area of the garden = Rs 40
Cost of planting grass in 784m2 area of the garden = 784 × 40 = Rs31360

Page : 226 , Block Name : Exercise 11.4

Q6 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

Page 16 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

area of the roads. Also �nd the area of the park excluding cross roads. Give the answer in
hectares.

Answer.

Length (l) of park = 700 m
Breadth (b) of park = 300 m
Area of park = 700 x 300 = 210000 m2
Length of road PQRS = 700 m
Length of road ABCD = 300 m
Width of each road = 10 m
Area of the roads = 10 m
Area of the roads = ar ( PQRS) + ar ( ABCD) - ar (KLMN)
= (700 × 10) + (300 × 10) − (10 × 10)

Length (l) of �eld = 90 m
Breadth (b) of �eld = 60 m
Area of �eld = 90 x 60 = 5400m2
Length of road PQRS = 90 m
Length of road ABCD = 60 m
Width of each road = 3m
Area of the roads = ar (PQRS) + ar ( ABCD) - ar ( KLMN)
= (90 × 3) + (60 × 3) − (3 × 3)
= 270 + 180 − 9 = 441m2
Cost for constructing 1m2 road = Rs 110
Cost for constructing 441m2 road = 110 × 441 = Rs48510

Page : 227 , Block Name : Exercise 11.4

Q8 Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining �gure) 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? (π = 3.14).

Page 17 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer. Perimeter of the square = 4 x Side of the square = 4 x 4 = 16 cm
Perimeter of circular pipe = 2πr = 2 × 3.14 × 4 = 25.12cm
Length of chord left with Pragya = 25.12 − 16 = 9.12cm

Page : 227 , Block Name : Exercise 11.4

Q9 The adjoining �gure represents a rectangular lawn with a circular �ower bed in the
middle. Find: (i) the area of the whole land (ii) the area of the �ower bed (iii) the area of
the lawn excluding the area of the �ower bed (iv) the circumference of the �ower bed.

Answer. (i) Area of whole land = Length × Breadth = 10 × 5 = 50m2
(ii) Area of �ower bed = πr2 = 3.14 × 2 × 2 = 12.56m2
(iii) Area of lawn excluding the �ower bed = Area of whole land - Area of �ower bed
= 50 − 12.56 = 37.44m2
(iv) Circumference of �ower bed = 2πr = 2 × 3.14 × 2 = 12.56m

Page : 227 , Block Name : Exercise 11.4

Q10 In the following �gures, �nd the area of the shaded portions:

Answer. (i) Area of EFDC = ar( ABCD) - ar (BCE) - ar ( AFE)
= (18 × 10) − 12 (10 × 8) − 12 (6 × 10)

Page 18 of 19 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-11 Maths

= 180 − 40 − 30 = 110cm2
(ii) ar (QTU) = ar (PQRS) - ar ( TSU) - ar (RUQ) - ar (PQT)
= (20 × 20) − 12 (10 × 10) − 12 (20 × 10) − 12 (20 × 10)
= 400 − 50 − 100 − 100 = 150cm2

Page : 227 , Block Name : Exercise 11.4

Q11 Find the area of the quadrilateral ABCD.
Here, AC = 22 cm, BM = 3 cm,
DN = 3 cm, and
BM ⊥ AC, DN ⊥ AC

Answer. ar ( ABCD ) = ar ( ABC) + ar ( ADC)
= 12 (3 × 22) + 12 (3 × 22)
= 33 + 33 = 66cm2

Page : 227 , Block Name : Exercise 11.4

Page 19 of 19 Aglasem Schools