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PERIMETER AND AREA
Exercise 11.1
Q.1. 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.
Ans. (i) The area of rectangular land = length × breadth
= l × b = 500 m × 300 m = 1,50,000 m2
(ii) The cost of 1 m2 of the land = Rs 10,000
So, the cost of 1,50,000 m2
= Rs 10,000 × 1,50,000 m2 = Rs 1,50,00,00,000
Q.2. Find the area of a square park whose perimeter is
320 m.
Ans. Perimeter of the square = 4 × side
⇒
320 m = 4 × side
320
m = 80 m
4
Area of the square = side × side
⇒
side =
= 80 m × 80 m = 6400 m2
Q.3. Find the breadth of a rectangular plot of land, if its area
is 440 m2 and the length is 22 m. Also, find its perimeter.
Ans. Area of rectangular plot of land
= length × breadth = l × b
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440 m2 = 22 m × breadth (b)
22 m × b = 440 m2
440 m 2
So,
b =
= 20 m
22 m
Perimeter of rectangular plot of land =
2 (l + b)
= 2 × (22 + 20) m = (2 × 42) m = 84 m
Hence,
Perimeter of rectangular plot of land is 84 m.
Q.4. The perimeter of a rectangular sheet is 100 cm. If the
length is 35 cm, find its breadth. Also, find the area.
Ans. Perimeter of rectangular sheet = 2 [length + breadth]
= 2 (l + b)
⇒
100 = 2 (35 + b)
⇒
100 = 70 + 2b
⇒
100 – 70 = 2b
⇒
30 = 2b
30
2
= 15 cm
So,
b =
Area = l × b = 35 × 15
= 525 cm2
Hence, the area of rectangular sheet is 525 cm2.
Q.5. 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, find the breadth
of the rectangular park.
Ans. The area of the square park = side × side
= 60 m × 60 m = 3600 m2
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As per condition,
Area of the square park = area of the rectangular park
∴
3600 m2 = area of the rectangular park
∵ Area of the rectangular park = length × breadth = l × b
∴
3600 m2 = 90 m × b
3600 m 2
So,
b =
= 40 m
90 m
Hence, the breadth of the rectangular park is 40 m.
Q.6. 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, find which shape enclose more area.
Ans. We have,
Perimeter of the rectangle = Perimeter of the square
⇒
2(l + b) = 4 × side
side =
2 (l + b )
=
2 × ( 40 cm + 22 cm )
4
2 × 62 cm
=
4
4
∴ Side of the square = 31 cm
∴ Area of the rectangle = l × b = 40 cm × 22 cm
= 880 cm2
∴
Area of the square = side × side
= 31 cm × 31 cm = 961 cm2
Hence, the square encloses more area.
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Q.7. The perimeter of a rectangle is 130 cm. If the breadth of
the rectangle is 30 cm, find its length. Also, find the area
of the rectangle.
Ans.
Perimeter of rectangle = 2 (l + b)
⇒
130 = 2 (l + 30)
⇒
130 = 2 l + 60
⇒
130 – 60 = 2 l ⇒ 70 = 2 l
So,
l =
70
= 35 cm
2
∴ Length of the rectangle = 35 cm
Hence, area of the rectangle = l × b = 35 × 30 = 1050 cm2
Q.8. A door of length 2 m and breadth 1 m is fitted in a wall.
The length of the wall is 4.5 m and the breadth is 3.6 m.
Find the cost of white washing the wall, if the rate of
white washing of the wall is Rs 20 per m2.
Ans.
Area of a door = l × b
= 2m×1m
= 2 m2
Area of the wall = l × b
= 4.5 m × 3.6 m
= 16.20 m2
So, Area of the rest of the wall = (16.20 – 2) m2 = 14.20 m2
∴ The rate of white washing of the 1 m2 wall = Rs 20
Hence, the rate of white washing of the 14.20 m2 wall
= 14.20 × Rs 20 = Rs 284
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Exercise 11.2
Q.1. Find the area of each of the following parallelograms :
Ans. (a) Length of the base (b) = 7 cm
Height (h) = 4 cm
So,
area of the parallelogram = b × h
= (7 × 4) cm2
= 28 cm2
(b) The length of the base (b) = 5 cm
Height (h) = 3 cm
So,
area of the parallelogram = b × h
= (5 × 3) cm2 = 15 cm2
(c) The length of the base (b) = 2.5 cm
Height (h) = 3.5 cm
So,
area of the parallelogram = b × h
= (2.5 × 3.5) cm2
= 8.75 cm2
(d) The length of the base (b) = 5 cm
Height (h) = 4.8 cm
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So,
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area of the parallelogram = b × h
= (5 × 4.8) cm2
= 24.0 cm2
(e) The length of the base (b) = 2 cm
Height (h) = 4.4 cm
So,
area of the parallelogram = b × h
= (2 × 4.4) cm2
= 8.8 cm2
Q.2. Find the area of each of the following triangles :
Ans. (a) The length of the base (b) = 4 cm
(b)
Height (h) = 3 cm
1
1
×b×h =
×4×3
Area of triangle =
2
2
12
= 6 cm2
=
2
Base (b) = 5 cm
Height (h) = 3.2 cm
1
Area of triangle =
×b×h
2
1
=
× 5 × 3.2
2
16
= 8 cm2
=
2
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(c)
(d)
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Base (b) = 3 cm
Height (h) = 4 cm
1
1
Area of triangle =
×b×h =
×3×4
2
2
12
= 6 cm2
=
2
Base (b) = 3 cm
Height (h) = 2 cm
1
1
×b×h =
×3×2
Area of triangle =
2
2
6
= 3 cm2
=
2
Q.3. Find the missing values :
Ans. (a) Area of the parallelogram = base × height
246 = 20 × height
246
= 12.3 cm
20
(b) Area of the parallelogram = base × height
So,
height =
154.5 = base × 15 cm
So,
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base =
7
154.5
= 10.3 cm
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(c) Area of the parallelogram = base × height
48.72 = 8.4 × base
48.72
= 5.8 cm
8.4
(d) Area of the parallelogram = base × height
So,
base =
16.38 = 15.6 × height
16.38
height =
= 1.05 cm
15.6
So,
Q.4. Find the missing values :
Area of Δ =
Ans. (a)
87 =
⇒
height =
Area of Δ =
(b)
1256 =
⇒
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base =
1
base × altitude (height)
2
1
× 15 × height
2
87 × 2
= 11.6 cm
15
1
× base × altitude (height)
2
1
× base × 31.4
2
1256 × 2
= 8 mm
31.4
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1
base × height
2
1
× 22 × height
170.5 =
2
170.5 × 2
height =
22
= 15.5 cm
Area of Δ =
(c)
⇒
Q.5. PQRS is a parallelogram.
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 parallelogram PQRS
(b) QN, if PS = 8 cm
Ans. (a)
QM (height) = 7.6 cm
SR (base) = 12 cm
So, area of the parallelogram PQRS = b × h
= (12 × 7.6) cm2
= 91.2 cm2
(b)
QN (height) = ?
PS (base) = 8 cm
So, area of the parallelogram PQRS = 91.2 cm2
Area of the parallelogram PQRS = base × height
91.2 cm2 = 8 cm × h
⇒
So,
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91.2 cm 2
h =
8 cm
(QN) height = 11.4 cm
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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 cm2, AB = 35 cm and
AD = 49 cm, find the length of BM
and DL.
Ans.
In parallelogram ABCD,
AB (base) = 35 cm
Area of the parallelogram = 1470 cm2
DL (height) = ?
Area of the parallelogram = b × h
⇒
1470 cm2 = 35 cm × h
1470
35
= 42 cm
⇒
So,
h =
height (DL) = 42 cm
height (DL) = 42 cm
AD (base) = 49 cm
BM (height) = ?
Area of the parallelogram = 1470 cm2
Area of the parallelogram = b × h
⇒
1470 = 49 × h
1470
49
= 30 cm
⇒
So,
h =
height (BM) = 30 cm
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Q.7. ΔABC 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
ΔABC. Also, find the length of AD.
Ans. In a right angled ΔABC,
base = 12 cm
height = 5 cm
Area of a right angled triangle
1
×b×h
=
2
1
=
× 12 × 5
2
= 30 cm2
1
×b×h
2
1
=
× 13 × AD
2
13 × AD
30 cm2 =
2
Area of ΔABC =
⇒
⇒
13 × AD = 30 × 2
= 60 cm2
60 cm 2
⇒
AD =
13 cm
60
=
cm
13
60
cm.
Hence, the length of AD is
13
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Q.8. ΔABC 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 ΔABC. What will be the height from C to AB
i.e., CE?
Ans. In a ΔABC,
BC = base
= 9 cm = b
AD = height = 6 cm = h
1
×b×h
Area of ΔABC =
2
1
×9×6
=
2
= 27 cm2
Now, in a ΔABC, if base = 7.5 cm and height
= ?
1
Area of a ΔABC =
× base × height
2
1
× 7.5 cm × CE
⇒
27 cm2 =
2
⇒
⇒
Hence,
7.5 cm × CE = 27 cm2 × 2
54 cm 2
CE =
7.5 cm
height, CE = 7.2 cm
Exercise 11.3
Q.1. Find the circumference of the circles with the following
22
radius : (Take π = )
7
(a) 14 cm
(b) 28 mm
(c) 21 cm
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Ans.
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Circumference of the circle = 2πr (where, r is radius)
or,
C = 2πr
(a)
r = 14 cm
So,
C = 2πr = 2 ×
22
× 14
7
= 88 cm
(b)
r = 28 mm
So,
C = 2πr = 2 ×
22
× 28
7
= 176 mm
(c)
r = 21 cm
22
× 21
7
= 132 cm
So,
C = 2πr = 2 ×
Q.2. Find the area of the following circles, given that :
22
(a) radius = 14 mm
[Take π =
]
7
(b) diameter = 49 m
(c) radius = 5 cm
Ans. Area of the circle = πr2, where π =
(a)
22
7
r = 14 mm
22
× (14)2
Hence, area of the circle =
7
22
=
× 14 × 14
7
= 616 mm2
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(b)
49
m
2
22 49 49
×
×
area of the circle =
7
2
2
= 1886.5 m2
d = 49 m,
Hence,
(c)
Hence,
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∴
r =
r = 5 cm
22
× (5)2
7
22
550
=
×5×5 =
cm2
7
7
area of the circle =
Q.3. If the circumference of a circular sheet is 154 m, find its
22
radius. Also, find the area of the sheet. [Take π = ]
7
Ans. Circumference of circular sheet = 2πr
So,
154 m = 2 ×
22
×r
7
∴
44
r = 154 m
7
⇒
44 r = 154 × 7
154 × 7
49
=
m
44
2
49
m
r =
2
⇒
r =
So,
Area of circular sheet = πr2
22 49 49
×
×
7
2
2
= 1886.5 m2
=
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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 costs of the
22
]
rope, if it cost Rs 4 per meter.
[Take π =
7
21
Ans.
We have, diameter = 21 m, so radius =
2
Circumference of circular garden = 2πr = length of the rope
22 21
×
= 66 m
7
2
So, length of the rope to make 2 rounds of the fence
= 2×
= 2 × 66 m
= 132 m
Hence,
the cost of 132 m rope = 132 × Rs 4 = 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 π = 3.14]
Ans.
Radius of the outer circle (R) = 4 cm
Area of the outer circular sheet = πR2
= 3.14 × 4 × 4
= 3.14 × 16 cm2
= 50.24 cm2
Radius of the inner circle (r) = 3 cm
Area of the inner circular sheet = πr2
= 3.14 × 32
= 3.14 × 9 cm2 = 28.26 cm2
∴ Area of the remaining sheet
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= Area of outer circular sheet – Area of inner circular sheet
= 50.24 cm2 – 28.26 cm2
= 21.98 cm2
Hence, area of the remaining sheet = 21.98 cm2
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 meter of the lace
costs Rs 15. [Take π = 3.14]
Ans.
Diameter of the table cover = 1.5 m
1.5
∴ Radius of the table cover (r) =
m
2
Circumference of the table cover = 2πr
= 2 × 3.14 ×
So,
1.5
= 4.71 m
2
length of the lace required = 4.71 m
Now, cost of one meter lace = Rs 15
∴
cost of 4.71 meter lace = Rs 15 × 4.71
= Rs 70.65
Hence, cost of the lace Rs 70.65
Q.7. Find the perimeter of the adjoining figure,
which is a semicircle including its diameter.
Ans. Perimeter of semicircle
= Half circumference of circle + diameter
= πr + d [where, d = 10 cm and r = 5 cm]
= 3.14 × 5 + 10 = 25.7 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 per m2
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[Take π = 3.14]
Ans.
Diameter of the table top = 1.6 m
⇒
Radius of the table top (r) =
1.6
m = 0.8 m
2
Area of circular table = πr2
= 3.14 × (0.8)2
= 3.14 × 0.64
= 2.0096 m2
∴ Cost of polishing 1m2 table top = Rs 15
∴ Cost of polishing 2.0096 m2 table top = Rs 15 × 2.0096
= Rs 30.144
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?
22
[Take π = ]
7
Ans.
Circumference of circle = 2πr
⇒
So,
∴
Length of wire = Circumference of circle
22
44 = 2 ×
×r
7
44 r = 44 × 7
44 × 7
r =
= 7 cm
44
radius of circle = 7 cm
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So,
22
× 7 × 7 = 154 cm2
7
Length of wire = Perimeter of square
area of circle = πr2 =
⇒
or,
∴
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44 = 4 × side
4 × side = 44
44
= 11 cm
side =
4
Area of square = side × side
= 11 cm × 11cm = 121 cm2
Hence, 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 adjoining figure).
Find the area of the remaining sheet.
[Take π =
Ans.
22
]
7
Radius of the circular card sheet (r) = 14 cm
Area of the circular card sheet = πr2
22
× 14 × 14 = 616 cm2
7
Area of two circles of radius 3.5 cm
=
22
× (3.5)2
7
22 35 35
= 2×
×
×
7
10 10
= 77 cm2
= 2×
Area of a rectangle of length 3 cm and breadth 1 cm
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= 3 × 1 = 3 cm2
Hence, area of the remaining sheet
= Area of circular card sheet– (Area of two circles
+ Area of rectangle)
= 616 cm2 – (77 cm2 + 3 cm2) = 616 cm2 – 80 cm2
= 536 cm2
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 π = 3.14]
Ans. Area of square piece = side × side = 6 × 6 = 36 cm2
Area of circle = πr2 = 3.14 × 2 × 2 = 12.56 cm2
Area of the left over aluminium sheet
= Area of square piece – Area of circle
= 36 cm2 – 12.56 cm2 = 23.44 cm2
Q.12. The circumference of a circle is 31.4 cm. Find the radius
and the area of the circle?
[Take π = 3.14]
Ans. Circumference of circle = 2πr
⇒
⇒
So,
(where, r = radius)
31.4 = 2 × 3.14 × r
31.4
3140
10
=
=
cm
2 × 3.14
2 × 314
2
r = 5 cm
r =
Hence, area of the circle = πr2
= 3.14 × 5 × 5
= 78.5 cm2
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 π = 3.14]
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Ans.
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Diameter of the flower bed = 66 m
⇒
66
m
2
= 33 m
Radius of the flower bed (r) =
∴
Area of the flower bed = πr2
= 3.14 × (33)2 m2
= 3.14 × 1089 m2
= 3419.46 m2
Radius of the flower bed with path (R) = 33 m + 4 m = 37m
∴ Area of the flower bed with path = πR2
= 3.14 × (37)2 m2
= 3.14 × 1369 m2
= 4298.66 m2
∴ Area of the path = Area of the flower bed with path
– Area of the flower bed
= 4298.66 m2 – 3419.46 m2
= 879.20 m2
Q.14. A circular flower garden has an area of 314 m2. 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]
Ans.
Area of circular flower garden = 314 m2
Radius (r) of sprinkler water = 12 m
∴
Area of sprinkler water = πr2
= 3.14 × 12 × 12
= 452.16 m2
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Yes, because the area of sprinkler water is greater than area
of circular flower garden.
Q.15. Find the circumference of the inner
and the outer circles, shown in the
adjoining figure?
[Take π = 3.14]
Ans.
Radius of the outer circle = 19 m
∴ Circumference of the outer circle = 2πr
= 2 × 3.14 × 19 m
= 119.32 m
Circumference of the inner circle = 2πr
= 2 × 3.14 × 9
= 56.52 m
Q.16. How many times a wheel of radius 28 cm must rotate to
22
go 352 m?
[Take π =
]
7
Ans.
Radius of the wheel (r) = 28 cm
Circumference of circle = 2πr
22
× 28
7
= 176 cm
= 2×
Wheel goes = 352 m
= 35200 cm (∴ 1 m = 100 cm)
35200
176
= 200 times
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.
Number of times the wheel must rotate to go =
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[Take π = 3.14]
Ans. Length of the minute hand of the circular clock = 15 cm
⇒
∴
Radius of the circular clock (r) = 15 cm
Circumference of circular clock = 2πr
= 2 × 3.14 × 15
= 94.20 cm
⇒ Length moved by the tip of the minute hand in 1 hour
= 94.2 cm [As the minute hand makes one complete
revolution round the clock in 1 hour].
Exercise 11.4
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.
Ans. Let us represent the garden by rectangle ABCD and the path
around it by shaded region as shown in figure.
Since, Area of path
= Area of rectangle EFGH
– Area of rectangle ABCD
So, In the figure, we have
EF = 90 m + 5 m + 5 m = 100 m = length
FG = 75 m + 5 m + 5 m = 85 m = breadth
∴
Area of rectangle EFGH = 100 m × 85 m = 8500 m2
Area of rectangle ABCD = 90 m × 75 m = 6750 m2
So,
area of path = Ar. of rectangle EFGH
– Ar. of rectangle ABCD
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= 8500 m2 – 6750 m2
= 1750 m2
Now,
the area of rectangle garden ABCD = 6750 m2
10000 m2 = 1 hectare
1
hectare
1 m2 =
10000
1
Hence, area of rectangular garden =
× 6750
10000
= 0.675 hectare
So,
Q.2. A 3 m wide path runs outside and around a rectangular
park of length 125 m and the breadth 65 m. Find the
area of the path.
Ans. For outside rectangle
Length = 125 + 3 + 3 = 131 m = l
Breadth = 65 + 3 + 3 = 71 m = b
Area of external rectangle = 131 × 71 = 9301 m2
For internal rectangle
Length = 125 m = l
Breadth = 65 m = b
Area of inner rectangle = 125 × 65 = 8125 m2
Hence, area of path = 9301 – 8125 = 1176 m2
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.
Ans. Let us represent the card board of length 8 cm
and width 5 cm by the rectangle ABCD,
and the painting by the rectangle EFGH.
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AB = 8 cm,
∴
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BC = 5 cm
EF = (8 – 2 × 1.5) cm = (8 – 3) cm = 5 cm
FG = (5 – 2 × 1.5) cm = (5 – 3) cm = 2 cm
Hence, area of margin = Area of rectangle ABCD
– Area of rectangle EFGH.
= 8 cm × 5 cm – 5 cm × 2 cm
= 40 cm2 – 10 cm2 = 30 cm2
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 m2.
Ans. Area of the rectangle ABCD = length × breadth
= 5.5 m × 4 m = 22 m2
For arc of rectangle EFGH,
Total length = 5.5 m + 2.25 m + 2.25 m
= 10 m
Total breadth = 4 m + 2.25 m + 2.25 m = 8.50 m
Area of EFGH = 10 m × 8.50 m = 85 m2
(i) Area of the verandah = Area of EFGH – Area of ABCD
= 85 m2 – 22 m2 = 63 m2
(ii) Cost of cementing the floor of verandah at the rate of
Rs 200 per m2
= Rs 200 × 63 m2
= Rs 12,600
Q.5. A path 1 m wide is built along the border and inside a
square garden of side 30 m. Find :
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(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.
Ans.
Area of square ABCD
= (side)2
= 30 m × 30 m
= 900 m2
Area of square EFGH = [30 – (1 + 1)]m × [30 – (1 + 1)]m
= 28 m × 28 m
= 784 m2
(i)
So, area of path = Area of ABCD – Area of EFGH
= 900 – 784 = 116 m2
(ii) Cost of the planting grass in the remaining portion of
the garden at the rate of Rs 40 per m2
= 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.
Ans.
Area of the roads
= Area of the rectangle PQRS + Area of the rectangle
EFGH – Area of the square KLMN
= (300 × 10 + 700 × 10 – 10 × 10) m2
= (3000 + 7000 – 100) m2 = 9900 m2 or
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9900
= 0.99 ha
10000
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Area of the rectangle ABCD = l × b
= (700 × 300) m2 = 210000 m2
∴
Area of the park excluding cross roads
= (210000 – 9900) m2 = 200100 m2
=
200100
ha = 20.01 ha.
10000
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 the rate of
Rs 110 per m2.
Ans.
For field length = l = 90 m
Breadth = b = 60 m
Width of road = 3 m
Area of road parallel to length = 90 × 3 = 270 m2
Area of road parallel to breadth = 60 × 3
= 180 m2
Area of common square = 3 × 3 = 9 m2
(i)
Area of roads = 270 + 180 – 9
= 441 m2
(ii)
Rate of constructing roads = Rs 110 per m2
Hence, cost of constructing roads = Rs 110 × 441
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= Rs 48,510.
Q.8. Pragya wrapped a cord around a circular pipe of radius
4 cm (adjoining figure) 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)
Ans.
Radius of the circular pipe (r) = 4 cm
Circumference of the circular pipe = 2πr
= 2 × 3.14 × 4
= 25.12 cm
Perimeter of the square box of the side 4 cm
= 4 × side = 4 × 4 cm
= 16 cm
So,
cord left = 25.12 – 16 = 9.12cm
Q.9. The Adjoining 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.
Ans. (i) The area of the whole rectangular lawn
= l × b = 5 × 10
= 50 m2
(ii) Area of circular flower bed
= πr2 = 3.14 × (2)2
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= 3.14 × 4
= 12.56 m2
(iii) Area of lawn excluding area of flower bed
= [(Area of rectangular whole land)
– (Area of circular flower bed)]
= (50 – 12.56)m2 = 37.44 m2
(iv) Circumference of flower bed = 2πr
= 2 × 3.14 × 2 = 12.56 m
Q.10. In the following figures, find the area of shaded
portions :
Ans. (i)
Area of rectangle ABCD = l × b
10 × 8 = 180 cm2
1
1
b×h =
× 6 × 10
Area of ΔEAF =
2
2
60 2
=
cm = 30 cm2
2
1
1
Area of ΔEBC =
b×h =
× 8 × 10
2
2
80
cm2 = 40 cm2
=
2
∴
Area of the shaded portion
= [(Area of rectangle ABCD)
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– (Area of ΔEAF + area of ΔEBC)
Hence, area of shaded portion = [(180) – (40 + 30)]
= (180 – 70)] = 110 cm2
(ii)
Area of square PQRS = side × side
= 20 × 20
= 400 cm2
1
Area of ΔTSU =
×b×h
2
1
× 10 × 10 cm2
=
2
100 2
=
cm
2
= 50 cm2
1
×b×h
Area of ΔQPT =
2
1
=
× 10 × 20 cm2
2
200
=
cm2 = 100 cm2
2
1
×b×h
Area of ΔQRU =
2
1
=
× 10 × 20
2
200
= 100 cm2
=
2
Hence, area of shaded portion
= (Area of square PQRS)
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– [(Area of ΔTSU)
+ (Area of ΔQPT)
+ (Area of ΔQRU)]
= [(400) – (50 + 100 + 100)] cm2
= (400 – 250) cm2
= 150 cm2
Q.11. Find the area of the quadrilateral ABCD. Here, AC = 22
cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC.
1
×b×h
Ans. Area of ΔABC =
2
1
× 22 × 3 cm2
=
2
66
=
cm2
2
= 33 cm2
1
Area of ΔADC =
×b×h
2
1
× 22 × 3 cm2
=
2
66
=
cm2 = 33 cm2
2
Hence, area of quadrilateral ABCD
= Area of ΔABC + Area of ΔADC
= (33 + 33) cm2
= 66 cm2
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