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CBSE Questions for Class 7 Maths Perimeter And Area Quiz 4 - MCQExams.com
CBSE
Class 7 Maths
Perimeter And Area
Quiz 4
A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running around it on the outside. Find the cost graveling the path at Rs 4 per square meter.
Report Question
0%
R
s
2002
0%
R
s
2003
0%
R
s
2004
0%
R
s
2000
Explanation
Given : Radius (
r
)
=
21
m
Radius (
R
)
=
24.5
m
Required area
=
π
(
R
2
−
r
2
)
Required area
=
π
(
24.5
2
−
21
2
)
=
500.5
Cost
=
500.5
×
4
=
2002
R
s
The sides of a triangle are (3p 4) cm, (2p 5) cm and (2p + 5) cm. Which of the following expressions gives its perimeter?
Report Question
0%
(7p 4) cm
0%
(5p + 5) cm
0%
5p 7 cm
0%
7p + 4cm
In fig. 3, the area of the shaded region is :
Report Question
0%
π
(
r
1
+
r
2
)
0%
π
(
r
2
1
+
r
2
2
)
0%
π
(
r
1
−
r
2
)
0%
π
(
r
2
2
−
r
2
1
)
Explanation
The area of the given shaded region will be :
The area of the smaller circle subtracted from the larger circle
Area of larger circle is
π
r
2
2
and
Area of smaller circle is
π
r
2
1
Then, the ares of shaded region is
π
r
2
2
−
π
r
2
1
=
π
(
r
2
2
−
r
2
1
)
If the base of a parallelogram is
8
c
m
and its altitude is
5
c
m
, then its area is equal to
Report Question
0%
15
c
m
2
0%
20
c
m
2
0%
40
c
m
2
0%
10
c
m
2
Explanation
Area of parallelogram
=
B
a
s
e
×
h
e
i
g
h
t
=
8
×
5
=
40
c
m
2
Report Question
0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
Explanation
Let the radius of the circle be
r
.
Circumference
=
2
×
π
×
r
⇒
176
=
2
×
22
7
×
r
⇒
r
=
176
×
7
2
×
22
⇒
r
=
28
c
m
We used the statement given in the reason to prove that the assertion is correct hence, option
A
is correct.
If a circular grass lawn of
35
m
in radius has a path
7
m
wide running around it on the outside, then the area of the path is
Report Question
0%
1450
m
2
0%
1576
m
2
0%
1694
m
2
0%
3368
m
2
Explanation
Radius of bigger circle(with the path) =
35
+
7
=
42
m
.
So,
R
=
42
c
m
&
r
=
35
c
m
Thus area of the path
=
Area of bigger circle
−
Area of smaller circle
∴
Required area
=
π
(
42
)
2
−
π
(
35
)
2
=
22
7
×
(
42
+
35
)
(
42
−
35
)
=
22
×
77
=
1694
m
2
A pond that is
100
m in diameter is surrounded by a circular grass walk that is
2
m wide. How many square meters of grass is there for the walk?
Report Question
0%
98
π
0%
100
π
0%
102
π
0%
204
π
Explanation
Area of grass on walk
=
Area of outer circle
−
Area of the pond
=
π
×
52
2
−
π
×
50
2
=
π
(
2704
−
2500
)
=
π
×
204
So, Area of grass for the walk
=
204
π
m
2
Find the area of a circular ring formed by two concentric circles whose radii are
5.7
c
m
and
4.3
c
m
respectively (Take
π
=
3.1416
)
Report Question
0%
43.98
sq. cm
0%
53.67
sq. cm
0%
47.24
sq. cm
0%
38.54
sq. cm
Explanation
Let the radii of the outer and inner circles be
r
2
and
r
1
respectively; we have
A
r
e
a
=
π
r
2
2
−
π
r
2
1
=
π
(
r
2
2
−
r
2
1
)
=
π
(
r
2
−
r
1
)
(
r
2
+
r
1
)
=
π
(
5.7
−
4.3
)
(
5.7
+
4.3
)
=
π
×
1.4
×
10
sq. cm
=
3.1416
×
14
s
q
.
c
m
.
=
43.98
sq. cm
If the radius of a circle is
7
√
π
cm, then the area of the circle is equal to
Report Question
0%
49
π
c
m
2
0%
154
π
c
m
2
0%
154
c
m
2
0%
49
c
m
2
Explanation
Area of the circle
=
π
(
7
√
π
)
2
=
π
(
49
)
π
=
49
c
m
2
Now,
154
π
=
154
×
7
22
=
49
c
m
2
A track is in the form of a ring whose inner circumference is
352
m
and the outer circumference is
396
m
. The width of the track is:
Report Question
0%
44
m
0%
14
m
0%
22
m
0%
7
m
Explanation
Let the radii of inner and outer circles be
r
and
R
, respectively.
Circumference of inner circle
=
352
m
∴
2
π
r
=
352
2
×
22
7
×
r
=
352
⇒
r
=
56
m
Similarly, Circumference of outer circle
=
396
m
∴
2
π
R
=
396
2
×
22
7
×
R
=
396
⇒
R
=
63
m
Now, Width of the track
=
R
−
r
=
63
−
56
=
7
m
So, width of the track
=
7
m
If area of circular field is
6.16
m
2
, then its diameter will be
Report Question
0%
6.8
m
0%
2.8
m
0%
12
m
0%
6.84
m
Explanation
Area of circular field
=
6.16
m
2
Area of a circle
=
π
r
2
⇒
6.16
=
22
7
×
r
2
⇒
6.16
×
7
22
=
r
2
⇒
r
2
=
43.12
22
⇒
r
2
=
1.96
⇒
r
=
1.4
m
⇒
d
=
2
r
=
1.4
×
2
=
2.8
m
∴
The diameter of a circular field is
2.8
m
.
A circular grassy plot of land,
42
m
in diameter, has a path
3.5
m
wide running round it on the outside. Find the cost of gravel ling the path at Rs.
4
per square metre.
Report Question
0%
Rs.
2002
0%
Rs.
2003
0%
Rs.
2004
0%
Rs.
2000
Explanation
We have, Radius of the plot
=
21
m
Radius of plot including the path
=
21
+
3.5
=
24.5
m
Area of the path
=
π
(
24.5
2
)
−
π
(
21
2
)
=
π
(
24.5
2
−
21
2
)
=
π
(
24.5
−
21
)
(
24.5
+
21
)
=
22
7
×
3.5
×
45.5
=
500.5
m
2
Total cost of gravelling the path
=
500.5
×
4
=
Rs.
2002
A certain right angled triangle has its area numerically equal to its perimeter. The length of its each side is an even integer. What is the perimeter?
Report Question
0%
24
u
n
i
t
s
0%
36
u
n
i
t
s
0%
32
u
n
i
t
s
0%
30
u
n
i
t
s
Explanation
Given that, the measure of each side of the right triangle is an even integer unit.
Also, its perimeter and area are numerically equal.
Let us start from even integers:
2
,
4
,
6
,
8
,
10
,
12
,
14
,
etc.
Out of them, only
(
6
,
8
,
10
)
make Pythagorean triplet
∵
8
2
+
6
2
=
10
2
Also,
8
+
6
+
10
=
24
(
P
e
r
i
m
e
t
e
r
)
And,
1
2
×
8
×
6
=
24
(
A
r
e
a
)
∴
This group of even integers comply with all the given conditions.
∴
Perimeter
=
24
units.
The adjacent sides of a parallelogram are 10 cm and 12 cm. The diagonal joining the ends of these sides is 14 cm. Its area is
Report Question
0%
48
√
6
c
m
2
0%
48
√
5
c
m
2
0%
48
√
3
c
m
2
0%
none of these
Explanation
Area of the parallelogram
A
B
C
D
=
2
×
area
△
B
C
D
Now, in
△
B
C
D
S
=
12
+
10
+
14
2
=
18
cm
Area of
△
B
C
D
=
√
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
√
18
(
18
−
12
)
(
18
−
10
)
(
18
−
14
)
=
√
18
×
6
×
8
×
4
=
√
3
×
6
×
6
×
2
×
4
×
4
=
4
×
6
√
6
=
24
√
6
cm
2
∴
Area of the parallelogram
A
B
C
D
=
2
×
24
√
6
=
48
√
6
cm
2
The circumference of a circular ground is
88
metres. A strip of land,
3
metres wide, inside and along circumference of the ground is to be levelled. What is the budgeted expenditure if the levelling costs Rs.
7
per square metre?
Report Question
0%
1050
0%
1125
0%
1325
0%
1650
Explanation
Circumference of ground
=
88
m
⇒
2
π
r
=
88
⇒
r
=
14
m
....take
π
=
22
7
Area of strip to be levelled
=
Area of ground
−
Area of inner circle
=
π
14
2
−
π
11
2
=
235.72
m
2
....take
π
=
22
7
Total cost of levelling the strip (if the cost of levelling
1
m
2
is Rs.
7
)
=
235.72
×
7
=
Rs.
1650
A flooring tile has the shape of a parallelogram whose base is
24
c
m
and the corresponding height is
10
c
m
. How many such tiles are required to cover a floor of area
1080
m
2
?
Report Question
0%
20000
0%
35000
0%
45000
0%
65000
Explanation
Area of the parallelogram
=
Base
×
Height
So area of each tile
=
24
×
10
=
240
c
m
2
Area of the floor
=
1080
m
2
=
(
1080
×
100
×
100
)
c
m
2
∴
Required number of tiles
=
Area of the floor
Area of each tile
=
10800000
240
=
45000
The cost of fencing a circular field at the rate of
R
s
.240
p
e
r
m
e
t
r
e
is
R
s
.
52
,
800
. The field is to be ploughed at the rate of
R
s
.
12.50
p
e
r
m
2
. Find the cost of ploughing the field.
Report Question
0%
R
s
.
48
,
125
0%
R
s
.
38
,
125
0%
R
s
.
10
,
125
0%
R
s
.
11
,
125
Explanation
Cost of fencing at the rate of Rs.
240
per meter=
R
s
.52800
Therefore,
Circumference=
52800
240
=
220
Let r be the radius of the field
Therefore,
2
π
r
=
220
⇒
r
=
220
×
7
2
×
22
⇒
r
=
35
Therefore,
Area of the field=
π
r
2
=
22
7
×
35
×
35
=
3850
m
2
Cost of fencing the field =
12.5
×
3850
=
R
s
.48125
The circumference of a circular field is
308
m
. Find its
area.
Report Question
0%
7546
m
2
0%
6546
m
2
0%
7046
m
2
0%
7846
m
2
Explanation
Let the radius of the circle be
r
cm
Circumference of the circle=
308
Therefore,
2
π
r
=
308
=>
r
=
308
×
7
2
×
22
=>
r
=
49
m
Therefore,
Area of the circle=
π
×
(
49
)
2
=
22
7
×
2401
=
7546
m
2
The diameter of a circle is
28
c
m
. Find its
area.
Report Question
0%
616
c
m
2
0%
516
c
m
2
0%
116
c
m
2
0%
216
c
m
2
Explanation
The diameter of a circle is
28
c
m
Therefore,
Radius of the circle=
28
2
=
14
c
m
Therefore,
Area of the circle=
π
(
14
)
2
=
22
7
×
196
=
616
c
m
2
The diagram, given below, shows two paths drawn inside a rectangular field 80 m long and 45 m wide. The widths of the two oaths are 8 m and 15 m as shown. Find the area of the shaded portion.
Report Question
0%
1195
m
2
0%
8895
m
2
0%
1155
m
2
0%
1895
m
2
Explanation
Given, rectangular field is 80 m long and 45 m wide.
Two paths are drawn inside the rectangular field. The widths of the two oaths are 8 m and 15 m.
The two paths intersect. The length and breath of the intersection is
8
m and
15
m.
Area of the shaded region
=
(
8
×
80
)
+
(
15
×
45
)
−
(
15
×
8
)
=
640
+
675
−
120
sq. m
=
1195
sq. m
Find the length and the breadth of the lawn.
Report Question
0%
26 m and 10 m
0%
35 m and 12 m
0%
41 m and 16 m
0%
49 m and 18 m
Explanation
Since, in the given figure
Outer length
=
30
m
and Outer breadth
=
12
m
As per given, On the three sides of the lawn there are flower-beds having a uniform width of 2 m.
∴
length of lawn
=
30
−
2
−
2
=
26
m
and
breadth of lawn
=
12
−
2
=
10
m
The shaded portion of the figure, given along side,shows two concentric
circles.
If the circumference of the two circles are
396
c
m
and
374
c
m
, find the area of the shaded portion.
Report Question
0%
1347.5
c
m
2
0%
1047.5
c
m
2
0%
2347.5
c
m
2
0%
1247.5
c
m
2
Explanation
Given,
2
π
r
=
374
c
m
2
π
R
=
396
c
m
Therefore,
2
π
r
=
374
c
m
=>
r
=
374
×
7
2
×
22
=
59.5
2
π
R
=
396
c
m
=>
R
=
396
×
7
2
×
22
=
63
Area of shaded portion=
(
π
R
2
−
π
r
2
)
c
m
2
=
π
[
(
63
)
2
−
(
59.5
)
2
]
c
m
2
=
π
(
3969
−
3540.25
)
=
1347.5
c
m
2
The area of a circle is
394.24
c
m
2
. What is t
he radius of the circle ?
Report Question
0%
11.2
c
m
0%
12.1
m
0%
13.5
m
0%
15.0
m
Explanation
Area of the circle
=
394.24
c
m
2
⟹
π
r
2
=
394.24
⟹
r
2
=
394.24
×
7
22
⟹
r
2
=
125.44
⟹
r
=
11.2
c
m
The circumference of a circular field is
308
m
, Find its
Area.
Report Question
0%
7050
m
2
0%
7546
m
2
0%
7946
m
2
0%
8129
m
2
Explanation
Let the radius of the circle be
r
cm
Circumference of the circle=
308
Therefore,
2
π
r
=
308
⇒
r
=
308
×
7
2
×
22
r
=
49
m
Therefore,
Area of the circle
=
π
×
(
49
)
2
=
22
7
×
2401
=
7546
m
2
The diameter of a circle is
28
c
m
. Find its c
ircumference
Report Question
0%
32
c
m
0%
48
c
m
0%
88
c
m
0%
98
c
m
Explanation
Diameter of a circle
=
28
cm
Thus, circumference
=
π
×
28
cm
=
22
7
×
28
cm
=
22
×
4
cm
=
88
cm
A floor which measures
15
m
×
8
m
is to be laid with tiles measuring
50
c
m
×
25
c
m
. Find the number of tiles required.
Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered.
Report Question
0%
960
;
7
20
0%
860
;
20
7
0%
760
;
7
13
0%
660
;
7
15
Explanation
We know,
50
c
m
=
0.5
m
and
25
c
m
=
0.25
m
Tiles along the length of the room=
15
0.5
=
30
tiles
Tiles along the length of the room=
8
0.25
=
32
tiles
Total number of tiles required =
30
×
32
=
960
tiles
To leave
1
m
between the carpet and wall on all sides, the carpet needs to be
2
m
shorter in each dimension.
Therefore,
(
15
−
2
)
m
(
8
−
2
)
m
=
(
13
×
6
)
m
=
78
m
2
Therefore, the room area is
15
×
8
=
120
m
2
The carpet area is
78
m
2
The fraction of the area uncovered
=
120
−
78
120
=
42
120
=
7
20
The radii of two circles are
25
c
m
and
18
c
m
. Find the radius of the circle which has a circumference equal to the sum of the circumference of these two circles.
Report Question
0%
43
c
m
0%
12
c
m
0%
34
c
m
0%
40
c
m
Explanation
The circumference of a circle
=
2
π
r
Let
R
be the radius of the
circle which has a circumference equal to the sum of the circumferences of the two circles.
2
π
R
=
2
π
×
25
+
2
π
×
18
⟹
R
=
25
+
18
=
43
c
m
The area enclosed by the circumferences of two concentric circles is
346.5
c
m
2
. If the circumference of the inner circle is
88
cm, then the radius of the outer circle is
Report Question
0%
88
cm
0%
17.5
cm
0%
35
cm
0%
9.3
cm
Explanation
G
i
v
e
n
−
O
i
s
t
h
e
c
e
n
t
r
e
o
f
t
w
o
c
o
n
c
e
n
t
r
i
c
c
i
r
c
l
e
s
w
i
t
h
o
u
t
e
r
r
a
d
i
u
s
O
A
=
r
2
&
i
n
n
e
r
c
i
c
u
m
f
e
r
e
n
c
e
=
C
=
88
c
m
.
a
r
.
r
i
n
g
b
e
t
w
e
e
n
t
h
e
c
i
r
c
l
e
s
=
346.5
c
m
2
.
T
o
f
i
n
d
o
u
t
−
r
2
=
?
S
o
l
u
t
i
o
n
−
i
n
n
e
r
c
i
c
u
m
f
e
r
e
n
c
e
=
88
c
m
.
L
e
t
t
h
e
r
a
d
i
u
s
o
f
t
h
e
i
n
n
e
r
c
i
r
c
l
e
b
e
r
1
.
∴
r
1
=
C
2
π
=
346.5
2
×
22
7
c
m
=
14
c
m
.
S
o
a
r
.
r
i
n
g
=
a
r
.
o
u
t
e
r
c
i
r
c
l
e
−
a
r
.
i
n
n
e
r
c
i
r
c
l
e
=
π
{
r
2
2
−
r
1
2
}
⟹
346.5
=
22
7
×
{
r
2
2
−
r
1
2
}
⟹
r
2
2
=
306.25
⟹
r
2
=
17.5
c
m
.
∴
T
h
e
r
a
d
i
u
s
o
f
t
h
e
o
u
t
e
r
c
i
r
c
l
e
=
17.5
c
m
.
A
n
s
−
O
p
t
i
o
n
B
.
The diameter of a circle is
28
c
m
. Find the o
ne-fourth of its area.
Report Question
0%
389
c
m
2
0%
512
c
m
2
0%
616
c
m
2
0%
154
c
m
2
Explanation
The diameter of a circle is
28
c
m
Therefore,
Radius of the circle=
28
2
=
14
c
m
Therefore,
Area of the circle
=
π
(
14
)
2
=
22
7
×
196
=
616
c
m
2
So,
1
4
×
616
=
154
c
m
2
Area of the shaded region is
Report Question
0%
96
m
2
0%
15
m
2
0%
81
m
2
0%
111
m
2
Explanation
We know that area of rectangle of sides
l
&
b
=
l
×
b
.
Area of the shaded region
=
Area of bigger rectangle
−
area of smaller rectangle
=
(
12
×
8
)
−
(
5
×
3
)
=
96
−
15
=
81
m
2
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