Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > NCERT Exemplar Solutions: Perimeter & Area
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Page 1
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Page 2
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Solution:-
(c) 24 cm
2
Area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm
2
So,
Area of sheet of paper = 6 cm × 5 cm
= 30 cm
2
Therefore,
Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm
2
3. 36 unit squares are joined to form a rectangle with the least perimeter.
Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
We have,
Area of rectangle = 36 units
2
36 = 6 × 6
= (2 × 3) × (2 × 3)
= 2
2
× 3
2
= 4 × 9
So, the sides of the rectangle are 4 cm and 9 cm.
Also,
Perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
4. A wire is bent to form a square of side 22 cm. If the wire is rebent to
form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
Page 3
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Solution:-
(c) 24 cm
2
Area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm
2
So,
Area of sheet of paper = 6 cm × 5 cm
= 30 cm
2
Therefore,
Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm
2
3. 36 unit squares are joined to form a rectangle with the least perimeter.
Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
We have,
Area of rectangle = 36 units
2
36 = 6 × 6
= (2 × 3) × (2 × 3)
= 2
2
× 3
2
= 4 × 9
So, the sides of the rectangle are 4 cm and 9 cm.
Also,
Perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
4. A wire is bent to form a square of side 22 cm. If the wire is rebent to
form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
We have,
Side of square = 22 cm.
And also, perimeter of a square and circumference of circle are equal, because the length of
the wire is same.
Perimeter of square = Circumference of circle
4 × side = 2 × p × r
4 × 22 = 2 ×
22
7
× r
r = 14 cm
Therefore, radius of circle is 14 cm.
5. Area of the circle obtained in Question 4 is
(a) 196 cm^2 (b) 212 cm^2 (c) 616 cm^2 (d) 644 cm^2
Solution:-
(c) 616 cm
2
Area of circle = pr
2
=
22
7
× 14 × 14
= 22 × 14 × 2
= 616 cm
2
6. Area of a rectangle and the area of a circle are equal. If the dimensions
of the rectangle are 14cm × 11 cm, then radius of the circle is
(a) 21 cm (b) 10.5 cm (c) 14 cm (d) 7 cm.
Solution:-
(d) 7 cm
We have,
length = 14 cm,
breadth = 11 cm
Also,
area of rectangle = area of circle
length × breadth = pr
2
14 × 11 =
22
7
× r
2
r
2
= 49
r = 7 cm
Page 4
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Solution:-
(c) 24 cm
2
Area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm
2
So,
Area of sheet of paper = 6 cm × 5 cm
= 30 cm
2
Therefore,
Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm
2
3. 36 unit squares are joined to form a rectangle with the least perimeter.
Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
We have,
Area of rectangle = 36 units
2
36 = 6 × 6
= (2 × 3) × (2 × 3)
= 2
2
× 3
2
= 4 × 9
So, the sides of the rectangle are 4 cm and 9 cm.
Also,
Perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
4. A wire is bent to form a square of side 22 cm. If the wire is rebent to
form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
We have,
Side of square = 22 cm.
And also, perimeter of a square and circumference of circle are equal, because the length of
the wire is same.
Perimeter of square = Circumference of circle
4 × side = 2 × p × r
4 × 22 = 2 ×
22
7
× r
r = 14 cm
Therefore, radius of circle is 14 cm.
5. Area of the circle obtained in Question 4 is
(a) 196 cm^2 (b) 212 cm^2 (c) 616 cm^2 (d) 644 cm^2
Solution:-
(c) 616 cm
2
Area of circle = pr
2
=
22
7
× 14 × 14
= 22 × 14 × 2
= 616 cm
2
6. Area of a rectangle and the area of a circle are equal. If the dimensions
of the rectangle are 14cm × 11 cm, then radius of the circle is
(a) 21 cm (b) 10.5 cm (c) 14 cm (d) 7 cm.
Solution:-
(d) 7 cm
We have,
length = 14 cm,
breadth = 11 cm
Also,
area of rectangle = area of circle
length × breadth = pr
2
14 × 11 =
22
7
× r
2
r
2
= 49
r = 7 cm
7. Area of shaded portion in Fig. 9.15 is
(a) 25 cm^2 (b) 15 cm^2 (c) 14 cm^2 (d) 10 cm^2
Solution:-
(d) 10 cm
2
We have,
length of rectangle = 5 cm,
breadth of the rectangle = 3 cm + 1 cm
= 4 cm
So,
Area of rectangle = length × breadth
= 5 × 4
= 20 cm
2
By observing the figure, the shaded part covered exactly half of the rectangle,
Therefore,
Area of shaded part is =
2
Area of rectangle
=
20
2
= 10 cm
2
8. Area of parallelogram ABCD (Fig. 9.16) is not equal to
(a) DE × DC (b) BE × AD (c) BF × DC (d) BE × BC
Page 5
Exercise
In the Questions 1 to 37, there are four options, out of which one
is correct. Choose the correct one.
1. Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following
statements is not correct?
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
(b) Shapes 1 and 4 have the same area as well as the same perimeter.
(c) Shapes 1, 2 and 4 have the same area.
(d) Shapes 1, 3 and 4 have the same perimeter.
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
As,
Shapes 1, 3 and 4 have same area and same perimeter.
2. A rectangular piece of dimensions 3 cm × 2 cm was cut from a
rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14). Area of
remaining sheet of paper is
(a) 30 cm
2
(b) 36 cm
2
(c) 24 cm
2
(d) 22 cm
2
Chapter - 9
Perimeter and Area
Solution:-
(c) 24 cm
2
Area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm
2
So,
Area of sheet of paper = 6 cm × 5 cm
= 30 cm
2
Therefore,
Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm
2
3. 36 unit squares are joined to form a rectangle with the least perimeter.
Perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
We have,
Area of rectangle = 36 units
2
36 = 6 × 6
= (2 × 3) × (2 × 3)
= 2
2
× 3
2
= 4 × 9
So, the sides of the rectangle are 4 cm and 9 cm.
Also,
Perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
4. A wire is bent to form a square of side 22 cm. If the wire is rebent to
form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
We have,
Side of square = 22 cm.
And also, perimeter of a square and circumference of circle are equal, because the length of
the wire is same.
Perimeter of square = Circumference of circle
4 × side = 2 × p × r
4 × 22 = 2 ×
22
7
× r
r = 14 cm
Therefore, radius of circle is 14 cm.
5. Area of the circle obtained in Question 4 is
(a) 196 cm^2 (b) 212 cm^2 (c) 616 cm^2 (d) 644 cm^2
Solution:-
(c) 616 cm
2
Area of circle = pr
2
=
22
7
× 14 × 14
= 22 × 14 × 2
= 616 cm
2
6. Area of a rectangle and the area of a circle are equal. If the dimensions
of the rectangle are 14cm × 11 cm, then radius of the circle is
(a) 21 cm (b) 10.5 cm (c) 14 cm (d) 7 cm.
Solution:-
(d) 7 cm
We have,
length = 14 cm,
breadth = 11 cm
Also,
area of rectangle = area of circle
length × breadth = pr
2
14 × 11 =
22
7
× r
2
r
2
= 49
r = 7 cm
7. Area of shaded portion in Fig. 9.15 is
(a) 25 cm^2 (b) 15 cm^2 (c) 14 cm^2 (d) 10 cm^2
Solution:-
(d) 10 cm
2
We have,
length of rectangle = 5 cm,
breadth of the rectangle = 3 cm + 1 cm
= 4 cm
So,
Area of rectangle = length × breadth
= 5 × 4
= 20 cm
2
By observing the figure, the shaded part covered exactly half of the rectangle,
Therefore,
Area of shaded part is =
2
Area of rectangle
=
20
2
= 10 cm
2
8. Area of parallelogram ABCD (Fig. 9.16) is not equal to
(a) DE × DC (b) BE × AD (c) BF × DC (d) BE × BC
Solution:-
(a) DE × DC
We have,
Area of parallelogram = base × corresponding height
So,
Area of parallelogram ABCD = DC × BF
AD × BE = BC × BE [as, AD = BC]
9. Area of triangle MNO of Fig. 9.17 is
(a) 1/2 MN × NO (b) 1/2 NO × MO (c) 1/2 MN × OQ (d) 1/2 NO
×OQ
Solution:-
(d)
1
2
NO × OQ
MNO is a triangle.
We have,
Area of triangle =
1
2
(base × height)
=
1
2
× NO × OQ
10. Ratio of area of ?MNO to the area of parallelogram MNOP in the same
figure 9.17 is
(a) 2 : 3 (b) 1 : 1 (c) 1 : 2 (d) 2 : 1
Solution:-
(c) 1 : 2
From figure, we have,
Area of ?MNO =
1
2
× base × height
=
1
2
× NO × OQ
Area of parallelogram MNOP = base × corresponding height
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