Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4)
Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
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Page 1
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
Page 2
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
4. Find the area of a right angled triangle whose sides containing the right angle are of
lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m
2
5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7
cm, is equal to area of a right triangle whose one of the sides containing the right
angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, given that
Base = 15 cm and altitude = 7 cm
We know that area of a triangle = ½ (Base x Altitude)
= ½ (15 x 7)
= 52.5 cm
2
It is also given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m
2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower
beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this
field?
Solution:
Given length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Page 3
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
4. Find the area of a right angled triangle whose sides containing the right angle are of
lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m
2
5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7
cm, is equal to area of a right triangle whose one of the sides containing the right
angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, given that
Base = 15 cm and altitude = 7 cm
We know that area of a triangle = ½ (Base x Altitude)
= ½ (15 x 7)
= 52.5 cm
2
It is also given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m
2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower
beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this
field?
Solution:
Given length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Area of the rectangular field = Length x Breadth
= 48 m x 20 m
= 960 m
2
Area of one right triangular flower bed = ½ (12 x 5) = 30 m
2
Therefore, required number of right triangular flower beds = area of the rectangular
field/ area of one right triangular flower bed.
= 960/30
Number of right triangular flower beds = 32
7. In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ?AC, BM ? AC,
DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.
Solution:
Given AC = 84 cm, DL = 16.5 cm and BM = 12 cm
We know that area of triangle = ½ x base x height
Area of triangle ADC = ½ (AC x DL)
= ½ (84 x 16.5)
= 693 cm
2
Area of triangle ABC = ½ (AC x BM)
= ½ (84 x 12) = 504 cm
2
Hence, Area of quadrilateral ABCD = Area of triangle ADC + Area of triangle ABC
= (693 + 504) cm
2
= 1197 cm
2
8. Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD
measure 48 m and 32 m respectively and are perpendicular to each other.
Page 4
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
4. Find the area of a right angled triangle whose sides containing the right angle are of
lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m
2
5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7
cm, is equal to area of a right triangle whose one of the sides containing the right
angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, given that
Base = 15 cm and altitude = 7 cm
We know that area of a triangle = ½ (Base x Altitude)
= ½ (15 x 7)
= 52.5 cm
2
It is also given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m
2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower
beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this
field?
Solution:
Given length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Area of the rectangular field = Length x Breadth
= 48 m x 20 m
= 960 m
2
Area of one right triangular flower bed = ½ (12 x 5) = 30 m
2
Therefore, required number of right triangular flower beds = area of the rectangular
field/ area of one right triangular flower bed.
= 960/30
Number of right triangular flower beds = 32
7. In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ?AC, BM ? AC,
DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.
Solution:
Given AC = 84 cm, DL = 16.5 cm and BM = 12 cm
We know that area of triangle = ½ x base x height
Area of triangle ADC = ½ (AC x DL)
= ½ (84 x 16.5)
= 693 cm
2
Area of triangle ABC = ½ (AC x BM)
= ½ (84 x 12) = 504 cm
2
Hence, Area of quadrilateral ABCD = Area of triangle ADC + Area of triangle ABC
= (693 + 504) cm
2
= 1197 cm
2
8. Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD
measure 48 m and 32 m respectively and are perpendicular to each other.
Solution:
Given diagonal AC = 48 m and diagonal BD = 32 m
Area of a quadrilateral = ½ (Product of diagonals)
= ½ (AC x BD)
= ½ (48 x 32) m
2
= 768 m
2
9. In Fig 31, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such
that EF ? AD and EF= 14 cm. Calculate the area of the shaded region.
Solution:
Given length of rectangle = 32m and breadth = 18m
We know that area of rectangle = length x breadth
Therefore area of the rectangle = AB x BC
= 32 m x 18 m
= 576 m
2
Page 5
1. Find the area in square centimetres of a triangle whose base and altitude are as
under:
(i) Base =18 cm, altitude = 3.5 cm
(ii) Base = 8 dm, altitude =15 cm
Solution:
(i) Given base = 18 cm and height = 3.5 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 18 x 3.5
= 31.5 cm
2
(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]
And height = 15 cm
We know that the area of a triangle = ½ (Base x Height)
Therefore area of the triangle = ½ x 80 x 15
= 600 cm
2
2. Find the altitude of a triangle whose area is 42 cm
2
and base is 12 cm.
Solution:
Given base = 12 cm and area = 42 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore altitude of a triangle = (2 x Area)/Base
Altitude = (2 x 42)/12
= 7 cm
3. The area of a triangle is 50 cm
2
. If the altitude is 8 cm, what is its base?
Solution:
Given, altitude = 8 cm and area = 50 cm
2
We know that the area of a triangle = ½ (Base x Height)
Therefore base of a triangle = (2 x Area)/ Altitude
Base = (2 x 50)/ 8
= 12.5 cm
4. Find the area of a right angled triangle whose sides containing the right angle are of
lengths 20.8 m and 14.7 m.
Solution:
In a right-angled triangle,
The sides containing the right angles are of lengths 20.8 m and 14.7 m.
Let the base be 20.8 m and the height be 14.7 m.
Then,
Area of a triangle = 1/2 (Base x Height)
= 1/2 (20.8 × 14.7)
= 152.88 m
2
5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7
cm, is equal to area of a right triangle whose one of the sides containing the right
angle is 10.5 cm. Find the other side of this triangle.
Solution:
For the first triangle, given that
Base = 15 cm and altitude = 7 cm
We know that area of a triangle = ½ (Base x Altitude)
= ½ (15 x 7)
= 52.5 cm
2
It is also given that the area of the first triangle and the second triangle are equal.
Area of the second triangle = 52.5 c m
2
One side of the second triangle = 10.5 cm
Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle
= (2x 52.5)/10.5
=10 cm
Hence, the other side of the second triangle will be 10 cm.
6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower
beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this
field?
Solution:
Given length of the rectangular field = 48 m
Breadth of the rectangular field = 20 m
Area of the rectangular field = Length x Breadth
= 48 m x 20 m
= 960 m
2
Area of one right triangular flower bed = ½ (12 x 5) = 30 m
2
Therefore, required number of right triangular flower beds = area of the rectangular
field/ area of one right triangular flower bed.
= 960/30
Number of right triangular flower beds = 32
7. In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ?AC, BM ? AC,
DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.
Solution:
Given AC = 84 cm, DL = 16.5 cm and BM = 12 cm
We know that area of triangle = ½ x base x height
Area of triangle ADC = ½ (AC x DL)
= ½ (84 x 16.5)
= 693 cm
2
Area of triangle ABC = ½ (AC x BM)
= ½ (84 x 12) = 504 cm
2
Hence, Area of quadrilateral ABCD = Area of triangle ADC + Area of triangle ABC
= (693 + 504) cm
2
= 1197 cm
2
8. Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD
measure 48 m and 32 m respectively and are perpendicular to each other.
Solution:
Given diagonal AC = 48 m and diagonal BD = 32 m
Area of a quadrilateral = ½ (Product of diagonals)
= ½ (AC x BD)
= ½ (48 x 32) m
2
= 768 m
2
9. In Fig 31, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such
that EF ? AD and EF= 14 cm. Calculate the area of the shaded region.
Solution:
Given length of rectangle = 32m and breadth = 18m
We know that area of rectangle = length x breadth
Therefore area of the rectangle = AB x BC
= 32 m x 18 m
= 576 m
2
Also given that base of triangle = 18m and height = 14m and EF ? AD
We know that area of triangle = ½ x base x height
Area of the triangle = ½ (AD x FE)
= ½ (BC x FE) [Since AD = BC]
= ½ (18 m x 14 m)
= 126 m
2
Area of the shaded region = Area of the rectangle – Area of the triangle
= (576 – 126) m
2
= 450 m
2
10. In Fig. 32, ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm. If P, Q,
R, S be the mid-points of the sides AB, BC, CD and DA respectively, find the area of the
shaded region.
Solution:
Given ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm.
Join PR and SQ so that these two lines bisect each other at point O
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Document Description: RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
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