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

Download, print and study this document offline
Please wait while the PDF view is loading
 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 
Read More
76 videos|344 docs|39 tests

Top Courses for Class 7

76 videos|344 docs|39 tests
Download as PDF
Explore Courses for Class 7 exam

Top Courses for Class 7

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

MCQs

,

Objective type Questions

,

study material

,

Important questions

,

Extra Questions

,

video lectures

,

pdf

,

ppt

,

Viva Questions

,

Exam

,

Summary

,

Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 7

,

practice quizzes

,

Free

,

Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 7

,

shortcuts and tricks

,

Semester Notes

,

mock tests for examination

,

Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.4) RD Sharma Solutions | Mathematics (Maths) Class 7

,

past year papers

,

Previous Year Questions with Solutions

,

Sample Paper

;