RS Aggarwal Solutions: Mensuration - 5 | Mathematics (Maths) Class 7 PDF Download

 Page 1


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Question 1.
  
 
 
 
Solution: 
 
We know: 
Area of a triangle = 1/2×Base×Height 
(i) Base = 42 cm 
Height = 25 cm 
     ? Area of the triangle = (1/2×42×25) cm
2
 = 525 cm
2 
 
(ii) Base = 16.8 m     
     Height = 75 cm = 0.75 m      [since 100 cm = 1 m] 
     ? Area of the triangle = (1/2×16.8×0.75) m
2
 = 6.3 m
2
 
(iii) Base = 8 dm = (8 × 10) cm = 80 cm     [since 1 dm = 10 cm] 
       Height = 35 cm 
     ? Area of the triangle = (1/2×80×35) cm
2
 = 1400 cm
2 
Question 2. 
 
Solution: 
Base of triangle = 16 cm 
Page 2


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Question 1.
  
 
 
 
Solution: 
 
We know: 
Area of a triangle = 1/2×Base×Height 
(i) Base = 42 cm 
Height = 25 cm 
     ? Area of the triangle = (1/2×42×25) cm
2
 = 525 cm
2 
 
(ii) Base = 16.8 m     
     Height = 75 cm = 0.75 m      [since 100 cm = 1 m] 
     ? Area of the triangle = (1/2×16.8×0.75) m
2
 = 6.3 m
2
 
(iii) Base = 8 dm = (8 × 10) cm = 80 cm     [since 1 dm = 10 cm] 
       Height = 35 cm 
     ? Area of the triangle = (1/2×80×35) cm
2
 = 1400 cm
2 
Question 2. 
 
Solution: 
Base of triangle = 16 cm 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
area of the triangle = 72 cm² 
 
Question 3. 
 
Solution: 
Area of triangular region = 224 m² 
Base = 28 m 
 
Question 4. 
 
Solution: 
Area of triangle = 90 cm² 
and height (h) = 12 cm 
 
 
 
Page 3


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Question 1.
  
 
 
 
Solution: 
 
We know: 
Area of a triangle = 1/2×Base×Height 
(i) Base = 42 cm 
Height = 25 cm 
     ? Area of the triangle = (1/2×42×25) cm
2
 = 525 cm
2 
 
(ii) Base = 16.8 m     
     Height = 75 cm = 0.75 m      [since 100 cm = 1 m] 
     ? Area of the triangle = (1/2×16.8×0.75) m
2
 = 6.3 m
2
 
(iii) Base = 8 dm = (8 × 10) cm = 80 cm     [since 1 dm = 10 cm] 
       Height = 35 cm 
     ? Area of the triangle = (1/2×80×35) cm
2
 = 1400 cm
2 
Question 2. 
 
Solution: 
Base of triangle = 16 cm 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
area of the triangle = 72 cm² 
 
Question 3. 
 
Solution: 
Area of triangular region = 224 m² 
Base = 28 m 
 
Question 4. 
 
Solution: 
Area of triangle = 90 cm² 
and height (h) = 12 cm 
 
 
 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Solution:
 
Let height of a triangular field = x m
 
Then base (b) = 3x m
 
and area =1/2 bh = ½ x 3x x x
 
 
 
 
Question 5.
Page 4


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Question 1.
  
 
 
 
Solution: 
 
We know: 
Area of a triangle = 1/2×Base×Height 
(i) Base = 42 cm 
Height = 25 cm 
     ? Area of the triangle = (1/2×42×25) cm
2
 = 525 cm
2 
 
(ii) Base = 16.8 m     
     Height = 75 cm = 0.75 m      [since 100 cm = 1 m] 
     ? Area of the triangle = (1/2×16.8×0.75) m
2
 = 6.3 m
2
 
(iii) Base = 8 dm = (8 × 10) cm = 80 cm     [since 1 dm = 10 cm] 
       Height = 35 cm 
     ? Area of the triangle = (1/2×80×35) cm
2
 = 1400 cm
2 
Question 2. 
 
Solution: 
Base of triangle = 16 cm 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
area of the triangle = 72 cm² 
 
Question 3. 
 
Solution: 
Area of triangular region = 224 m² 
Base = 28 m 
 
Question 4. 
 
Solution: 
Area of triangle = 90 cm² 
and height (h) = 12 cm 
 
 
 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Solution:
 
Let height of a triangular field = x m
 
Then base (b) = 3x m
 
and area =1/2 bh = ½ x 3x x x
 
 
 
 
Question 5.
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Solution: 
Area of the right angled triangle = 129.5 cm² 
 
Question 7. 
 
Solution: 
In right angled ?ABC, 
Base BC = 1.2 m 
 
Question 6.
Page 5


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Question 1.
  
 
 
 
Solution: 
 
We know: 
Area of a triangle = 1/2×Base×Height 
(i) Base = 42 cm 
Height = 25 cm 
     ? Area of the triangle = (1/2×42×25) cm
2
 = 525 cm
2 
 
(ii) Base = 16.8 m     
     Height = 75 cm = 0.75 m      [since 100 cm = 1 m] 
     ? Area of the triangle = (1/2×16.8×0.75) m
2
 = 6.3 m
2
 
(iii) Base = 8 dm = (8 × 10) cm = 80 cm     [since 1 dm = 10 cm] 
       Height = 35 cm 
     ? Area of the triangle = (1/2×80×35) cm
2
 = 1400 cm
2 
Question 2. 
 
Solution: 
Base of triangle = 16 cm 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
area of the triangle = 72 cm² 
 
Question 3. 
 
Solution: 
Area of triangular region = 224 m² 
Base = 28 m 
 
Question 4. 
 
Solution: 
Area of triangle = 90 cm² 
and height (h) = 12 cm 
 
 
 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Solution:
 
Let height of a triangular field = x m
 
Then base (b) = 3x m
 
and area =1/2 bh = ½ x 3x x x
 
 
 
 
Question 5.
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
Solution: 
Area of the right angled triangle = 129.5 cm² 
 
Question 7. 
 
Solution: 
In right angled ?ABC, 
Base BC = 1.2 m 
 
Question 6.
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20E) Exercise 20E 
  
 
  
   
and hypotenuse AC = 3.7 m 
But AC² = AB² + BC² (Pythagoras Theorem) 
? (3.7)² = AB² + (1.2)² 
? 13.69 = AB² + 1.44 
? AB² = 13.69 – 1.44 
? AB² = 12.25 = (3.5)² 
? AB = 3.5 m 
Now, area of ?ABC = ½ x base x altitude 
= ½ x 1.2 x 3.5 m² = 2.1 m² 
Question 8. 
 
Solution: 
Legs of a right angled triangle = 3 : 4 
Let one leg (base) = 3x 
 
Then second leg (altitude) = 4x 
Area = ½ x base x altitude 
= ½ x 3x x 4x = 6x² 
6x² = 1014 
? x² = 1014/6 = 169 = (13)² 
x = 13 
one leg'(Base) = 3x = 3 x 13 = 39 cm 
and second leg (altitude) = 4x = 4 x 13 = 52 cm