RS Aggarwal Solutions: Mensuration - 4 | Mathematics (Maths) Class 7 PDF Download

 Page 1


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
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
 
area of the triangle = 72 cm²
Page 2


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
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
 
area of the triangle = 72 cm²
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
 
 
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
 
 
Question 5.
 
 
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area =1/2 bh = ½ x 3x x x
Page 3


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
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
 
area of the triangle = 72 cm²
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
 
 
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
 
 
Question 5.
 
 
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area =1/2 bh = ½ x 3x x x
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
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 4


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
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
 
area of the triangle = 72 cm²
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
 
 
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
 
 
Question 5.
 
 
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area =1/2 bh = ½ x 3x x x
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
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 20D) Exercise 20D 
  
 
  
   
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 
Page 5


RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
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
 
area of the triangle = 72 cm²
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
 
 
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
 
 
Question 5.
 
 
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area =1/2 bh = ½ x 3x x x
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
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 20D) Exercise 20D 
  
 
  
   
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 
RS Aggarwal Solutions Class 7 Chapter 20 - Mensuration  (Ex 20D) Exercise 20D 
  
 
  
   
Question 9. 
 
Solution: 
One side BC of a right triangular scarf = 80 cm 
 
and longest side AC = 1 m = 100 cm 
By Pythagoras Theorem, 
AC² = AB² + BC² 
? (100)² = AB² + (80)² 
? 10000 = AB² + 6400 
? AB² = 10000 – 6400 
? AB² = 3600 = (60)² 
? AB = 60 
Second side = 60 cm 
Area of the scarf = ½ x b x h 
= ½ x 80 x 60 cm
2
 = 2400 cm² 
Rate of cost = Rs. 250 per m² 
Total cost =2400/200  x 250 = Rs. 60