Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  RD Sharma (Very Short Answer Questions): Heron's Formula

Class 9 Maths Chapter 10 Question Answers - Heron`s Formula

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Question:26
Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.
Solution:
Given, base = 5 cm; height = 4 cm
Area of the triangle = 
1
2
×Base ×Height
Question:27
Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where
Therefore the area of a triangle, say having sides 3 cm, 4 cm and 5 cm is given by
a = 3 cm ; b = 4 cm ; c = 5 cm
Now, area 
Page 2


        
     
                                    
            
               
                                    
                
 
 
                     
             
 
 
 
 
        
Question:26
Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.
Solution:
Given, base = 5 cm; height = 4 cm
Area of the triangle = 
1
2
×Base ×Height
Question:27
Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where
Therefore the area of a triangle, say having sides 3 cm, 4 cm and 5 cm is given by
a = 3 cm ; b = 4 cm ; c = 5 cm
Now, area 
= v 6(6 -3)(6 -4)(6 -5) = v 6 ×3 ×2 ×1 = v 36 = 6 cm
2
Question:28
Find the area of an isosceles triangle having the base x cm and one side y cm.
Solution:
Let us assume triangle ABC be the given isosceles triangle having sides AB = AC and base BC. The area of a triangle ABC, say A having given sides AB and AC equals to y cm and
given base BC equals to x cm is given by
Where,
Base = BC = x cm; Height = 
y
2
-
x
2
4
A =
1
2
(Base ×Height) =
1
2
×x y
2
-
x
2
4
=
x
2
y
2
-
x
2
4
Question:29
Find the area of an equilateral triangle having each side 4 cm.
Solution:
Area of an equilateral triangle having each side a cm is given by 
Area of the given equilateral triangle having each equal side equal to 4 cm is given by
a = 4 cm
Question:30
Find the area of an equilateral triangle having each side x cm.
Solution:
Area of an equilateral triangle, say A having each side a cm is given by 
Area of the given equilateral triangle having each equal side equal to x cm is given by
a = x cm
Question:31
The perimeter of a triangullar field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where,
 
It is given the sides of a triangular field are in the ratio 3:4:5 and perimeter=144 m
Therefore, a: b: c = 3:4:5
We will assume the sides of triangular field as
 
Substituting the value of x in, we get sides of the triangle as
v
(v ) (v )
Page 3


        
     
                                    
            
               
                                    
                
 
 
                     
             
 
 
 
 
        
Question:26
Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.
Solution:
Given, base = 5 cm; height = 4 cm
Area of the triangle = 
1
2
×Base ×Height
Question:27
Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where
Therefore the area of a triangle, say having sides 3 cm, 4 cm and 5 cm is given by
a = 3 cm ; b = 4 cm ; c = 5 cm
Now, area 
= v 6(6 -3)(6 -4)(6 -5) = v 6 ×3 ×2 ×1 = v 36 = 6 cm
2
Question:28
Find the area of an isosceles triangle having the base x cm and one side y cm.
Solution:
Let us assume triangle ABC be the given isosceles triangle having sides AB = AC and base BC. The area of a triangle ABC, say A having given sides AB and AC equals to y cm and
given base BC equals to x cm is given by
Where,
Base = BC = x cm; Height = 
y
2
-
x
2
4
A =
1
2
(Base ×Height) =
1
2
×x y
2
-
x
2
4
=
x
2
y
2
-
x
2
4
Question:29
Find the area of an equilateral triangle having each side 4 cm.
Solution:
Area of an equilateral triangle having each side a cm is given by 
Area of the given equilateral triangle having each equal side equal to 4 cm is given by
a = 4 cm
Question:30
Find the area of an equilateral triangle having each side x cm.
Solution:
Area of an equilateral triangle, say A having each side a cm is given by 
Area of the given equilateral triangle having each equal side equal to x cm is given by
a = x cm
Question:31
The perimeter of a triangullar field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where,
 
It is given the sides of a triangular field are in the ratio 3:4:5 and perimeter=144 m
Therefore, a: b: c = 3:4:5
We will assume the sides of triangular field as
 
Substituting the value of x in, we get sides of the triangle as
v
(v ) (v )
Area of a triangular field, say A having sides a, b , c and s as semi-perimeter is given by
Question:32
Find the area of an equilateral triangle having altitude h cm.
Solution:
Altitude of a equilateral triangle, having side a is given by
Substituting the given value of altitude h cm, we get
 
Area of a equilateral triangle, say A having each side a cm is given by 
Area of the given equilateral triangle having each equal side equal to  is given by; 
Question:33
Let ? be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.
Solution:
We are given assumed value  is the area of a given triangle ABC
We assume the sides of the given triangle ABC be a, b, c
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
Where,
We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and  as semi-perimeter is given by,
, where
Now,
Question:34
If each side of a triangle is doubled, the find percentage increase in its area.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
Page 4


        
     
                                    
            
               
                                    
                
 
 
                     
             
 
 
 
 
        
Question:26
Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.
Solution:
Given, base = 5 cm; height = 4 cm
Area of the triangle = 
1
2
×Base ×Height
Question:27
Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where
Therefore the area of a triangle, say having sides 3 cm, 4 cm and 5 cm is given by
a = 3 cm ; b = 4 cm ; c = 5 cm
Now, area 
= v 6(6 -3)(6 -4)(6 -5) = v 6 ×3 ×2 ×1 = v 36 = 6 cm
2
Question:28
Find the area of an isosceles triangle having the base x cm and one side y cm.
Solution:
Let us assume triangle ABC be the given isosceles triangle having sides AB = AC and base BC. The area of a triangle ABC, say A having given sides AB and AC equals to y cm and
given base BC equals to x cm is given by
Where,
Base = BC = x cm; Height = 
y
2
-
x
2
4
A =
1
2
(Base ×Height) =
1
2
×x y
2
-
x
2
4
=
x
2
y
2
-
x
2
4
Question:29
Find the area of an equilateral triangle having each side 4 cm.
Solution:
Area of an equilateral triangle having each side a cm is given by 
Area of the given equilateral triangle having each equal side equal to 4 cm is given by
a = 4 cm
Question:30
Find the area of an equilateral triangle having each side x cm.
Solution:
Area of an equilateral triangle, say A having each side a cm is given by 
Area of the given equilateral triangle having each equal side equal to x cm is given by
a = x cm
Question:31
The perimeter of a triangullar field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where,
 
It is given the sides of a triangular field are in the ratio 3:4:5 and perimeter=144 m
Therefore, a: b: c = 3:4:5
We will assume the sides of triangular field as
 
Substituting the value of x in, we get sides of the triangle as
v
(v ) (v )
Area of a triangular field, say A having sides a, b , c and s as semi-perimeter is given by
Question:32
Find the area of an equilateral triangle having altitude h cm.
Solution:
Altitude of a equilateral triangle, having side a is given by
Substituting the given value of altitude h cm, we get
 
Area of a equilateral triangle, say A having each side a cm is given by 
Area of the given equilateral triangle having each equal side equal to  is given by; 
Question:33
Let ? be the area of a triangle. Find the area of a triangle whose each side is twice the side of the given triangle.
Solution:
We are given assumed value  is the area of a given triangle ABC
We assume the sides of the given triangle ABC be a, b, c
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
Where,
We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and  as semi-perimeter is given by,
, where
Now,
Question:34
If each side of a triangle is doubled, the find percentage increase in its area.
Solution:
The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
  
Where,
We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and  as semi-perimeter is given by,
Where,
Now,
Therefore, increase in the area of the triangle
Percentage increase in area 
Question:35
If each side of a equilateral triangle is tripled then what is the percentage increase in the area of the triangle?
Solution:
Area of an equilateral triangle having each side a cm is given by 
Now, Area of an equilateral triangle, say  if each side is tripled is given by
a = 3a
Therefore, increase in area of triangle
Percentage increase in area
        
              
 
 
 
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FAQs on Class 9 Maths Chapter 10 Question Answers - Heron`s Formula

1. What is Heron's Formula?
Ans. Heron's Formula is a mathematical formula used to find the area of a triangle when the lengths of its sides are known. It is named after Hero of Alexandria who discovered it.
2. How is Heron's Formula derived?
Ans. Heron's Formula is derived by using the concept of semi-perimeter of a triangle. It states that the area of a triangle with sides of lengths a, b, and c is given by the formula: Area = √(s(s-a)(s-b)(s-c)) where s is the semi-perimeter of the triangle, given by s = (a + b + c)/2.
3. What is the significance of Heron's Formula?
Ans. Heron's Formula is significant as it provides a simple and effective method to find the area of a triangle without needing its height. It is particularly useful when the height is not given or is difficult to measure.
4. How is Heron's Formula used in real-life situations?
Ans. Heron's Formula is used in various real-life situations where triangles are involved, such as in architecture, construction, and design. It helps in calculating the area of irregularly shaped triangles or triangles with sides of different lengths.
5. Can Heron's Formula be used for any type of triangle?
Ans. Yes, Heron's Formula can be used for any type of triangle, including equilateral, isosceles, and scalene triangles. As long as the lengths of the triangle's sides are known, Heron's Formula can be applied to find its area.
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