Solved Question and Answers: Heron's Formula | Mathematics (Maths) Class 9 PDF Download

Q1: The area of a triangle with base 8 cm and height 10 cm is
(a) 20 cm²
(b) 18 cm²
(c) 80 cm²
(d) 40 cm²
Ans: d
Sol: are of triangle = 1/2 x base x height
= 1/2 x 8 x 10
=  40

Q2: The sides of a triangle are in the ratio of 3 : 4 : 5. If its perimeter is 36 cm, then what is its area?
(a) 32 cm²
(b) 54 cm²
(c) 67 cm²
(d) 72cm²
Ans: b
Sol: It is given that the sides of a triangle are in the ratio 3 : 4 : 5.
Let the length of sides are 3x, 4x and 5x.
It's perimeter is 36 cm.
3x+4x+5x = 36
12x = 36
x = 3
The value of x is 3. The length of sides are 9, 12, 15.
It is an right angled triangle because the sum of squares of two smaller sides is equal to the square of larger sides.
9² + 12² = 15²
81 + 144 = 225
Here, the length of hypotenuse is 15 cm.
The area of triangle is
1/2 x 9 x12
54 cm²

Q3: An isosceles right triangle has area 8 cm². The length of its hypotenuse is
(a) √32 cm
(b) √16 cm
(c) √48 cm    
(d) √24
Ans: a
Sol: area of a triangle = base x height/2
for a right angled isosceles triangle, base = height
so area = base²/2
8 x2 = base²
base = height = 4 cm
hypotnuese = √base² + height² = √2base²
= √2 x 8 x 2
= √32 cm

Q4: The cost of turfing a triangular field at the rate of Rs. 45 per 100 m² is Rs. 900. If the double the base of the triangle is 5 times its height, then its height is
(a) 42 cm
(b) 40 cm
(c) 32 cm
(d) 44 cm
Ans: b
Sol: Let the height of triangular field be h metres.
It is given that 2 x (base) = 5 × (Height)

Q5: The area of the the triangle having sides 1 m, 2 m and 2 m is :
(a) 
(b) 
(c) 
(d) 
Ans: a
Sol: Through Heron’s formula =
S=Perimeter/2
S=5/2
Let a=1m, b=2m ,c=2m
Formulae= Root(s(s-a)(s-b)(s-c))
=root(5/2(5/2 - 1)(5/2 - 2)(5/2 - 2))
=root(5/2*3/2*1/2*1/2)
=root(15/16)
=root(15)/4

Q6: The base of a right triangle is 8 cm and hypotenuse is 10 cm. Its area will be
(a) 24 cm²
(b) 40 cm²
(c) 80 cm²
(d) 48 cm²
Ans: a
Sol: 

• Given that the base of the right triangle is 8 cm and the hypotenuse is 10 cm.
• Using the Pythagorean theorem, we can find the height of the triangle:
• Height = sqrt(hypotenuse² - base²) = sqrt(10² - 8²) = sqrt(100 - 64) = sqrt(36) = 6 cm
• Now, we can calculate the area of the triangle using the formula: Area = 0.5 * base * height
• Area = 0.5 * 8 * 6 = 24 cm² 

Therefore, the area of the right triangle is 24 cm², which corresponds to option A.

Q7: The area of a triangle whose sides are 12 cm, 16 cm and 20 cm is
(a) 96 cm²
(b) 320 cm²
(c) 240 cm²
(d) 72 cm²
Ans: a
Sol: Area of a triangle of sides a , b & c is : √( s(s-a)(s-b)(s-c) )
where s=(a+b+c)/2
ln given triangle , s=(20+12+16)/2=48/2=24
Thus area of given triangle = √(24(24-20)(24-12)(24-16))
=√(24×4×12×8)
=√9216
=96
Thus area of triangle of sides 20cm ,12cm and 16cm is 96cm²

Q8: The difference of semi-perimeter and the sides of △ABC are 8, 7 and 5 cm respectively. Its semi-perimeter ‘s’ is
(a) 20 cm
(b) 5 cm
(c) 15 cm
(d) 10 cm
Ans: a
Sol: 

Q9: The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is
(a) 1322 cm²
(b) 1392 cm²
(c) 1344 cm²
(d) 1311 cm²
Ans: c

Q10: The sides of a triangular flower bed are 5 m, 8 m and 11 m. the area of the flower bed is
(a) √300 m²
(b) √330 m²
(c) 21√4 m²
(d) 4√21 m²
Ans: d

Q11: A triangle ABC in which AB = AC = 4 cm and ∠A = 90^o, has an area of
(a) 4 cm²
(b) 16 cm²
(c) 8 cm²
(d) 12 cm²
Ans: c

Q12: The perimeter and area of a triangle whose sides are of lengths 3 cm, 4 cm and 5 cm respectively are
(a) 12 cm, 12 cm²
(b) 12 cm, 6 cm²
(c) 6 cm, 12 cm²
(d) 6 cm, 6 cm²
Ans: b
Sol: Area of  triangle:

Perimeter of Triangle:
Perimeter of a closed figure is the sum of lengths of all of its sides.
Given, sides of a triangle are 3 cm, 4 cm and 5 cm.
Then, its perimeter = (3 + 4 + 5) cm
= 12 cm

Q13: The perimeter of a rhombus is 20 cm. One of its diagonals is 8 cm. Then area of the rhombus is
(a) 24 cm²
(b) 42 cm²
(c) 18 cm²
(d) 36 cm²
Ans: a
Sol: Given, Perimeter of a rhombus = 20 cm
Perimeter of a rhombus = 4*side
Hence, side = 20/4 = 5 cm.
Now, we know that the diagonals of a rhombus bisect each other at right angles (90 degree).
Hence 'a right angled triangle can be visualised with 'side' as the hypotenuse'.\
diagonal length = 8 cm
Half the length (since diagonal bisects each other) = 8/2 = 4 cm
(d/2)² + (d1/2)² = 5²
4² + (d1/2)² = 5²
(d1/2)² = 9
d1/2 = 3
d1 = 3*2 = 6 cm
Hence other diagonal = 6 cm.
Area = 1/2 * d1*d = 1/2 * 8 * 6 = 24 cm²

Q14: The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is
(a) √15 cm²
(b) 4√15 cm²
(c) 
(d) 2√15 cm²
Ans: a
Sol: 

Q15: The length of the sides of a triangle are 5 cm, 7 cm and 8 cm. Area of the triangle is :
(a) 10√3 cm²
(b) 100√3 cm²
(c) 300 cm²
(d) 50√3 cm²
Ans: a

Q16: If the side of an equilateral triangle is 4 cm, then its area is
(a) 8√3 m²
(b) 4√3 cm²
(c) 12√3 cm²
(d) 16√3 cm²
Ans: b

Q17: The area of a triangle whose sides are 15 cm, 8 cm and 19 cm is
(a) 6√91 cm²
(b) 19√91 cm²
(c) 8√91 cm²
(d) 4√91 cm²
Ans: a

Q18: If one side and one diagonal of a rhombus and 20 m and 24 m, then its area =
(a) 96 m²
(b) 192 m²
(c) 384 m²
(d) 284 m²
Ans: c
Sol: The rhombus has a perimeter of 80m.
A rhombus has 4 equal sides, each would be 20m.
One of the diagonals is 24m...this is bisected by the other diagonal into 12m segs.
The diagonals of a rhombus intersect at right angles. Use the Pythagorean Theorem to find the length of the other diagonal...
12² + b² = 20²
144 + b² = 400
b² = 256
b = 16
This is of course the length of just one segment, the diagonal is 32m.
Now we can use our area formula with the diagonals.
A = ½d¹d²
A = ½(24)(32)
A = 384 m²

Q19: The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm² is
(a) Rs. 2.00
(b) Rs. 2.16
(c) Rs. 2.48
(d) Rs. 3.00
Ans: b

Q20: Area of an equilateral triangle of side 2 cm is :
(a) 1 cm²
(b) √2 cm²
(c) √5 cm²
(d) √3 cm²
Ans: d

Q21: If the perimeter of an equilateral triangle is 24 m, then its area is
(a) 24√3 m²
(b) 16√3 m²
(c) 8√3 m²
(d) 20√3 m²
Ans: b
Sol: Given: Area of equilateral ∆ = 16√3 m²
Let the side of equilateral ∆ = a m
Area of equilateral ∆ = (√3/4)a²
16√3 = (√3/4)a²
16√3 × (4/√3) = a²
64 = a²
a = √64
a =√ 8 × 8 = 8 m
Side of equilateral ∆ = 8 m
Perimeter of equilateral ∆ = 3 × side
Perimeter of equilateral ∆ = 3 × 8 = 24 m.
Hence, the perimeter of equilateral ∆ is 24 m.

Q22: Each of the equal sides of an isosceles triangle is 2 cm greater than its height. If the base of the triangle is 12 cm, then its area is
(a) 40 cm²
(b) 24 cm²
(c) 48 cm²
(d) 36 cm²
Ans: c
Sol: 

and, equating this with the basic area formula,

i.e. A = 1/2.h.b

Q23: The product of difference of semi-perimeter and respective sides of △ABC are given as 13200 m³. The area of △ABC, if its semi-perimeter is 132m, is given by
(a) 132 m²
(b) 13200 m²
(c) 1320 m²
(d) 20√33 m²
Ans: c

Q24: The area of equilateral triangle of side ‘a’ is 4√3 cm². Its height is given by
(a) 
(b) 2√3 cm
(c) 
(d) 
Ans: b

Q25: The base of a right triangle is 8 cm and hypotenuse is 10 cm. Its area will be :
(a) 40 cm²
(b) 48 cm²
(c) 24 cm²
(d) 80 cm²
Ans: c
Sol: Given,
base= 8cm
hypotenuse= 10cm  
First, let us find the height of the triangle:  
Let height = h  
By applying pythagorean theorem  
h² + 8² = 10²
h² + 64 = 100
h² = 100 - 64
h² = 36
h = 6cm  
Now let us find the area,  
Area = 1/2 x b x h    
= 1/2 x 8 x 6  
= 1/2 x 48  
= 24cm²
Therefore, area of the right triangle = 24cm²