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Heron of Alexandria was a brilliant mathematician who was the first to discover the formula for calculating the area of a triangle using all three sides' lengths. Heron's Formula was also used by Heron of Alexandria to calculate the area of quadrilaterals and high-order polygons.
This formula is commonly used in trigonometry to prove the laws of cosines and cotangents, among other things. Its area, on the other hand, does not require angle measurement. For example, if we have a triangle XYZ and the sides are x, y, and z respectively, then the area of the triangle will be as follows-
Area=√s(s−x)(s−y)(s−z)
Area=√s(s−x)(s−y)(s−z)
Heres refers to the semi-perimeter of the triangle i.e. s = (x + y + z)/2.
Ques 1. The area of a triangle is equal to:
- Base x Height
- 2(Base x Height)
- ½(Base x Height)
- ½ (Base + Height)
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Ans. ½(Base x Height)
Explanation: As shown in the diagram below, the area of the triangle is given by 1/2 * base * height where base = BC and height is AD.
Ques 2. If the perimeter of an equilateral triangle is 180 cm. Then its area will be:
- 900 cm2
- 900√3 cm2
- 300√3 cm2
- 600√3 cm2
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Ans. 900√3 cm2
Explanation:
Given, Perimeter = 180 cm
3a = 180 (Equilateral triangle)
a = 60 cm
Semi-perimeter = 180/2 = 90cm
Now as per Heron’s formula,
A=√s(s−a)(s−b)(s−c)
Hence, if we put the values here, we get:
A = 900√3
Read More: Area of a Triangle: Formula, Types and Solved Examples
Ques 3. The sides of a triangle are 122 m, 22 m, and 120 m respectively. The area of the triangle is:
- 1320 sq.m
- 1300 sq.m
- 1400 sq.m
- 1420 sq.m
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Ans. 1320 sq.m.
Explanation: Given,
a = 122 m
b = 22 m
c = 120 m
Semi-perimeter, s = (122+22+120)/2 = 132 m
Using heron’s formula:
A=√s(s−a)(s−b)(s−c)
Put the values of s, a, b and c, to get the answer equal to 1320 sq.m.
Ques 4. The area of a triangle with given two sides 18cm and 10cm respectively and perimeter equal to 42 cm is:
- 20√11 cm2
- 19√11 cm2
- 22√11 cm2
- 21√11 cm2
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Ans. 21√11 cm2
Explanation: Perimeter = 42
a+b+c=42
18+10+c=42
c=42-28=14 cm
Semiperimeter, s = 42/2 = 21cm
Using Heron’s formula:
A=√s(s−a)(s−b)(s−c)
Put the values of s, a, b and c, to get the answer equal to 21√11 cm2
Read More: NCERT Solutions for Class 9 Mathematics Chapter 12: Heron's Formula
Ques 5. The sides of a triangle are in the ratio 12: 17: 25 and its perimeter is 540cm. The area is:
- 1000 sq.cm.
- 5000 sq.cm.
- 9000 sq.cm.
- 8000 sq.cm.
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Ans. 9000 sq.cm.
Explanation: The ratio of the sides are 12:17:25
Perimeter = 540 cm
Let the sides of the triangle be 12x, 17x and 25x.
Hence,
12x+17x+25x = 540 cm
54x = 540 cm
x = 10
Therefore,
a = 12x=12 x 10 = 120
b = 17x = 17 x 10 = 170
c = 25x = 25 x 10 = 250
Semi-perimeter, s = 540/2 = 270 cm
Putting the values of s, a, b and c in Heron’s formula, we will get the area equal to 9000 sq. cm.
Ques 6. The equal sides of the isosceles triangle are 12 cm and the perimeter is 30 cm. The area of this triangle is:
- 9√15 sq.cm
- 6√15 sq.cm
- 3√15 sq.cm
- √15 sq.cm.
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Ans. 9√15 sq.cm
Explanation:
Given, Perimeter = 30cm
Semiperimeter, s = 30/2 = 15cm
a = b = 12cm
c=?
a+b+c = 30
12 +12+c=30
c=30-24 = 6cm
By putting the values of s, a, b and c in Heron’s formula, we can get the value of area.
Read More: Scalene Triangle: Definition, Area, Perimeter and Examples
Ques 7. A quadrilateral whose sides are 3cm, 4cm, 4cm, 5cm, and one of the diagonal is equal to 5cm as per the below figure. The area of the quadrilateral is:
- 19.17 sq.cm
- 15.17 sq.cm
- 20.17 sq.cm
- 22.17 sq.cm.
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Ans. 15.17 sq.cm
Explanation: Using Pythagoras theorem, in ΔABC,
AC2 = AB2 + BC2
⇒ 52 = 32 + 42
⇒ 25 = 25
Hence, ABC is a right triangle.
Area of ΔABC = ½ x 3 x 4 = 6 sq.cm
Semi-perimeter of ΔACD = (5+5+4)/2 = 14/2 = 7cm
The area of ΔACD can be determined by using Heron’s formula.
Therefore, the area of quad.ABCD = Area of ΔABC + Area of ΔACD
Ques 8. The area of an equilateral triangle having a side length equal to √3/4cm is:
- 2/27 sq.cm
- 2/15 sq.cm
- 3/16 sq.cm
- 3/14 sq.cm
Clcik Here for Answer
Ans. 3/16 sq. cm
Explanation: Here, a = b = c = √3/4
Find the semi-perimeter of the triangle and use Heron’s formula to find the answer.
Ques 9. The sides of a parallelogram are 100 m each and the length of the longest diagonal is 160m. The area of a parallelogram is:
- 9600 sq.m
- 9000 sq.m
- 9200 sq.m
- 8800 sq.m
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Ans. 9600 sq. m
Explanation: The diagonal divides the parallelogram into two equivalent triangles. Hence, its area will be equal to the sum of the area of the two triangles.
Hence, we can determine the area of the two triangles using Heron’s formula.
Ques 10. The sides of a triangle are in the ratio of 3: 5: 7 and its perimeter is 300 cm. Its area will be:
- 1000√3 sq.cm
- 1500√3 sq.cm
- 1700√3 sq.cm
- 1900√3 sq.cm
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Ans. 1500√3 sq.cm
Explanation: Perimeter = 300
Semiperimeter = 300/2=150
Sides of the triangle are in ratio 3: 5: 7
3x+5x+7x=300
15x=300
x=20
Sides values are
3 x 20= 60
5 x 20= 100
7 x 20= 140
Area of triangle = √s(s−a)(s−b)(s−c)
√150(150-60)(150-100)(150-140)
After solving this you will get 1500√3 sq.cm
Read More: Binary Operations: Definition, Characteristics and Examples
Ques 11. In heron’s formula √s(s−a)∗(s−b)∗(s−c), what is the value of s if a, b, and c are sides of the triangle?
- a+b+c/4
- a+b+c
- a+b+c/2
- 2a+2b+2c
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Ans. a+b+c/2
Explanation: In heron’s formula √s(s−a)∗(s−b)∗(s−c), s is the half perimeter of the triangle.
The perimeter of the triangle having sides a, b and c is a + b + c.
Hence, s = half perimeter of the triangle = a+b+c/2.
Ques 12. What is the area of the triangle having sides equal to 10 cm, 16 cm, and 24 cm?
- 25√15cm2
- 35√17cm2
- 15√13cm2
- 15√15cm2
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Ans. 15√15cm2
Explanation: a = 10, b = 16 and c = 24
s = a+b+c/2=10+16+242=502 = 25
According to heron’s formula, the area of the triangle = √s∗(s−a)∗(s−b)∗(s−c)
= √25∗(25−10)∗(25−16)∗(25−24)
= √25∗15∗9∗1
= 15√15cm2
Read More: Derivative of Inverse Trigonometric Functions
Ques 13. The sides of a triangle are in the proportion of 2 : 3: 5 and its perimeter is 200 cm. The area of this triangle is __________ cm2.
- 375√23
- 375√21
- 345√23
- 345√21
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Ans. 375√23
Explanation: Let the side of a triangle be a = 2x, b = 3x and c = 5x.
The perimeter of the triangle is 200 cm.
Therefore a + b + c = 2x + 3x + 5x = 200
10x = 200
Hence, x = 20
a = 2x = 2(20) = 40 cm
b = 3x = 3(30) = 90 cm
c = 5x = 5(20) = 100 cm
Now, s = a+b+c2=40+90+1002=2302 = 115
According to heron’s formula, the area of the triangle = √s∗(s−a)∗(s−b)∗(s−c) = √115∗(115−40)∗(115−90)∗(115−100)
= √115∗75∗25∗15
= 375√23
Read More: Pythagoras Theorem
Ques 14. A triangular garden has sides 90m, 140m, and 80m. A fence is to be put all around the garden. What will be the total cost of fencing at the rate of Rs 15 per meter? A 5m wide space is to be left on one side for gate opening.
- Rs 4525
- RS 4975
- Rs 4575
- Rs 4230
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Ans. Rs 4575
Explanation: 90 + 140 + 80 = 310m
Considering 5m space for gate opening, length of fencing = 310 – 5 = 305m.
Cost of 1m fencing = Rs 15,
Then the cost for 305m fencing = 305 * 15
= Rs 4575.
Read More: Triangles
Ques 15. A triangular board having sides 45m, 30m, and 35m is used for advertising. One company uses this board for its advertisement for 4 months. How much rent will the company have to pay if the rent is Rs 3500 per m2?
- Rs 611800
- Rs 1835400
- Rs 611900
- Rs 1835500
Click Here for Answer
Ans. Rs 611800
Explanation: a = 45, b = 30 and c = 35
We know that s = a+b+c2=45+30+352=1102 = 55
According to heron’s formula, area of the triangle = √s∗(s−a)∗(s−b)∗(s−c)
= √55∗(55−45)∗(55−30)∗(55−35)
= √55∗10∗25∗20
= 524.40 m2
Now the rent for 1 m2 = Rs 3500,
Then rent for 992.15 m2 = 524.40 * 3500 = Rs 18,35,400
The rent which we got above is for one year.
The company used the board for 4 months.
Therefore rent for 4 months = Rs 18,35,400∗4/12
= Rs 6,11,800.
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