Heron's formula - Class 9 Maths
Heron's formula or Hero's formula, named after Hero of Alexandria, gives the area of a triangle when the length of all three sides are known.
Some important points that needs to be known
- Perimeter of triangle whose sides are a,b,c is a+b+c
- Semi-perimeter of triangle whose sides are a,b,c is (a+b+c)/2
- Pythogora's theorem is H 2=P 2+B 2, here H is hypotenuse, P is perpendicular and B is base
- Area of right angled triangle is 1/2*base*altitude
- Equilateral triangle has all the three sides equal
- Isosceles triangle has two sides equal
According to Heron's formula if we have the three sides of triangle then we can find the area by the below formula :
Area of triangle = √ s(s-a)(s-b)(s-c), here s is the semi-perimeter
By Heron's method finding area of equilateral triangle Three sides of equilateral triangle are equal Now s = (a+a+a)/2 = 3a/2 Putting the value of s in √ s(s-a)(s-a)(s-a) we will get √ 3a/2 (3a/2-a)(3a/2-a)(3a/2-a) = √ 3a/2 * a/2 * a/2 * a/2 = √ 3 /4 a 2
There are other ways by which we can find the area of triangle if we have the three sides
- Area = √ 4a 2b 2-(a 2+b 2-c 2)/4
- Area = √ 4b 2c 2-(b 2+c 2-a 2)/4
- Area = √ 4a 2c 2-(a 2+c 2-b 2)/4
Trick to remember the above 3 formulas if we take the first two sides then in bracket the third side will be subtracted
We will do a few problems for understanding -
Exercise
Q1.Find the area of triangle, the lengths of whose sides are 18cm, 24cm and 30cm
Solution -
Given lengths are 18cm, 24cm and 30cm
s = (18 + 24 + 30)cm/2
= 36cm
Area of triangle = √ s(s-a)(s-b)(s-c)
Putting the values in the above formula we get
Area of triangle = √ 36(36-18)(36-24)(36-30)
= √ 36 * 18 * 12 * 6
= 216 cm 2
Q2.Find the are of triangle whose sides are 120cm, 150cm and 200cm.
Solution -
Given lengths are 120cm, 150cm and 200cm
s = (120 + 150 + 200)cm/2
= 235cm
Area of triangle = √ s(s-a)(s-b)(s-c)
Putting the values in the above formula we get
Area of triangle = √ 235(235-120)(235-150)(235-200)
= √ 235 * 85 * 115 * 35
= 8966.57 cm 2 [Find the square root using long division method]
Q3.The base of a right-angled triangle measures 48cm and its hypotenuse measures 50cm. Find the area of the triangle.
Solution -
Given base is 48 cm and hypotenuse is 50
Area of right angled triangle is 1/2*base*altitude
But we do not know the altitude(perpendicular/height), to find the altitude we will use Pythogora's theorem - H 2=P 2+B 2
P 2 = H 2 - B 2
Putting the values we get - P = √ 50 2 - 48 2
Trick - As 48 2 may take some time we can use the formula a 2-b 2 = (a+b)(a-b)
Now P = √ (98)(2)
or P = √ 7 2*2 2
or P = 14cm
Now putting the values in the Area formula for right angled triangle we have
Area = 1/2 * 48 * 14
= 336 cm 2
Q4.Find the area of triangle whose sides are 91cm, 96 cm and 105 cm in length. Find the height corresponding to this greater side.
Solution -
Given lengths are 91cm, 96cm, 105cm
s = (91 + 96 + 105)/2 = 146
Area of triangle = √ s(s-a)(s-b)(s-c)
Putting the values in the above formula we get
Area of triangle = √ 146(146-91)(146-96)(146-105)
= √ 146 * 55 * 50 * 41
= √ 2 * 73 * 5 * 11 * 2 * 5 * 5 * 41
= 10 √ 164615
= 10 * 405.728 [Find the square root using long division method]
= 4057.28 cm 2
Now, height corresponding to the greater side can be found if consider that in a right angled triangle resting on the hypotenuse, the base is the longest side with the altitude on the base being short
Area of right angled triangle is 1/2*base*altitude
Putting the values we get 4057.28 = 1/2*105*H or H = 77.28 cm
Q5. Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.
Solution -
For an isosceles triangle we know two sides are equal.
Given lengths are 13cm, 13cm and 20cm
s = (13 + 13 + 20)cm/2
= 23cm
Area of triangle = √ s(s-a)(s-b)(s-c)
Putting the values in the above formula we get
Area of triangle = √ 23(23-13)(23-13)(23-20)
= √ 23 * 10 * 10 * 3
= 10 √ 23*3 [Find the square root using long division method]
= 83.07 cm 2
Q6.An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution -
For an isosceles triangle we know two sides are equal so we have two sides as 12cm, 12cm
Given perimeter = 30cm
Perimeter of a triangle = a + b + c (where a,b and c are the sides)
Let the third side be x, now to find x we put the values as
30 = 12 + 12 + x
or x = 6
As perimeter = 30 cm so semi-perimeter(s) = 30/2 cm = 15 cm
Area of triangle = √ s(s-a)(s-b)(s-c)
Putting the values in the above formula we get
Area of triangle = √ 15(15-12)(15-12)(15-6)
= √ 15 * 3 * 3 * 9
= 9 √ 15 [Find the square root using long division method]
= 34.83 cm2
Q7.The base of an isosceles triangle measures 80cm and its area is 360sq.cm. Find the perimeter of the triangle (in cm).
Q8 (i).If the area of an equilateral triangle is 36 √ 3 cm 2, find its perimeter.
Solution - Area of equilateral triangle is √ 3/4 a 2 Putting the values we get 36 √ 3 cm 2 = √ 3/4 a 2 or a 2 = 36 * 4 or a = 12 cm Perimeter of a triangle is a + b + c, so the perimeter of the equilateral triangle is (12 + 12 + 12)cm = 36 cm
Q8 (ii). If the area of an equilateral triangle is 8 √ 3 cm 2 find its height
Solution - Area of equilateral triangle is √ 3/4 a 2 Putting the values we get 8 √ 3 cm 2 = √ 3/4 a 2 or a 2 = 32 or a = √ 32 or a = 4 √ 2 As, the altitude(height) bisects the opposite side in perpendicular Now, area of right angled triangle is 1/2 * base * height Substituting the values we get 8 √ 3 = 1/2 * 8 * height or height = 2 √ 6 cm
Q9.The sides of a triangle are in the ratio 3:5:7 and its perimeter is 300m. find its area
Solution - Let the coefficient of the ratios be x Now, perimeter of a triangle is a + b + c Putting the values we get 300 = 3x + 5x + 7x or x = 300/15 = 20 Now the sides are 3*20 cm, 5*20 cm, 7*20 cm or the sides are 60 cm, 100 cm, 140 cm As perimeter = 300 cm, then semi perimeter will be 300cm/2 = 150 cm By Heron's formula the area of triangle is √ s(s-a)(s-b)(s-c) Putting the values we get √ 150(150-60)(150-100)(150-140) or √ 150*90*50*10 or √ (10*3*5)*(10*9)*(10*5)*(10) or √ 104 * 32 * 52 *3 or 102 * 3 * 5 √ 3 or 1500 √ 3 cm2
Q10.The height of an equilateral triangle measures 9cm. Find its area, take √ 3 = 1.732
Solution -Height of an equilateral triangle = √ 3 a/2 Given height = 9, so a = 18/√ 3 Now, area of equilateral triangle is √ 3 /4 a 2 Putting the value of a we get Area = √ 3 / 4 * (18/√ 3) 2 or Area = 3 √ 3 * 9 = 3 * 1.732 * 9 = 46.764 cm 2
Quadrilaterals
Quadrilateral – A quadrilateral can be defined as a closed two dimensional figure which has four sides.
Diagonal – A diagonal is a straight line which joins the two opposite corners of a quadrilateral.
Types of Quadrilaterals
| Quadrilateral Type | Area | Perimeter |
|---|---|---|
| Rectangle | l*b [l-length, b-breadth] | 2(l+b) [l-length, b-breadth] |
| Square | s2 [s-Side] | 4*s [s-Side] |
| Rhombus | 1/2*d1*d2 [d-Diagonal] | 4*s [s-Side] |
| Parallelogram | b*h [b-Base, h-Height] | 2(l+b) [l-length, b-breadth] |
| Trapezoid | (b1*b2)/2*h [b1-Base1,b2-Base2, h-Height ] | s1+s2+s3+s4 [s-Side] |
