Heron’s formula grade 9- NCERT-01

introduction

We all ready know about how to find out the area, perimeter of different shapes like rectangle and square. In this chapter we will study details about how we can find the area of various type of triangle.

area of the triangle

Area of the triangle = ½ * base * height

We can see that if the triangle is right angle triangle, we can directly take two side, which contain the right angle, as base and height. see the the fig.12.1 in p-197. So in this triangle ABC, BC and CA are the height and base ( any one can select as base or height).  There for the

Area of the triangle ABC = ½ * 5 * 12  =30 cm².

Now we can consider an equilateral triangle PQR as shown in fig 12.2 p-198. In this we can know all the side of the triangle. But how we can find the height. We can find the midpoint of QR and mark as M and join it to P. Now we can see that ΔPMQ is aright angle triangle. And we can find the PQ using the pythagorean theorem.

PQ ² = PM² + QM²

10²=PM²+5²

PM²= 75

PM= √75=5√3

Area of the triangle = ½ * 10 * 5√3 =25√3 cm².

Now we can consider an isosceles triangle. In this two sides is same. How we can find the height. We can consider the triangle XYZ in the fig 12.3 p-198. Find and mark the midpoint P on YZ and joint it with point x. Now we can see the triangle XYP ia right angle triangle.  So

XP² = XY² – YP² = 5² -4² = 25-16=9

XP= √9 = 3

Area of the triangle = ½ * 8 * 3 =12 cm².

Now we can consider the scalene triangle. In this type of triangle all sides are different. So we cannot find the height of the triangle as described in the above method. Her we can use the Heron’s formula. Heron is the famous mathematician (p-199). The formula given by him can use to find the area of any type of triangle. And it is called hero’s  formula

Area of a triangle = √(s(s-a)(s-b)(s-c))

Where a,b,c are the three side of the triangle. ANd s is the semi perimeter of the triangle. So  s= (a+b+c)/2.

Now consider the triangle ABC in the fig 12.5 p-200. Here a= 40, b=24,c=32. so

s= (40+24+32)/2=  48

s-a= 48-40=8

s-b=48-24=24

s-c=48-32=16

Area of the triangle = √(48*8*24*16) =384 m² see page 200,201,203 for exercise

we can also use this formula for finding the area of quadrilateral by dividing it in to two triangle, see page 206 for exercises.

 

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