How do you find the area of a triangle?

Hint: We will apply different methods to find the area of a triangle depending up on what is the data given in the question about the triangle. If the one side and the height to the side is given in the question then we can write the area of the triangle as $\dfrac{1}{2}\times b\times h$ where b is base and h is height of triangle. If 2 sides and angle between those 2 sides are given then then the area of triangle is equal to $\dfrac{1}{2}ab\sin \theta $ where a and b are given 2 sides and $\theta $ is the angle between them

Complete answer:
Let’s assume base and height of triangle are given, we have to find the area of triangle
We will find the area by drawing a figure.


Let’s assume in the above figure base BC and height AD are given
We can write area of triangle ABC= area of triangle ABD + area of triangle ADC
We can see from the above figure that area of ABD is half of the rectangle AEBD, and area of triangle ADC is half of the area of ADCF
So we can write area of triangle ABD + area of triangle ADC is equal to $\dfrac{1}{2}$ (area of AEBD + area of ADCF)
area of AEBD + area of ADCF= area of EFCB
So area of triangle ABC= $\dfrac{1}{2}$ (area of rectangle EFCB)
Area of EFCB = length $\times $ breadth
We can see length of EFCB =BC which is base of ABC
Breadth of EFCB= EB =AD which is height of ABC
So area of triangle = $\dfrac{1}{2}$ $\times $ BC $\times $ AD
We can say the area of the triangle is $\dfrac{1}{2}\times b\times h$ where b is base and h is height of triangle.

Note: We can see that height AD in triangle ABC is equal to $AB\sin B$ so area of triangle $\dfrac{1}{2}\times AB\times BC\times \sin B$ so we can say the area of triangle is equal to $\dfrac{1}{2}ab\sin \theta $ where a and b are 2 sides and $\theta $ is the angle between them. Another formula for area of triangle is $\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ where a, b and c are sides of triangle and s is half of perimeter of triangle.