Most of us are already acquainted with the term area. It is defined as the region occupied inside the boundary of a flat object or figure. The measurement is done in square units with the standard unit being square meters (m^2). For the computation of area, there are predefined formulas for squares, rectangles, circle, triangles, etc. In this article, we would discuss the area of a triangle. For this, we would use the formula for the area of triangle on different types of triangles and see if we are able to obtain their area.
Formula for Triangle
Area of Triangle is given by the formula mentioned below:
Area of Triangle = \(A=\frac{1}{2} (bh)\)
Rightangled Triangle
When we consider a rightangled triangle, its area could be computed directly. Two of its sides that we consider is the base and the height. So, the formula for area is calculated as:
Area of a triangle = \(\frac{1}{2}\) x base x height = \(\frac{1}{2}\) x 30 x 40 = 600 \(cm^{2}\)
Example 1:  Find the area of an acute triangle with a base of 15 inches and a height of 4 inches.  
Solution:  \(A=\frac{1}{2}.b.h\)
\(A=\frac{1}{2}.(15in).(4in)\) 

\(A=\frac{1}{2} .(60\;in^{2})\) A = 30 \(in^{2}\) 

Equilateral Triangle
For an equilateral triangle, the height of the triangle is to be found before computing its area. For doing so, we drop a perpendicular from one of its vertices (we know that height of an equilateral triangle is same as the perpendicular bisector of one of its sides). In the ∆ABC, CD represents the height.
For calculating CD, we consider the rightangled ∆ACD. We have,
\(AC^{2}=AD^{2}+CD^{2}\)Or \(CD^{2}=AC^{2}AD^{2}\)
\(CD^{2}\) = 82 – 42 = 64 – 16 = 48
CD = \(\sqrt{48}\) = \(4\sqrt{3}\) cm
Thus, area of ∆ABC= \(\frac{1}{2}\) x base x height = \(\frac{1}{2}\) x 8 x \(4\sqrt{3}\) = \(4\sqrt{3}\) \(cm^{2}\)
Isosceles Triangle
Similar to the equilateral triangle, we find the height of the triangle first to compute its area. Note that the height of the ABC could be obtained by drawing the perpendicular AD from A which bisects the base AC at right angles. Now, in the rightangled ∆ABD, we find AD by,
\(AD^{2}=AB^{2}BD^{2}\) = 102 – 82 = 100 – 64= 36
AD = \(\sqrt{36}\) = 6 cm
Thus, area of ∆ABC = \(\frac{1}{2}\) x base x height = ½ x 16 x 6 = 48 \(cm^{2}\)
We have seen that the area of special triangles could be obtained using the formula. However, for a triangle with the sides being given usually, calculation of height would not be simple. For the same reason, we rely on the Heron’s Formula.
More Examples
Q1) Find the area of the acute triangle with a base of 13 inches and height of 5 inches.
Sol.) \(A=\frac{1}{2}.b.h\)
\(A=\frac{1}{2}.(13\;in).(5\;in)\) \(A=\frac{1}{2}.(65\;in^{2})\)A = \(32.5 in^{2}\)
Q2) Find the area of the right angled triangle with a base of 7cm and a height of 8cm.
Sol.) \(A=\frac{1}{2}.b.h\)
\(A=\frac{1}{2}.(7\;cm).(8\;cm)\) \(A=\frac{1}{2}.(56\;cm)\)A = \(28 cm^{2}\)
Q3) Find the area of obtuse angled triangle with a base of 4cm and a height 7cm.
Sol.) \(A=\frac{1}{2}.b.h\)
\(A=\frac{1}{2}.(4\;cm).(7\;cm)\) \(A=\frac{1}{2}.(28\;cm)\)A = \(14 cm^{2}\)
Q4) The area of triangular shaped field is 24 square feet and its height is 6 feet. Find the base.
Sol.) \(A=\frac{1}{2}.b.h\)
\(24\;ft^{2}=\frac{1}{2} \cdot b\cdot 6\;ft\)On multiplying both sides of the equation by 2, we get:
\(48\;ft^{2}=b\cdot 6\;ft\)Dividing both sides of the equation by 6 ft, we get:
4 ft = b
Commuting this equation, we get:
b = 4 ft
Therefore, Base = 4 feet
To learn more about the computation of the area of the triangle using Heron’s Formula, visit Byju’s.