how to find area of triangle

The most common way to find the area of a triangle is to use base and height: take half of the base multiplied by the corresponding height.

Quick Scoop

1. Basic school formula

If you know a triangle’s base and its height (the perpendicular distance to that base), use:

Area=12×base×height\text{Area}=\tfrac12 \times \text{base}\times \text{height}Area=21​×base×height

• In symbols: Area=12bh\text{Area}=\tfrac12 bhArea=21​bh.

• Example: base =8=8=8 cm, height =5=5=5 cm
  • Area =12×8×5=20textcm2=\tfrac12 \times 8\times 5=20\\text{cm}^2=21​×8×5=20textcm2.

This works for any triangle type as long as the “height” is perpendicular to the chosen base.

2. When you know three sides (Heron’s formula)

If you only know the three sides a,b,ca,b,ca,b,c, you can still find the area using Heron’s formula.

1. Compute the semi‑perimeter sss:

s=a+b+c2s=\tfrac{a+b+c}{2}s=2a+b+c​

2. Then use:

Area=s(s−a)(s−b)(s−c)\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}Area=s(s−a)(s−b)(s−c)​

Example: sides 333, 444, 555 cm.

• s=3+4+52=6s=\tfrac{3+4+5}{2}=6s=23+4+5​=6.
• Area =6(6−3)(6−4)(6−5)=6⋅3⋅2⋅1=36=6textcm2=\sqrt{6(6-3)(6-4)(6-5)}=\sqrt{6\cdot 3\cdot 2\cdot 1}=\sqrt{36}=6\\text{cm}^2=6(6−3)(6−4)(6−5)​=6⋅3⋅2⋅1​=36​=6textcm2.

3. When you know two sides and the included angle

If you know two sides and the angle between them (say sides a,ba,ba,b and included angle θ\theta θ), use a trigonometric version:

Area=12absin⁡(θ)\text{Area}=\tfrac12 ab\sin(\theta)Area=21​absin(θ)

• This is handy when you’re given side–side–angle situations in trigonometry and no height.

• Example: sides 777 cm and 101010 cm with included angle 30∘30^\circ 30∘:
  • Area =12×7×10×sin⁡(30∘)=35×0.5=17.5textcm2=\tfrac12 \times 7\times 10\times \sin(30^\circ)=35\times 0.5=17.5\\text{cm}^2=21​×7×10×sin(30∘)=35×0.5=17.5textcm2.

4. Special triangle shortcuts (optional but useful)

Some special triangles have neat direct formulas.

• Equilateral triangle (all sides =a=a=a):

Area=34a2\text{Area}=\tfrac{\sqrt{3}}{4}a^2Area=43​​a2

• Right‑angled triangle (legs as base and height):

Area=12×leg1×leg2\text{Area}=\tfrac12 \times \text{leg}_1\times \text{leg}_2Area=21​×leg1​×leg2​

These are just special cases of the same ideas above: base–height or sine formulas.

5. Tiny story to remember it

Imagine cutting a rectangle in half along a diagonal: each half is a triangle.

• A rectangle’s area is base×height\text{base}\times \text{height}base×height.
• Each triangle is exactly half of that rectangle, so its area must be 12×base×height\tfrac12 \times \text{base}\times \text{height}21​×base×height.

That picture is essentially the reason why the base–height formula for triangle area works.

TL;DR:

• Most of the time: Area=12×base×height\text{Area}=\tfrac12 \times \text{base}\times \text{height}Area=21​×base×height.

• With three sides: Heron’s formula Area=s(s−a)(s−b)(s−c)\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}Area=s(s−a)(s−b)(s−c)​.

• With two sides and angle between: Area=12absin⁡(θ)\text{Area}=\tfrac12 ab\sin(\theta)Area=21​absin(θ).

Information gathered from public forums or data available on the internet and portrayed here.