We all know that a triangle is a polygon, which has three sides. The area of a triangle is a measurement of the area covered by the triangle. We can express the area of a triangle in the square units. The area of a triangle is determined by two formulas i.e. the base multiplies by the height of a triangle divided by 2 and second is Heron’s formula. Let us discuss the Area of a Triangle formula in detail.

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**Area of Triangle a Formula**

**What is an Area of a triangle?**

The area of a polygon is the number of square units covered by the polygon. The area of a triangle is determined by multiplying the base of the triangle and the height of the triangle and then divides it by 2. The division by 2 is done because the triangle is a part of a parallelogram that can be divided into 2 triangles.

Area of a parallelogram = B × H

Where,

B | the base of the parallelogram |

H | the height of the parallelogram |

As triangle is the one-half of the parallelogram, so the area of a triangle is:

A= \( \frac{1}{2} \times b \times h \)

Where,

B | the base of the triangle |

H | the height of the triangle |

**Heron’s Formula for Area of a Triangle**

Herons formula is a method for calculating the area of a triangle when the lengths of all three sides of the triangle are given.

Let a, b, c are the lengths of the sides of a triangle.

The area of the triangle is:

Area=\( \sqrt{s(s−a)(s−b)(s−c)}\)

Where, s is half the perimeter,

s= \( \frac{a+ b+ c}{2} \)

We can also determine the area of a triangle by the following methods:

- In this method two Sides, one included Angle is given

Area= \( \frac{1}{2} \times a \times b \times \sin c \)

Where a, b, c are the lengths of the sides of a triangle

- In this method we find area of an Equilateral Triangle

Area= \( \frac{\sqrt{3} \times a^{2}}{4} \)

- In this method we find area of a triangle on a coordinate plane by Matrices

$$\frac{1}{2} \times \begin{bmatrix}

x_1 & y_1 & 1 \\

x_2 & y_2 & 1 \\

x_3 & y_3 & 1

\end{bmatrix}$$

Where, *(x*_{1}*, y*_{1}*), (x _{2}, y_{2}), (x_{3}, y_{3})* are the coordinates of the three vertices

- In this method, we find area of a triangle in which two vectors from one vertex is there.

Area of triangle = \( \frac{1}{2} \left( \overrightarrow{u}\times \overrightarrow{v}\right) \)

**Solved Examples**

**Q.1: **The sides of a right triangle ABC are 5 cm, 12 cm, and 13 cm.

**Solution: **In \(\bigtriangleup \)ABC in which base= 12 cm and height= 5 cm

Area of \( \bigtriangleup ABC = \frac{1}{2} \times B \times H \)

A = \( \frac{1}{2} \times 12 \times 5 \)

A = 30 cm^{2}

**Q.2:** Find the area of a triangle, which has two sides 12 cm and 11 cm and the perimeter is 36 cm.

** Solution: **Here we have perimeter of the triangle = 36 cm, a = 12 cm and b = 11 cm.

Third side c = 36 cm – (12 + 11) cm = 13 cm

So, 2s = 36, i.e., s = 18 cm,

s – a = (18 – 12) cm = 6 cm,

s – b = (18 – 11) cm = 7 cm,

and, s – c = (18 – 13) cm = 5 cm.

Area of the triangle = \( \sqrt{s(s−a)(s−b)(s−c)} \)

A= \(\sqrt{18\times 6\times 7\times 5} \)

A= \( 6\sqrt{105}\) cm^{2}

I get a different answer for first example.

I got Q1 as 20.5

median 23 and

Q3 26