• For any triangle with base, \(b\) and height, \(h\):

\(\text{Area}=\dfrac{1}{2}\times\text{base}\times \text{height}=\dfrac{1}{2}\times b\times h\)

• The base can be any one of the sides of the triangle, and the height is the perpendicular distance form the opposite vertex to that base.

• Based on the triangle in the figure above, formula for area of the triangle given by:

\(\begin{aligned} \text{Area}&=\dfrac{1}{2}ab\sin{C} \\\\ &=\dfrac{1}{2}ac\sin{B} \\\\ &=\dfrac{1}{2}bc\sin{A} \end{aligned}\)

• This formula is useful when two sides and the included angle are known.

• For a triangle with sides \(a\), \(b\), and \(c\), and semi-perimeter, \(s=\dfrac{a+b+c}{2}\):

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

• Useful when all three sides of the triangle are known.

Based on the triangle \(ABC\) above, find the area of the triangle.

Based on the question above, given:

\(AB=13\) cm,
\(BC=16\) cm,
\(\angle{ABC}=36^\circ\).

Using the trigonometry formula:

​\(\begin{aligned} \text{Area of }\Delta ABC&=\dfrac{1}{2}ac\sin{B} \\\\ &=\dfrac{1}{2}(16)(13)\sin{36^\circ} \\\\ &=61.13\text{ cm}^2. \end{aligned}\)​

The diagram above shows a triangle \(ABC\).

By using Heron's formula, calculate the area of the triangle.

Based on the question, given:

\(a=1.8\) cm,
\(b=3\) cm,
\(c=3.8\) cm.

Calculate the semi-perimeter, \(s\):

\(\begin{aligned} s&=\dfrac{a+b+c}{2} \\\\ &=\dfrac{1.8+3+3.8}{2} \\\\ &=4.3. \end{aligned}\)

Substitute the value of \(a\), \(b\), \(c\), and \(s\) into the Heron's formula:

\(\begin{aligned} \text{Area}&=\sqrt{s(s-a)(s-b)(s-c)} \\ &=\sqrt{4.3(4.3-1.8)(4.3-3)(4.3-3.8)} \\ &=\sqrt{4.3(2.5)(1.3)(0.5)}\\ &=\sqrt{6.9875}\\ &=2.6434\text{ cm}^2. \end{aligned}\)

• For any triangle with base, \(b\) and height, \(h\):

\(\text{Area}=\dfrac{1}{2}\times\text{base}\times \text{height}=\dfrac{1}{2}\times b\times h\)

• The base can be any one of the sides of the triangle, and the height is the perpendicular distance form the opposite vertex to that base.

• Based on the triangle in the figure above, formula for area of the triangle given by:

\(\begin{aligned} \text{Area}&=\dfrac{1}{2}ab\sin{C} \\\\ &=\dfrac{1}{2}ac\sin{B} \\\\ &=\dfrac{1}{2}bc\sin{A} \end{aligned}\)

• This formula is useful when two sides and the included angle are known.

• For a triangle with sides \(a\), \(b\), and \(c\), and semi-perimeter, \(s=\dfrac{a+b+c}{2}\):

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

• Useful when all three sides of the triangle are known.

Based on the triangle \(ABC\) above, find the area of the triangle.

Based on the question above, given:

\(AB=13\) cm,
\(BC=16\) cm,
\(\angle{ABC}=36^\circ\).

Using the trigonometry formula:

​\(\begin{aligned} \text{Area of }\Delta ABC&=\dfrac{1}{2}ac\sin{B} \\\\ &=\dfrac{1}{2}(16)(13)\sin{36^\circ} \\\\ &=61.13\text{ cm}^2. \end{aligned}\)​

The diagram above shows a triangle \(ABC\).

By using Heron's formula, calculate the area of the triangle.

Based on the question, given:

\(a=1.8\) cm,
\(b=3\) cm,
\(c=3.8\) cm.

Calculate the semi-perimeter, \(s\):

\(\begin{aligned} s&=\dfrac{a+b+c}{2} \\\\ &=\dfrac{1.8+3+3.8}{2} \\\\ &=4.3. \end{aligned}\)

Substitute the value of \(a\), \(b\), \(c\), and \(s\) into the Heron's formula:

\(\begin{aligned} \text{Area}&=\sqrt{s(s-a)(s-b)(s-c)} \\ &=\sqrt{4.3(4.3-1.8)(4.3-3)(4.3-3.8)} \\ &=\sqrt{4.3(2.5)(1.3)(0.5)}\\ &=\sqrt{6.9875}\\ &=2.6434\text{ cm}^2. \end{aligned}\)