Area of Triangles.

Subsection 2.5.1 Area of Triangles.

Activity 2.5.2.

Building a Triangle Garden.

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Working im groups of groups of 3–5 students.

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$\textbf{Materials needed:}$

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Cereal boxes or recycled cardboard
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String or yarn
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Pencil and Scissors
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Chalk (if outdoors) or masking tape (if indoors)
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Manila paper or newspaper
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A few small stones or bottle caps (for weights or markers)
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Preferably conuct this activity outside on sand or cement, or inside with tape/string.
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- Ensure you have a rectangle (or square) drawn with chalk/tape—e.g. 60 cm × 40 cm.
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- Can you use string to form a triangle that perfectly fits inside this rectangle?
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- Can you make the triangle cover exactly half of the space?
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- Use a string from corner to corner, or make triangles by connecting midpoints.
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- Take a manila paper or cardboard. Draw a rectangle (e.g. 10 cm by 6 cm).
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- Cut along a diagonal to get two right triangles. Rearrange the two triangles to form a rectangle again.
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- Now try : Drawing a triangle with the same base and height, but different shape (scalene, obtuse). Cut and rearrange it against the original rectangle.
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On the board or chart, formulate the rule:
- If the area of a rectangle is base × height, and the triangle is always half of it, What will be the formula of a triangle?
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- Write it down together in your groups.
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$\textbf{Review of properties of a triangle}.$

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A triangle is a three-sided polygon with three angles which add up to $180^\circ $ and three vertices.
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$\textbf {Key Properties of a Triangle}$
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♦ A triangle has three Sides and Three Angles.
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♦ A triangle has three edges and three vertices (corner points).
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♦ Angle Sum Property - The sum of all three interior angles of a triangle adds up to 180°.
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$\displaystyle \textbf{Types of Triangles (by Sides).}$
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🔹 $\textbf {Equilateral Triangle}$ - All sides of the triangle are equal and have an angle of 60° each.
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🔹 $\textbf {Isosceles Triangle}$ - Two sides of the triangle are equal, and the angles opposite these sides are also equal.
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🔹 $\textbf {Scalene Triangle}$ - All three sides of the triangle have different lengths, and all angles are different.
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$\text{Types of Triangles (by Angles).}$
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🔹 $\textbf {Acute-Angled Triangle} (T_1)$ - All three angles are less than 90°.
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🔹 $\textbf {Right-Angled Triangle}(T_2)$ - One angle is exactly 90°.
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🔹 $\textbf {Obtuse-Angled Triangle}(T_3)$ - One angle is greater than 90° but less than $180 \circ .$
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Example 2.5.1.

Calculate the unknown variables in each of the following figures

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Solution.

(a) An isosceles triangle has two angles that are equal. Sum of interior angles of triangle is $180 ^\circ$

\begin{align*} = \amp 180^\circ - 36^\circ \\ = \amp 144^\circ \\ = \amp \frac{144^ \circ}{2} \\ = \amp 72^ \circ \end{align*}

Therefore $x = 72 ^\circ $ , $y = 72^ \circ $

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(b) To find $x$ we use pythagorean relationship.

\begin{equation*} a^2 + b^2 = c^2 \end{equation*}

In our case our $c$ is $x$

\begin{equation*} 20^2 + 15^2 = c^2 \end{equation*}

\begin{equation*} 400 + 225 = 625 \end{equation*}

\begin{equation*} =\sqrt{625} \end{equation*}

\begin{equation*} = 25 \,\text{cm} \end{equation*}

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$\textbf{Exercise}$

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Determine the unknown angles and sides in the figures below

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Subsubsection 2.5.1.1 Area of a Triangle using Heron’s Formula

Activity 2.5.3.

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Pair up students and give each pair a triangular surface to measure. This could be:

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A triangular classroom table
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A triangular signboard
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A triangular cutout from paper
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A drawn triangle on graph paper
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$\textbf{Materials needed.}$
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Ruler
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Graph paper
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Calculator
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String or measuring tape
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Pre-cut paper triangles (optional)
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1. Recall the Standard Triangle Area Formula
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  - How do we find the area of a triangle?
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    Expected answer ; $a = \frac{1}{2} \times \text{base} \times \text{height}$
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  - This method works only when we know the height.
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  - $\textbf{What if we don’t know the height?}$ Today, they will discover how to find the area $\textbf{only using side lengths}\text{.}$
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2. Construct a Triangle and Introduce Semi-Perimeter
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  - Draw a triangle with sides labeled as $\textbf{a, b} \, \text{and} \, \textbf{c}\text{.}$ Using a ruler or measuring tape, measure the $\textbf{three sides of the triangle}$ and record the lengths.
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  - Measure and calculate the semi-perimeter using:
    
    \begin{align*} S = \amp \frac{a+b+c} {2} \end{align*}
    
    This helps divide the triangle into manageable parts for the area calculation.
    
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3. Express the$\textbf{ Area} $ in Terms of the $\textbf{Semi-Perimeter.}$
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  - Use the right-angle criteria.i.e:
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    $\textbf{Drop a perpendicular}$ from $\textbf{one vertex to the opposite side}$ (splitting the triangle into two right-angled triangles).
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    Let’s call the base $\textbf{b}$ and split it into two parts using the perpendicular.
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  - Use the $\textbf{Pythagorean Theorem}$ to express the height $h$ in terms of $\textbf{ a, b, c}.$
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    Instead of solving fully, guide students to realize that the height can be found using algebraic manipulation.
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  - Introduce the $\textbf{squared form}$ and take the $\textbf{square root}\text{:}$
    
    \begin{align*} A= \amp \sqrt{S(S-a)(S-b)(S-c)} \end{align*}
    
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  - This is the final area using Heron’s Formula.
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  - In conclusion, Heron’s Formula is finding the area of any triangle using only its sides.
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$\textbf{Key Takeaway}$

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Given three unequal sides of a triangle, we can calculate the area using a formula called Heron’s Formula.

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Follow the simple guidelines below and let’s explore on how to get to the formula.

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Consider the triangle below.

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Perimeter refers to adding all round.
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\begin{equation*} \textbf{P} = a+b+c \end{equation*}

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Let’s find the Semi perimeter of the triangle by the perimeter diving by half.
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\begin{equation*} \textbf{S} = \frac{a+b+c}{2} \end{equation*}

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Then find the area using Heron’s formula:
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\begin{equation*} \textbf{A} = \sqrt{S(S-a)(S-b)(S-c)} \end{equation*}

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Calculate $\textbf{S}$ (half of the triangle’s perimeter):
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Then calculate the Area:
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Example 2.5.2.

If the length of the sides of a triangle $ABC$ are $4$ inches, $3$ inches and $5$ inches. Calculate its area using the heron’s formula.

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$triangle$

Solution.

To find: Area of the triangle $ABC\text{.}$

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Given that AB $= 4$in and BC $= 3$in, AC $= 5$ in.

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Using Heron’s Formula,

\begin{align*} A \amp = \sqrt{(s(s - a)(s - b)(s - c)) } \end{align*}

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\begin{align*} s \amp = \frac{(a + b + c)}{2} \end{align*}

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\begin{align*} s \amp = \frac{(4+3+5)}{2} \end{align*}

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\begin{align*} \amp = 6 \, \text{units} \end{align*}

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Substitute in the values,

\begin{align*} A \amp = \sqrt{(6(6 - 4)(6 - 3)(6 - 5))} \end{align*}

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\begin{align*} A \amp = \sqrt{(6 \text (2)\times(3)\times(1))} \end{align*}

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\begin{align*} A \amp = \sqrt{36} \end{align*}

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\begin{align*} \amp = 6 \, \text{in}^2 \end{align*}

The area of the triangle is $6 \, \text{in}^2\text{.}$

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$\textbf{Exercise}$

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1. In the diagram below $\angle \, ABC $ is right-angled at B. Complete the table below.

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Table 2.5.3. Trigonometry question test

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Trigonometic ratios	Solution
sin C	=
=	$\frac{BC}{AC}$
tan $\theta$	=
Value of AC using pythagorean relationship	=
Area given a, b and sin $\theta$	=

2.The area of a triangle with sides 5 cm, 12 cm, and $x$ cm is 30 cm². Find the possible values of $x\text{.}$

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3. A right-angled triangle has sides 9 cm, 12 cm and 15 cm.

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(a) Verify that the triangle satisfies the Pythagorean relationship.

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(b) Find its area using both the standard formula and Heron’s formula.

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4. A triangular plot of land has sides measuring 50 m, 65 m and 75 m. Find the area of the land using Heron’s formula.

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5. An aeroplane at $J$ is flying directly over a point $D$ on the ground at a height of 5 kilometres. It is heading to land at point $K\text{.}$ The angle of depression from $J$ to $K$ is $8^\circ\text{.}$ $S $ is a point along the route from $D$ to $K\text{.}$

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I) What’s the size of $\angle \, JKD? $

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II) Calculate the distance $DK\text{,}$ correct to the nearest metre

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III) If the distance $SK$ is $8$ kilometres, calculate the distance $DS\text{.}$

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IV) Calculate the area of $\triangle \, JKD\text{.}$

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Subsubsection 2.5.1.2 Area of a triangle given two sides and an included angle

Activity 2.5.4.

What you require:
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- A tall object (flagpole, tree or building)
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- Measuring tape or ruler.
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- A sunny day ☀
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$\textbf {Identify the Tall Object }\text{.}$
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- Choose a tree, flagpole, or lamp post as the vertical height (like $A$ in the diagram below).
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- The ground acts as the base $(BC)\text{.}$
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$\textbf {Find the Shadow Length}$
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- Measure the length of the shadow cast by the object on the ground ($BC$).
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- This represents the long horizontal base in the diagram.
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$\textbf {Measure the Angle of Elevation}$
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- Stand some distance away and use a protractor or a phone app to measure the angle of elevation from your eyes to the top of the object.
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- This represents the $30^ \circ $ angle at $B$ in the diagram.
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- If no protractor is available, use similar triangles by measuring the shadow of a known object (like a stick) and comparing proportions.
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Discuss your findings with your group members.
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- $\textbf{Apply Trigonometry Using Sine}$
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  Using the sine function to calculate the height of the object $AN\text{.}$
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  Drop a perpendicular line from Point A to meet line BC at N .
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- Formula:
  
  \begin{align*} Height (AN) = \amp \text{Hypotenuse(AB)} \times \text{ sin }30^ \circ \end{align*}
  
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- AN is therefore the height for triangle ABC.
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- In a triangle $\text{sin} \, \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\text{.}$
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  From the figure shown above, Opposite = AN, the height of triangle ABN whereas the hypotenuse i $AB$ is the longest side of a triangle.
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  Where $AN$ = $\textbf{h}$ and $BA$ = $\textbf{a}$
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  \begin{align*} \text{In triangle ABN, sin}30^ \circ = \amp \frac{AN}{a} \\ \text{sin } 30^ \circ = \amp \frac{\textbf{h}}{\textbf{a}} \\ \text{ height of }\angle \text{ABC} \, \textbf{h} = \amp \textbf{a} \, \text{sin } 30^ \circ \\ \text{ Since sin } 30^ \circ = 1/2 \textbf{a} \end{align*}
  
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- Area of Triangle ABC will be:
  
  \begin{align*} \triangle \, ABC = \amp \frac{1}{2} \times \text {Base}\, \textbf{ (b)} \times \text { Height}\, \textbf{(h)} =\textbf{a} \,\text{sin}\theta \\ = \amp \frac{1}{2} \, \textbf{ ab } \, \textbf{ sin } \, \theta \end{align*}
  
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♦ Generally, if the lengths of the two sides and an included angle of a triangle are given, then the area of the triangle is $\text{A} = \frac{1}{2}\textbf{ab}\textbf{sin}\theta\text{.}$ Where $\textbf{ b } \text{is the base and } \textbf{ h } \text{ height.}$

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Exploration 2.5.5.

Deriving the Formula: $\frac {1}{2}ab\, \text{sin C} $
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- Step 1: Let’s recall the Basic Formula for finding area of a triangle.
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- Area = $\frac {1}{2} \times \text {Base} \times \text {Height}$
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$\textbf {Step 2: Consider a Triangle with an Angle}\text{.}$
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- Let’s take a triangle ABC with sides $a \text{,}$ $b $ and included angle $C\text{.}$
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- Side$\, a$ and Side $b$ form the triangle.
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- The height $h$ is perpendicular from the top vertex to the base.
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$\textbf {Step 3: Express Height in Terms of Sin}$
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- Using trigonometry, we know that in a right-angled triangle:
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- $\displaystyle \text{sin} \,\theta = \frac{\text{Hypotenuse}}{\text{Opposite}}$
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- In our case, the height $h$ is the $\textbf{opposite side}$ of angle $C\text{,}$ and side $b$ is the $\textbf{hypotenuse}\text{.}$
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- So, we can write it as:
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  $h = b \text{ sin} \, C$
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$\textbf {Step 4: Substitute into the Area Formula}$
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- Now, take the basic Formula for finding area of a triangle;
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- Substituting $\textbf{Base}$ and $\textbf{Height} = \text{b sin C} $ .
  
  \begin{align*} \text{Area} =\amp \frac {1}{2} \times a \times \text{(b sin C)}\\ \text{Area} = \amp \frac {1}{2} ab \, \text{ sin C} \end{align*}
  
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$\textbf{NOTE:}$
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- \begin{align*} \text{Area} = \amp \frac {1}{2} ab \, \text{ sin C} \end{align*}
  
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- This formula is useful when we know two sides of a triangle and the angle between them instead of the height..
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Example 2.5.4.

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1.A triangle HFG has sides $10 \text{ cm}, 7 \text{ cm and } 9 \text{ cm.}$ Find:

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(a) Its area.

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(b) The sizes of its angles.

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2. The area of triangle ABC is $28.1 \text{cm}^2$ . It’s side AB $= 7.2 \text{cm} $ and $\angle$ ABC $= 48.6^\circ\text{.}$ Find:

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(a) The length of the perpendicular from A to BC.

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(b) The length of BC.

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Solution 1.

(a) Find the area using Heron’s formula.

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\begin{align*} A \amp = \sqrt{(s(s - a)(s - b)(s - c))} \end{align*}

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\begin{align*} s \amp = \frac{(a + b + c)}{2} \end{align*}

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\begin{align*} s \amp = \frac{(10+7+9)}{2}\, cm \end{align*}

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\begin{align*} \amp = 13 \, \text{cm} \end{align*}

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Substitute in the values.

\begin{align*} A \amp = \sqrt{(13(13 -10)(13 - 7)(13 - 9))} \end{align*}

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\begin{align*} A \amp = \sqrt{(13 \text (6)\times(3)\times(4))} \end{align*}

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\begin{align*} A \amp = \sqrt{936} \end{align*}

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\begin{align*} \amp = 30.6 \, \text{cm}^2 \end{align*}

The area of the triangle is $\, 30.6 \text{cm}^2\text{.}$

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(b) Find the angles using the sine area formula.

\begin{equation*} A = \frac{1}{2} \,\text{ab sin C} \end{equation*}

\begin{equation*} A = 30.6 \, \text{ cm}^2 \end{equation*}

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To find $\angle HFG$ Using sides $a = 10$ and $b =7 $ and area$= 30.6 \,\text{cm}^2$

\begin{align*} 30.6 \, \text{cm}^2 = \amp \frac {1}{\cancel{2}} \times \cancel{10} \, 5\, \text{cm} \times 7\, \text{cm} \times \text{sin} \, A\\ \text{Sin A} = \amp \frac{30.6 \cancel{\text{cm}}^2 \,\text{cm}}{35 \cancel{\text{cm}} }\\ \text{sin A} = \amp 0.8743\\ A = \amp sin^{-1} \, {0.8743} \\ = \amp 60.96^\circ \end{align*}

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To find $\angle FGH$ Using sides $a = 10$ and $b =9 $and area$= 30.6 \,\text{cm}^2$

\begin{align*} 30.6 \, \text{cm}^2 = \amp \frac {1}{\cancel{2}} \times \cancel{10} \, 5\, \text{cm} \times 9\, \text{cm} \times \text{sin} \, C \\ \text{Sin C} = \amp \frac{30.6 \cancel{\text{cm}}^2 \,\text{cm}}{45 \cancel{\text{cm}} }\\ \text{sin C} = \amp 0.68\\ C = \amp \text{sin}^{-1} {0.68} \\ = \amp 42.84 ^\circ \end{align*}

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To find the $\angle GHF $ we use the angle sum property.

\begin{align*} \angle GHF = \amp 180^\circ - (\angle FGH + \angle HFG)\\ = \amp (180 - 103.80)^\circ\\ = \amp 76.2^\circ \end{align*}

Similarly we can use the Sin A rule which states that :

\begin{equation*} \frac{ \text{sin A}}{a} = \frac{ \text{sin B}}{b} = \frac{ \text{sin C}}{c} \end{equation*}

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\begin{align*} \frac{ \text{sin B}}{b} \amp = \frac{ \text{sin C}}{c} \\ \frac{ \text{sin B}}{10 \,\text{cm}} = \amp \frac{ \text{sin 0.8743}}{9} \\ \text{sin} \, B = \amp 0.9714\\ = \amp \text{sin}^{-1} \, 0.9714\\ B = \amp 76.2^\circ \end{align*}

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Solution 2.

(a) Find the length of the perpendicular from A to BC.

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The formula for area using base and height.

\begin{align*} A = \amp\frac{1}{2} \times base \times height \\ 28.1 \text{cm}^2 = \amp \frac{1}{2}\times b \times h \\ h = \amp \frac {2 \times 28.1}{BC} \end{align*}

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(b) Finding BC using sine rule

\begin{align*} A = \amp \frac{1}{2} \times AB \times BC \times \text{sin} \, \theta \\ 28.1 \,\text{cm}^2=\amp \frac{1}{2} \times 7.2\,\text{cm} \times BC \times \text{sin} \, \, 48.6^\circ \\ \text{sin} \, 48.6^\circ = \amp \, 0.7501\\ 28.1 \,\text{cm}^2=\amp \frac{1}{2} \times 7.2\, \text{cm}\times BC \times \text{sin} \, 0.7501 \\ BC = \amp \frac{28.1 \times 2}{7.2 \times 0.7501} \\ = \amp 10.4 \, \text{cm} \end{align*}

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Now substituting BC into the height equation :

\begin{align*} h = \amp \frac{2 \times 28.1}{10.4}\\ =\amp 5.4 \,\text{cm} \end{align*}

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$\textbf{Exercise2}$

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$1.$ In a triangle $\triangle\textbf{QRS}\text{,}$ $\textbf{QR}$ = 10 cm, $\textbf{RS}$ = 24 cm and $\textbf{QS}$ = 26 cm. Find the length of the perpendicular from vertex $\textbf{Q}$ to side $\textbf{RS}$ .Therefore find it’s area.

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$2. $ In $\triangle \textbf{ABC}\text{,}$ $\angle \textbf{BAC} = 40^\circ\text{,}$ $\angle \textbf{ABC} = 65^\circ\text{,}$ and side $\textbf{BC} = 8 $cm. Find it’s area.

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$3. \triangle \textbf{PQR} \, \text{is} \, \textbf{isosceles} \, \text{with} \, \textbf{PQ} = \textbf{PR} = 10$ cm. The base angle is $48^\circ. \text{Find its area}.$

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$4. \text{Triangle} \textbf{XYZ}$ has side $\textbf{XY} = 15$ cm, $\textbf{YZ} = 20$ cm, and $\textbf{XZ} = 25 cm. \text{Show that this is a right-angled triangle and find it's area.}$

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$5. \text {Triangle} \textbf{MNO}$ \, has $\angle \textbf{NMO} = 50^\circ$, $\angle \textbf{MNO} = 60^\circ$, and $\textbf{MO} = 18$ cm. Find the length of $\textbf{MN} \, \text{using the sine rule } \text{thus find the triangle's area.}$

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$6. \triangle \textbf{PQR}$ has $\angle \textbf{RPQ} = 35^\circ$, $\angle \textbf{PQR} = 75^\circ$, and $\textbf{QR} = 15$ cm. Find the area of triangle PQR.\)

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Trigonometic ratios	Solution
sin C	=
=	\(\frac{BC}{AC}\)
tan \(\theta\)	=
Value of AC using pythagorean relationship	=
Area given a, b and sin \(\theta\)	=