Real-World Applications

This final lesson of the chapter shows Heron's formula at work on real problems. Land plots, gardens, parking lots, flags, banners , anything with a triangular or polygonal boundary can have its area calculated by these methods. The mathematics is identical to what we have practised, but the setting up requires careful reading.

A six-step approach to area word problems

Sketch the figure. Draw the plot, field, or shape from the problem.
Label all known lengths. Mark sides and any diagonals or heights given.
Identify the shape. Triangle? Rectangle? Quadrilateral? Composite (multiple shapes joined)?
Choose the right formula or triangulation.
Compute. Apply Heron's formula or the appropriate area formula.
Interpret. Convert units if needed; state the final area in the right units (m $^2$ , hectares, etc.).

Five typical situations

Situation 1: Triangular plot. A farmer has a triangular field with sides $50$ , $80$ , and $90$ m. To find the area for fencing or pricing, use Heron's formula directly.

Situation 2: Quadrilateral plot. A plot has four sides $30, 40, 50, 60$ m and a diagonal of length $50$ m. Triangulate using the diagonal, apply Heron's formula twice, add.

Situation 3: Composite shape. A garden is a rectangle of $30 \times 20$ m with a triangular extension of base $20$ m and height $10$ m on one side. Total area $=$ rectangle area $+$ triangle area $= 600 + 100 = 700$ m $^2$ .

Situation 4: Surveying irregular land. An L-shaped plot is divided into two rectangles, or a triangular plot into two right triangles. Find each piece, add.

Situation 5: Unit conversion. A field of $1$ hectare = $10000$ m $^2$ . Always convert at the end to the unit requested.

A complete worked problem

A triangular park has sides $50$ m, $60$ m, and $70$ m. The municipality wants to surface it with grass costing $\text{Rs.}~20$ per square metre. Find the cost.

Step 1. The park is a triangle with sides $50, 60, 70$ .

Step 2. Compute the area using Heron's formula. $s = \tfrac{50+60+70}{2} = 90$ . $s - 50 = 40, s - 60 = 30, s - 70 = 20$ . Product: $90 \times 40 \times 30 \times 20 = 2160000$ . Area $= \sqrt{2160000} = 600\sqrt{6}$ m $^2$ $\approx 1469.69$ m $^2$ .

Step 3. Cost $= 1469.69 \times 20 \approx \text{Rs.}~29393.88$ .

Another worked problem

A kite is to be made from coloured paper. The kite has two pairs of sides: $25$ cm each on top and $35$ cm each on bottom, with diagonals $30$ cm and $48$ cm. How much paper is needed?

Step 1. The kite has diagonals $d_1 = 30$ and $d_2 = 48$ .

Step 2. For a kite, area $= \tfrac{1}{2} d_1 d_2 = \tfrac{1}{2} \cdot 30 \cdot 48 = 720$ cm $^2$ .

Step 3. So $720$ cm $^2$ of paper is needed.

(Equivalently, you could split the kite by its long diagonal into two triangles with sides $25, 25, 30$ and $35, 35, 48$ , find each area by Heron's, and add , but the diagonal formula is faster.)

A quadrilateral plot problem

A quadrilateral plot has sides $26$ , $28$ , $26$ , $24$ m and a diagonal of length $30$ m connecting the first and third vertices.

Step 1. Diagonal $AC = 30$ m divides the plot into $\triangle ABC$ with sides $26, 28, 30$ and $\triangle ACD$ with sides $30, 26, 24$ .

Step 2. $\triangle ABC$ : $s = 42$ , $s - 26 = 16, s - 28 = 14, s - 30 = 12$ . Product: $42 \cdot 16 \cdot 14 \cdot 12 = 112896$ . Area $= \sqrt{112896} = 336$ m $^2$ .

Step 3. $\triangle ACD$ : $s = 40$ , $s - 30 = 10, s - 26 = 14, s - 24 = 16$ . Product: $40 \cdot 10 \cdot 14 \cdot 16 = 89600$ . Area $= \sqrt{89600} \approx 299.33$ m $^2$ .

Step 4. Total area $\approx 635.33$ m $^2$ .

Worked examples

Example 1. A triangular flower bed has sides $9, 12, 15$ m. Find the area.

Check: $9^2 + 12^2 = 81 + 144 = 225 = 15^2$ . Right triangle. Area $= \tfrac{1}{2} \cdot 9 \cdot 12 = 54$ m $^2$ .

Example 2. A field is a parallelogram with base $50$ m and height $20$ m. Find the area.

Area $= 50 \cdot 20 = 1000$ m $^2$ $= 0.1$ hectare.

Example 3. A trapezium-shaped field has parallel sides $80$ m and $100$ m with the perpendicular distance between them $50$ m. Find the area.

Area $= \tfrac{1}{2}(80 + 100) \cdot 50 = 90 \cdot 50 = 4500$ m $^2$ $= 0.45$ hectare.

Example 4. A triangular signboard has sides $11, 60, 61$ cm. Find the area and the cost of painting it at $\text{Rs.}~5$ per cm $^2$ .

Check: $11^2 + 60^2 = 121 + 3600 = 3721 = 61^2$ . Right triangle. Area $= \tfrac{1}{2} \cdot 11 \cdot 60 = 330$ cm $^2$ . Cost $= 330 \cdot 5 = \text{Rs.}~1650$ .

Example 5. A quadrilateral park has sides $40, 50, 60, 75$ m and diagonal $70$ m. Find its area.

Triangle 1 (sides $40, 50, 70$ ): $s = 80$ . $s - 40 = 40, s - 50 = 30, s - 70 = 10$ . Product: $80 \cdot 40 \cdot 30 \cdot 10 = 960000$ . Area $= \sqrt{960000} \approx 979.80$ .

Triangle 2 (sides $70, 60, 75$ ): $s = 102.5$ . $s - 70 = 32.5, s - 60 = 42.5, s - 75 = 27.5$ . Product: $102.5 \cdot 32.5 \cdot 42.5 \cdot 27.5 \approx 3897851.6$ . Area $\approx \sqrt{3897851.6} \approx 1974.32$ .

Total $\approx 2954.12$ m $^2$ .

Try it yourself

A triangular plot has sides $30, 40, 50$ m. Find the area.
A trapezium has parallel sides $24$ and $36$ cm and height $20$ cm. Find the area.
A rhombus has diagonals $36$ and $48$ cm. Find the area.
A quadrilateral field has sides $40, 32, 24, 18$ m and one diagonal $30$ m. Find the area.
A triangular banner has sides $13, 14, 15$ m. Find the cost of painting it at $\text{Rs.}~5$ per m $^2$ .
A garden in the shape of an L can be split into a rectangle of $30 \times 20$ m and a rectangle of $20 \times 10$ m. Find the total area.
A kite has diagonals $20$ and $40$ cm. Find the area.
An isosceles triangular flower bed has sides $30, 30, 36$ cm. Find the area.
An equilateral triangular plot of side $10$ m needs grass at $\text{Rs.}~50$ per m $^2$ . Find the cost.
A rectangular field is $80 \times 50$ m. Find the area in hectares.

Pitfalls / Insight

Read the problem carefully. Identify the shape, then choose the technique.
Always state units in the final answer. m $^2$ , cm $^2$ , hectares , match the question.
For composite shapes, split into known sub-shapes. Add the parts.

Insight. Heron's formula and triangulation, combined with the standard formulas for rectangles, parallelograms, trapezia, and rhombi, give you a complete toolkit for any plane-area problem. With this toolkit, you can compute the area of any polygonal land plot , exactly what land surveyors did before satellite imagery existed.