Applications of Heron's formula

Any quadrilateral $ABCD$ can be cut into two triangles by drawing a single diagonal — say $AC$. This produces $\triangle ABC$ and $\triangle ACD$, which together make up the whole quadrilateral and share the diagonal $AC$ as a common side. Because the two triangles exactly tile the quadrilateral with no overlap and no gap, their areas simply add.

where: $\text{Area}(ABCD)$ is the area of the whole quadrilateral, and $\triangle ABC$, $\triangle ACD$ are the two triangles formed by the diagonal $AC$. Each triangle's area is then found with Heron's formula, using its own semi-perimeter:

where: $a, b, c$ are the three sides of one triangle (two sides of the quadrilateral plus the shared diagonal), $s$ is that triangle's semi-perimeter, and $A$ is its area. The one real skill here is choosing the right diagonal. A quadrilateral has two diagonals, but only one of them may give you two triangles whose three sides are all known (or easily computed). If a problem states the diagonal explicitly, that choice is already made for you; if not, look for the diagonal that turns both halves into fully specified triangles — often the one that, combined with the given sides, lets you spot a familiar triple such as 3-4-5. This method is not restricted to fields and parks in word problems — it is how the area of any irregular four-sided plot is found in practice, since a general quadrilateral has no simple "base times height" shortcut the way a rectangle or parallelogram does. The two special sides that matter most are the four boundary sides (which are usually all you can physically measure with a tape around the plot) and the one diagonal (which requires a measurement across the interior). Once those five lengths are known, the shape is completely fixed — no other quadrilateral has the same four sides and the same diagonal — so its area is determined exactly, with no need for angles or heights at all. This is one reason the diagonal-and-Heron approach is so useful for real survey work: it needs only lengths, which are far easier to measure accurately in the field than angles.

The goal is to find the area of quadrilateral $ABCD$ when the four sides and one diagonal are known, by splitting it into two triangles and applying Heron's formula independently to each one before adding the two areas together.

Step 1. Identify the diagonal that has been given (or that you can compute), and use it to split the quadrilateral into two triangles. Diagonal $AC$ splits $ABCD$ into $\triangle ABC$ with sides $AB$, $BC$, $CA$, and $\triangle ACD$ with sides $AC$, $CD$, $DA$. The diagonal $AC$ appears as a side in both triangles — it is shared, not doubled.

Step 2. For $\triangle ABC$, with sides $a_1, b_1, c_1$, compute the semi-perimeter $s_1 = \frac{a_1+b_1+c_1}{2}$, then apply Heron's formula: $\text{Area}_1 = \sqrt{s_1(s_1-a_1)(s_1-b_1)(s_1-c_1)}$.

Step 3. For $\triangle ACD$, with sides $a_2, b_2, c_2$ (one of which is the same diagonal length used in Step 2), compute its own semi-perimeter $s_2 = \frac{a_2+b_2+c_2}{2}$, then $\text{Area}_2 = \sqrt{s_2(s_2-a_2)(s_2-b_2)(s_2-c_2)}$.

Step 4. Add the two areas: $\text{Area}(ABCD) = \text{Area}_1 + \text{Area}_2$. Because the diagonal lies inside the (convex) quadrilateral, the two triangles neither overlap nor leave a gap, so this simple sum gives the exact total area.

Problem. A quadrilateral-shaped field $ABCD$ has $AB = 9\text{ m}$, $BC = 40\text{ m}$, $CD = 28\text{ m}$, $DA = 15\text{ m}$, and diagonal $AC = 41\text{ m}$. Find the area of the field.

Step 1. Diagonal $AC$ splits the field into $\triangle ABC$ with sides $9, 40, 41$ and $\triangle ACD$ with sides $41, 28, 15$.

Step 2. For $\triangle ABC$: $s_1 = \frac{9+40+41}{2} = 45$, so $s_1-9=36$, $s_1-40=5$, $s_1-41=4$. Then

(since $180\times180=32400$).

Step 3. For $\triangle ACD$: $s_2 = \frac{41+28+15}{2} = 42$, so $s_2-41=1$, $s_2-28=14$, $s_2-15=27$. Then

(since $126\times126=15876$).

Step 4. Add the two areas: $\text{Area}(ABCD) = 180 + 126 = \mathbf{306\ m^2}$.

Competitive stretch

A similar park $PQRS$ has $PQ=15\text{ m}$, $QR=20\text{ m}$, $RS=7\text{ m}$, $SP=24\text{ m}$, diagonal $PR=25\text{ m}$. Diagonal $PR$ gives $\triangle PQR$ (sides $15,20,25$) with $s_1=30$ and $\text{Area}_1=\sqrt{30\times15\times10\times5}=\sqrt{22500}=150\text{ m}^2$, and $\triangle PRS$ (sides $25,7,24$) with $s_2=28$ and $\text{Area}_2=\sqrt{28\times3\times21\times4}=\sqrt{7056}=84\text{ m}^2$. Total area $= 150+84=\mathbf{234\ m^2}$. Notice $15$-$20$-$25$ is a multiple of the $3$-$4$-$5$ right triangle — spotting such triples is a fast arithmetic check, though Heron's formula works regardless of whether a triangle is right-angled.

• Splitting along the wrong diagonal. Only one diagonal may give two triangles whose three sides are all known — check both before committing to one.
• Forgetting the diagonal is shared. The diagonal's length is used as one side in both triangles; it is never added again as an extra quadrilateral side.
• Assuming a non-convex shape still just adds. For the standard Class 9 problems the quadrilateral is convex and the diagonal lies fully inside it, so the two triangle areas add directly — this reasoning would need care for a non-convex figure.
• Mixing up which sides belong to which triangle. Re-list each triangle's three sides carefully before applying Heron's formula; substituting a side from the wrong triangle is a frequent slip.

Ask the tutor: Want to try splitting a trickier quadrilateral yourself, or check whether you picked the right diagonal for a given field problem — ask the AI tutor to walk through another composite-figure example with you.