Area of a triangle

Every triangle secretly wants to be half of a rectangle. That single idea — cut, rearrange, compare to a shape whose area you already trust — is the whole story of this topic in the Area chapter of Class 8 Mathematics. Once rectangle area is settled, the natural next question is: what about the enormous family of shapes with three straight sides and no right angles guaranteed anywhere? This section shows that a triangle's area is always half of base times the perpendicular height to that base, proves why the half appears using a parallelogram-completion argument, and shows that the answer never depends on which of the three sides you happen to call the base. That last fact — same triangle, same area, different base-height pairs — is the part students most often get wrong, and it is where most of the worked practice below is aimed.

The core insight

A triangle's area is always $\frac12 \times \text{base} \times \text{perpendicular height}$ , and this stays true no matter which side is chosen as the base, because every triangle is exactly half of some parallelogram sharing that base and height.

The core idea

The formula. For a triangle with base $b$ and the perpendicular height $h$ measured from the opposite vertex to that base (or to its extension):

A_{\triangle} = \frac12 \times b \times h

where: $b$ is the length of the chosen base and $h$ is the perpendicular distance from the opposite vertex to the line containing that base — never a slanted side length.

Why the half appears. Draw a line through the vertex opposite the base, parallel to the base, and complete a parallelogram by attaching a congruent copy of the triangle, rotated to fit. This parallelogram has the same base $b$ and the same height $h$ as the original triangle, and its diagonal splits it into exactly two congruent triangles. A parallelogram of base $b$ and height $h$ has area $b \times h$ (itself provable by rearranging it into a rectangle), so each half — our original triangle — has area $\frac12 bh$ .

The base can be any side. Whichever of the three sides you pick as base, pair it with its own perpendicular height, and the area comes out the same:

\frac12 \times b_1 \times h_1 = \frac12 \times b_2 \times h_2 = \frac12 \times b_3 \times h_3

For a right-angled triangle the two legs are automatically a valid base-height pair since they meet at $90^\circ$ . In an obtuse triangle, the height for some sides lands outside the triangle, on the extended base — the formula still applies exactly, but you must extend the base line to find the foot of the perpendicular.

Figure

A triangle with its base at the bottom and a dashed perpendicular height dropped from the top vertex, shown alongside the same triangle completed into a parallelogram by a congruent rotated copy

Derivation AThe parallelogram-halving proof

Prove, using only the already-established area of a parallelogram and one congruence argument, that the area of any triangle equals exactly half of base times perpendicular height for every one of its three possible base choices, so that the formula $A = \frac12 bh$ can be trusted no matter which side a problem happens to label as the base.

Step 1. Start with triangle $ABC$ , base $BC = b$ , and drop the perpendicular height $h$ from vertex $A$ to line $BC$ (extending $BC$ if needed for an obtuse triangle). Through $A$ , draw a line parallel to $BC$ .

Step 2. On that parallel line, mark point $D$ so that $BCDA$ ... more precisely, take a second copy of triangle $ABC$ , rotate it $180^\circ$ about the midpoint of $AC$ , and place it so its image sits exactly against the original along side $AC$ . The two triangles together now form quadrilateral $ABCD$ .

Step 3. Show $ABCD$ is a parallelogram. Rotating $180^\circ$ about the midpoint of a segment sends that segment to itself and swaps the two triangles, so $AD$ is parallel and equal to $BC$ , and $AB$ is parallel and equal to $DC$ — the two defining conditions of a parallelogram. This parallelogram has base $BC = b$ and the same height $h$ , since $A$ and $D$ both lie on the line through $A$ parallel to $BC$ .

Step 4. Apply the known parallelogram-area result. A parallelogram of base $b$ and height $h$ has area $b \times h$ (established earlier in the Area chapter by cutting off a triangular end and reattaching it to form a rectangle). Since the diagonal $AC$ splits parallelogram $ABCD$ into two congruent triangles — the original and its rotated copy — each triangle has exactly half that area: $A_{\triangle} = \frac12 bh$ .

Step 5. Confirm this holds for any of the three sides as base. The same construction works starting from side $CA$ or side $AB$ instead of $BC$ , each time producing a parallelogram of the corresponding base and its own perpendicular height. Since all three constructions describe the same fixed triangle, all three computations $\frac12 b_1 h_1$ , $\frac12 b_2 h_2$ , $\frac12 b_3 h_3$ must equal the same single number — the triangle's one true area.

Worked example

A right-angled triangle has legs $9\ \text{cm}$ and $12\ \text{cm}$ and hypotenuse $15\ \text{cm}$ . Find its area, then find the height corresponding to the hypotenuse as base, and finally verify the parallelogram check by doubling the triangle.

Step 1. The two legs are perpendicular, so they form a ready-made base-height pair: $A = \frac12 \times 9 \times 12 = \frac{108}{2} = 54\ \text{cm}^2$ .

Step 2. Since the area is fixed at $54\ \text{cm}^2$ regardless of which side is used as base, set up the same area using the hypotenuse $15\ \text{cm}$ as base with unknown height $h$ : $54 = \frac12 \times 15 \times h$ .

Step 3. Solve for $h$ : $h = \dfrac{54 \times 2}{15} = \dfrac{108}{15} = 7.2\ \text{cm}$ . This height lies inside the triangle, perpendicular to the hypotenuse, and — sensibly — is shorter than either leg.

Step 4. Parallelogram check: doubling this triangle (rotating a copy about the midpoint of the hypotenuse) gives a parallelogram of base $15\ \text{cm}$ and height $7.2\ \text{cm}$ , with area $15 \times 7.2 = 108\ \text{cm}^2$ — exactly double the triangle's area, confirming the halving relationship.

The triangle's area is $54\ \text{cm}^2$ , and the height corresponding to the $15\ \text{cm}$ hypotenuse is $7.2\ \text{cm}$ .

Competitive stretch

If instead only the area ( $54\ \text{cm}^2$ ) and one leg-as-base ( $9\ \text{cm}$ ) were given, the corresponding height is $h = \frac{54 \times 2}{9} = 12\ \text{cm}$ — exactly the other leg, since legs of a right triangle are always mutual base-height pairs.

Common traps

Using a slanted side length in place of the true perpendicular height — height must be measured at right angles to whichever side is the base.
Dropping the factor of $\frac12$ and computing base $\times$ height as though finding a parallelogram's area instead.
Assuming the height must be one of the triangle's own sides — in scalene or obtuse triangles it is usually a separate segment drawn to the base or its extension.
Believing that switching the base changes the triangle's actual area — the base-height pair changes, the answer never does.

Ask the tutor: Not sure which segment counts as the height in an obtuse or scalene triangle? Ask the AI tutor to draw a few labelled triangles with you and identify the height in each.

Why does the height sometimes fall outside the triangle?

This happens in obtuse triangles when the base you choose is not adjacent to the obtuse angle. The perpendicular from the opposite vertex then meets the extension of the base rather than the base segment itself. The formula $A = \frac12 bh$ is still exactly correct; you just have to imagine or draw the base line stretched beyond the triangle to locate where the perpendicular foot lands.

Can I use any two sides of a triangle as base and height, like I would with a rectangle?

No — only a side and its own true perpendicular height work, and for a general triangle no two sides are perpendicular to each other. The one convenient exception is a right-angled triangle, where the two legs happen to meet at $90^\circ$ , so they can be used directly as a base-height pair without drawing any extra line.

Do I have to redraw the parallelogram every time to find a triangle's area?

No — the parallelogram-completion picture is only there to justify why the formula has a $\frac12$ in it. Once you trust $A = \frac12 bh$ , you just measure or are given a base and its perpendicular height and substitute directly; the parallelogram argument does not need to be repeated for every problem, only understood once.

Area of a triangle

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads