Two Triangles That Look Nothing Alike Can Have Equal Area

Because area does not care about shape — only about base and height.

Slide the top vertex of a triangle sideways as far as you like. As long as it stays on the same line parallel to the base, the height never changes, so the area never changes. The triangle can end up looking severely lopsided and its area stays put.

That single idea runs through the whole first part of the CBSE Class 8 Mathematics chapter on area. This page covers it, the parallelogram it comes from, and the medians it explains.

Two figures are on the same base when they share one whole side, and between the same parallels when their opposite vertices or sides both lie on one line parallel to that base.

When both conditions hold, the figures have the same height, because parallel lines stay a constant distance apart.

The triangle result. Triangles on the same base and between the same parallels have equal area.

Take base $BC=10\text{cm}$ on one line, with a parallel line $4\text{cm}$ above it. Put the apex anywhere on that upper line — call three choices $A$, $P$ and $Q$. Every one of those triangles has

$$\text{Area}=\frac{1}{2}\times 10\times 4=20{\text{cm}}^2$$

The triangles look completely different. One may be isosceles, another badly stretched with its apex far off to the right, but the height from the upper line down to $BC$ is $4\text{cm}$ in every case.

The parallelogram result. Parallelograms on the same base and between the same parallels have equal area too, for the same reason: base $10\text{cm}$, height $4\text{cm}$, so area $40{\text{cm}}^2$ regardless of how far the top side is pushed sideways.

The converse is also useful. If two triangles on the same base have equal areas, then their apexes lie on a line parallel to that base. Equal area plus equal base forces equal height, and equal height means parallel.

A median joins a vertex to the midpoint of the opposite side. It creates two triangles with equal bases and the same height, so their areas must be equal.

Worked example. In triangle $ABC$, $BC=14\text{cm}$ and the height from $A$ to $BC$ is $6\text{cm}$:

$$\text{Area of }ABC=\frac{1}{2}\times 14\times 6=42{\text{cm}}^2$$

Let $D$ be the midpoint of $BC$, so $BD=DC=7\text{cm}$. Both smaller triangles keep the same apex $A$, so both keep the height $6\text{cm}$:

$$\text{Area of }ABD=\frac{1}{2}\times 7\times 6=21{\text{cm}}^2$$

$$\text{Area of }ADC=\frac{1}{2}\times 7\times 6=21{\text{cm}}^2$$

Equal halves, and together $21+21=42{\text{cm}}^2$. Correct.

Pushing it further. All three medians of a triangle meet at one point, and they cut the triangle into six small triangles of equal area. For the triangle above, each of the six has area

$$\frac{42}{6}=7{\text{cm}}^2$$

Notice that the two halves are usually not congruent — they need not have the same shape at all. Equal area is a weaker statement than congruence, and this is one of the clearest places to see the difference.

Substitute what you know into the area formula and solve for the unknown. Working backwards is the commonest way these results are tested.

Example 1 — a triangle's height. A triangle has area $54{\text{cm}}^2$ and base $12\text{cm}$:

$$54=\frac{1}{2}\times 12\times h=6h,h=\frac{54}{6}=9\text{cm}$$

Example 2 — two bases, two heights. A parallelogram has adjacent sides $8\text{cm}$ and $5\text{cm}$. The height on the $8\text{cm}$ side is $4.5\text{cm}$. Find the height on the $5\text{cm}$ side.

The area, computed with the matching pair:

$$\text{Area}=8\times 4.5=36{\text{cm}}^2$$

Now use the other side as the base:

$$5\times h_2=36,h_2=\frac{36}{5}=7.2\text{cm}$$

A sensible check: the shorter side needs the taller height, and $7.2>4.5$. That matches.

Example 3 — a side from the area. A triangle has area $30{\text{cm}}^2$ and height $5\text{cm}$:

$$30=\frac{1}{2}\times b\times 5=2.5b,b=\frac{30}{2.5}=12\text{cm}$$

In every case the method is the same: write the formula, substitute, solve. What changes is only which letter is missing.