
# 20.7: Area of solid triangles


Theorem $$\PageIndex{1}$$

Let $$a=BC$$ and $$h_A$$ to be the altitude from $$A$$ in $$\triangle ABC$$. Then

$$\text{area }(\blacktriangle ABC)=\tfrac12\cdot a\cdot h_A.$$

Proof

Draw the line $$m$$ thru $$A$$ that is parallel to $$(BC)$$ and line $$n$$ thru $$C$$ parallel to $$(AB)$$. Note that the lines $$m$$ and $$n$$ are not parallel; denote by $$D$$ their point of intersection. By construction, $$\square ABCD$$ is a parallelogram.

Note that $$\blacksquare ABCD$$ admits a subdivision into $$\blacktriangle ABC$$ and $$\blacktriangle CDA$$. Therefore, $\text{area }(\blacksquare ABCD) = \text{area }(\blacktriangle ABC) + \text{area }(\blacktriangle CDA)$

Since $$\square ABCD$$ is a parallelogram, Lemma 7.5.1 implies that

$$AB=CD \quad \text{and} \quad BC=DA.$$

Therefore, by the SSS congruence condition, we have $$\triangle ABC\cong\triangle CDA$$. In particular

$$\text{area }(\blacktriangle ABC) = \text{area }(\blacktriangle CDA).$$

From above and Proposition 20.6.1, we get that

\begin{aligned} \text{area }(\blacktriangle ABC) &=\tfrac12\cdot\text{area }(\blacksquare ABCD)= \\ &=\tfrac12\cdot h_A\cdot a\end{aligned}

Exercise $$\PageIndex{1}$$

Let $$h_A$$, $$h_B$$, and $$h_C$$ denote the altitudes of $$\triangle ABC$$ from vertices $$A$$, $$B$$ and $$C$$ respectively. Note that from Theorem $$\PageIndex{1}$$, it follows that

$$h_A\cdot BC=h_B\cdot CA=h_C\cdot AB.$$

Give a proof of this statement without using area.

Hint

Without loss of generality, we may assume that the angles $$ABC$$ and $$BCA$$ are acute.

Let $$A'$$ and $$B'$$ denote the foot points of $$A$$ and $$B$$ on $$(BC)$$ and $$(AC)$$ respectively. Note that $$h_A = AA'$$ and $$h_B = BB'$$.

Note that $$\triangle AA'C \sim \triangle BB'C$$; indeed the angle at $$C$$ is shared and the angles at $$A'$$ and $$B'$$ are right. In particular $$\dfrac{AA'}{BB'} = \dfrac{AC}{BC}$$ or, equivalently, $$h_A \cdot BC = h_B \cdot AC$$.

Exercise $$\PageIndex{2}$$

Assume $$M$$ lies inside the parallelogram $$ABCD$$; that is, $$M$$ belongs to the solid parallelogram $$\blacksquare ABCD$$, but does not lie on its sides. Show that

$$\text{area }(\blacktriangle ABM)+\text{area }(\blacktriangle CDM) =\tfrac12\cdot \text{area }(\blacksquare ABCD).$$

Hint

Draw the line $$\ell$$ thru $$M$$ parallel to $$[AB]$$ and $$[CD]$$; it subdivides $$\blacksquare ABCD$$ into two solid parallelograms which will be denoted by $$\blacksquare ABEF$$ and $$\blacksquare CDFE$$. In particular,

$$\text{area } (\blacksquare ABCD) = \text{area } (\blacksquare ABEF) + \text{area } (\blacksquare CDFE)$$.

By Proposition 20.6.1 and Theorem $$\PageIndex{1}$$ we get that

$$\begin{array} {l} {\text{area } (\blacktriangle ABM) = \dfrac{1}{2} \cdot \text{area } (\blacksquare ABEF),} \\ {\text{area } (\blacktriangle CDM) = \dfrac{1}{2} \cdot \text{area } (\blacksquare CDFE)} \end{array}$$

and hence the result.

Exercise $$\PageIndex{3}$$

Assume that diagonals of a nondegenerate quadrangle $$ABCD$$ intersect at point $$M$$. Show that

$$\text{area }(\blacktriangle ABM)\cdot\text{area }(\blacktriangle CDM) = \text{area }(\blacktriangle BCM)\cdot\text{area }(\blacktriangle DAM).$$

Hint

Let $$h_A$$ and $$h_C$$ denote the distances from $$A$$ and $$C$$ to the line $$(BD)$$ respectively. According to Theorem $$\PageIndex{1}$$,

$$\text{area }(\blacktriangle ABM) = \dfrac{1}{2} \cdot h_A \cdot BM$$;                $$\text{area }(\blacktriangle BCM) = \dfrac{1}{2} \cdot h_C \cdot BM$$;
$$\text{area }(\blacktriangle CDM) = \dfrac{1}{2} \cdot h_C \cdot DM$$;                $$\text{area }(\blacktriangle ABM) = \dfrac{1}{2} \cdot h_A \cdot DM$$.

Therefore

$$\begin{array} {rcl} {\text{area } (\blacktriangle ABM) \cdot \text{area } (\blacktriangle CDM)} & = & {\dfrac{1}{4} \cdot h_A \cdot h_C \cdot DM \cdot BM =} \\ {} & = & {\text{area } (\blacktriangle BCM) \cdot \text{area } (\blacktriangle DAM)} \end{array}$$

Exercise $$\PageIndex{4}$$

Let $$r$$ be the inradius of $$\triangle ABC$$ and $$p$$ be its semiperimeter; that is,$$p=\tfrac12\cdot(AB+BC+CA)$$. Show that

$$\text{area }(\blacktriangle ABC)=p\cdot r.$$

Hint

Let $$I$$ be the incenter of $$\triangle ABC$$. Note that $$\blacktriangle ABC$$ can be subdivided into $$\blacktriangle IAB, \blacktriangle IBC$$ and $$\blacktriangle ICA$$.

It remains to apply Theorem $$\PageIndex{1}$$ to each of these triangles and sum up the results.

Exercise $$\PageIndex{5}$$

Show that any polygonal set admits a subdivision into a finite collection of solid triangles and a degenerate set. Conclude that for any polygonal set, its area is uniquely defined.

Hint

Fix a polygonal set $$\mathcal{P}$$. Without loss of generality we may assume that $$\mathcal{P}$$ is a union of a finite collection of solid triangles. Cut $$\mathcal{P}$$ along the extensions of the sides of all the triangles, it subdivides $$\mathcal{P}$$ into convex polygons. Cutting each polygon by diagonals from one vertex produce a subdivision into solid triangles.