Area of Triangle in Terms of Side and Altitude/Proof 1
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Theorem
The area of a triangle $\triangle ABC$ is given by:
- $\dfrac {c \cdot h_c} 2 = \dfrac {b \cdot h_b} 2 = \dfrac {a \cdot h_a} 2$
where:
Proof
Construct a point $D$ so that $\Box ABDC$ is a parallelogram.
From Opposite Sides and Angles of Parallelogram are Equal:
- $\triangle ABC \cong \triangle DCB$
hence their areas are equal.
The Area of Parallelogram is equal to the product of one of its bases and the associated altitude.
Thus
\(\ds \paren {ABCD}\) | \(=\) | \(\ds c \cdot h_c\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \paren {ABC}\) | \(=\) | \(\ds c \cdot h_c\) | because congruent surfaces have equal areas | ||||||||||
\(\ds \paren {ABC}\) | \(=\) | \(\ds \frac {c \cdot h_c} 2\) |
where $\paren {XYZ}$ is the area of the plane figure $XYZ$.
A similar argument can be used to show that the statement holds for the other sides.
$\blacksquare$
Motivation
This formula is perhaps the best-known and most useful for determining a triangle's area.
It is usually remembered, and quoted, as half base times height.