Perimeter of Trapezium
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Theorem
Let $ABCD$ be a trapezium:
and
The perimeter $P$ of $ABCD$ is given by:
- $P = a + b + h \paren {\csc \theta + \csc \phi}$
where $\csc$ denotes cosecant.
Proof
The perimeter $P$ of $ABCD$ is given by:
- $P = AB + BC + CD + AD$
where the lines are used to indicate their length.
Thus:
\(\text {(1)}: \quad\) | \(\ds AB\) | \(=\) | \(\ds b\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds CD\) | \(=\) | \(\ds a\) |
\(\ds h\) | \(=\) | \(\ds AD \sin \theta\) | Definition of Sine of Angle | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds AD\) | \(=\) | \(\ds \frac h {\sin \theta}\) | |||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(=\) | \(\ds h \csc \theta\) | Cosecant is Reciprocal of Sine |
\(\ds h\) | \(=\) | \(\ds BC \sin \phi\) | Definition of Sine of Angle | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds BC\) | \(=\) | \(\ds \frac h {\sin \phi}\) | |||||||||||
\(\text {(4)}: \quad\) | \(\ds \) | \(=\) | \(\ds h \csc \phi\) | Cosecant is Reciprocal of Sine |
Hence:
\(\ds P\) | \(=\) | \(\ds AB + BC + CD + AD\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b + h \csc \phi + a + h \csc \theta\) | from $(1)$, $(2)$, $(3)$ and $(4)$ |
Hence the result.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 4$: Geometric Formulas: $4.8$