Let $ABC$ be a triangle with $AC = 18$ and $D$ is the point on the segment $AC$ such that $AD = 5$. Draw perpendiculars from $D$ on $AB$ and $BC$ which have lengths $4$ and $5$ respectively. Find the area of the triangle $ABC$.
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Tags: geometry, area of a triangle
Desimathematics
18.09.2021 17:29
Let the feet of perpendicular from D to AB , BC be X and Y then from pythagoras theorem we get AX = 3 and CY = 12 this gives $tanA = \frac{4}{3}$ and $tanB = \frac{5}{12}$ Let the feet of perpendicular from B upon AC be Z then we have $\frac{4}{3} = \frac{CZ}{AZ}$ and $\frac{5}{12} = \frac{CZ}{ZC}$ Solving this we get $AZ = \frac{30}{7}$ and $CZ = \frac{40}{7}$ area of triangle $ABC = \frac{CZ \cdot AC}{2} = \frac{360}{7}$
Grolarbear
18.09.2021 20:47
^^ Just to clarify, similar triangle relations between ADX ABZ and DYC BZC
OlympusHero
18.09.2021 21:09
Let the foot of the perpendicular from $D$ to $AB$ be $E$ and let the foot of the perpendicular from $D$ to $BC$ be $F$. Then $\sin \angle ADE = \frac{3}{5}, \cos \angle ADE = \frac{4}{5}, \sin \angle CDF = \frac{12}{13}, \cos \angle CDF = \frac{5}{13}$, so $\sin(\angle ADE + \angle CDF)=\frac{63}{65}$. But $\angle ADE + \angle CDF = \angle B$, so $\sin \angle B = \frac{63}{65}$. Then LoS gives $\frac{18}{\frac{63}{65}}=\frac{BC}{\frac{4}{5}}$, so $BC=\frac{104}{7}$. Our answer is hence $\frac{\frac{104}{7} \cdot 18 \cdot \frac{5}{13}}{2}=\boxed{\frac{360}{7}}$.