Properties of Triangles 2 Question 12
12. In a triangle of base $a$, the ratio of the other two sides is $r(<1)$. Show that the altitude of the triangle is less than or equal to $\frac{a r}{1-r^{2}}$.
(1991, 4M)
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Solution:
- Let $A B C$ be a triangle with base $B C=a$ and altitude $A D=p$, then
Area of $\triangle A B C=\frac{1}{2} b c \sin A$
Also, area of $\triangle A B C=\frac{1}{2} a p$
$\therefore \quad \frac{1}{2} a p=\frac{1}{2} b c \sin A$
$\Rightarrow \quad p=\frac{b c \sin A}{a}$
$\Rightarrow \quad p=\frac{a b c \sin A}{a^{2}}$
$\Rightarrow \quad p=\frac{a b c \sin A \cdot\left(\sin ^{2} B-\sin ^{2} C\right)}{a^{2}\left(\sin ^{2} B-\sin ^{2} C\right)}$
$$ \begin{aligned} & =\frac{a b c \sin A \cdot \sin (B+C) \sin (B-C)}{\left(b^{2} \sin ^{2} A-c^{2} \sin ^{2} A\right)} \\ & \text { using sine rule, } \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ & =\frac{a b c \sin ^{2} A \cdot \sin (B-C)}{\left(b^{2}-c^{2}\right) \cdot \sin ^{2} A}=\frac{a b c \sin (B-C)}{b^{2}-c^{2}} \\ & =\frac{a b^{2} r \sin (B-C)}{b^{2}-b^{2} r^{2}}=\frac{a r \sin (B-C)}{1-r^{2}} \\ \Rightarrow \quad p & \leq \frac{a r}{1-r^{2}} \quad[\because \sin (B-C) \leq 1] \end{aligned} $$
