A triangle; given side and two heights

Question

A triangle $ABC$ is given $BC=8$ and $AE=4$ and $BK=6,4$ are heights. Find $AC$ and $AB$.

So we can find the area of $ABC$: $$S=\dfrac{BC\cdot AE}{2}=16.$$ So $AC=\dfrac{2S}{BK}=\dfrac{32}{6,4}=5.$ How can I find $AB$ from here?

Answer 1

Note that the area can also be expressed as $$S= 16= \frac12 AC \cdot BC \sin C= \frac12 \cdot 5 \cdot 8 \sin C $$ which leads to $$ \sin C = \frac45\implies \cos C= \pm \sqrt{1-\sin^2 C}=\pm \frac35$$ Per the cosine rule $$AB=\sqrt{ AC^2 +BC^2 -2AC\cdot BC \cos C } = \sqrt{ 25+64\pm 48}=\sqrt{41},\>\sqrt{137} $$

Answer 2

Just use Pythagorean theorem. To be specific:

1. we have $AC=5, AE=4$, use Pythagorean theorem for triangle AEC,so $EC=3\rightarrow BE=BC-EC=5$
2. use Pythagorean theorem for triangle AEB, we have $AB=\sqrt{4^{2}+5^{2}}=\sqrt{41}$

Answer 3

Let $|BC|=a=8$, $|AE|=h_a=4$, $|BK|=h_b=6.4$, $|AC|=b$, $|AB|=c$.

As you already found, $|AC|=b=5$, we can use Heron’s formula for the area in terms of squared lengths of the sides $a,b,c$:

\begin{align} 16\,S^2&= 4a^2 b^2-(a^2+b^2-c^2)^2 , \end{align}

which provides two valid solutions,

\begin{align} c^2&=a^2+b^2-\sqrt{4a^2 b^2-16S^2} \\ \text{and }\quad c^2&=a^2+b^2+\sqrt{4a^2 b^2-16S^2} . \end{align}

Using known value for $S=16$, we arrive at two solutions, \begin{align} c&=\sqrt{41} \\ \text{and }\quad c&=\sqrt{137} , \end{align}

both fit the given conditions, see the images: