Geometry problem related to right angled triangles.

Question

In the given figure $AC = 12 cm, AE = 6 cm$ and $CD = 8 cm$. CD is perpendicular to $AD$ and $BE$ is perpendicular to $AC$. How can we find the value of $BE$?

Answer 1

Consider that triangle $ABE$ is similar to triangle $ACD$, therefore $$\begin{align}&\frac{BE}{AE}=\frac{8cm}{12cm}\\\to&\frac{BE}{6cm}=\frac{8cm}{12cm}\\\to&BE=\frac{48}{12}cm=4cm\end{align}$$

Answer 2

Area of ACE = $ 6*x $

Area of CED is also known(Get length of ED as you know other two sides).Add this to equate to total area of right triangle

Answer 3

This is a method without using similar triangles:

First of all, we can find $\text{AD}$ by using the Pythagorean Theorem:

$$\text{AD}=\sqrt{\text{AC}^2-\text{CD}^2}\to\text{AD}=\sqrt{12^2-8^2}=4\sqrt{5}\space\text{cm}$$

So, for $\text{DE}$:

$$\text{DE}=\text{AD}-\text{AE}\to\text{DE}=\left(4\sqrt{5}-6\right)\space\text{cm}$$

Now, we can find $\text{CE}$ using again the Pythagorean Theorem:

$$\text{CE}=\sqrt{\text{CD}^2+\text{DE}^2}\to\text{CE}=\sqrt{8^2+\left(4\sqrt{5}-6\right)^2}=2\sqrt{45-12\sqrt{5}}\space\text{cm}$$

So, we also can say that:

$$ \begin{cases} \text{BE}^2+\left(12-\text{AB}\right)^2=180-48\sqrt{5}\\ \text{BE}^2+\text{AB}^2=36 \end{cases}\Longleftrightarrow \begin{cases} \text{BE}^2=180-48\sqrt{5}-\left(12-\text{AB}\right)^2\\ \text{BE}^2+\text{AB}^2=36 \end{cases} $$

Now, we can conclude:

$$\left(180-48\sqrt{5}-\left(12-\text{AB}\right)^2\right)+\text{AB}^2=36\Longleftrightarrow\text{AB}=2\sqrt{5}\space\text{cm}$$

So, for $\text{BE}$ we find:

$$\text{BE}^2=180-48\sqrt{5}-\left(12-2\sqrt{5}\right)^2=16\to\text{BE}=4\space\text{cm}$$