Area of a quadrilateral inside right angled triangle

Question

$ABC$ is right angle triangle. $AB=24 cm$, $BC=10 cm$, $AC=26 cm$. Point $D$ on $AC$ (hypotenuse) bisects $AC$ and connects point $E$ on side $AB$ such that $ED$ is perpendicular to $AC$. Side $AC$ is folded into half so that angle $A$ falls on angle $C$, creating line $ED$ perpendicular to hypotenuse $AC$ and bisecting side $AC$. What is the area of quadrilateral $BEDC$?

Answer 1

The area of the quadrilateral $BEDC$ is what's leftover if you subtract the triangle $ADE$ from triangle $ABC$. Since both triangles are right triangles,

$$area(BEDC)=area(ABC)-area(ADE)\\ =\frac12\|AB\|\cdot\|BC\|-\frac12\|AD\|\cdot\|DE\|\\ =\frac{24\cdot10}{2}-\frac{13}{2}\|DE\|.$$

To find $\|DE\|$, note that $\triangle ADE$ is similar to the $5,12,13$ right triangle, so

$$\frac{\|DE\|}{\|DA\|}=\frac{\|DE\|}{13}=\frac{5}{12}\implies \|DE\|=\frac{65}{12}.$$

Now just complete the arithmetic.

Answer 2

Let us look at $\triangle AED$ and $\triangle CED$:

• $\angle EDA = \angle EDC = 90\,^{\circ}$
• $DC= DA = 13$ cm
• $ED$ is common

So, by $SAS$ congruence, those two triangles are congruent, and hence $EC = AE$. Now let $BE$ be $x$. So, by Pythagoras Theorem, $(EC)^2=x^2+100$. But, $(EC)^2= (AE)^2 = (24-x)^2$. Therefore, we get an equation, $(24-x)^2= 576 + x^2 - 48x = x^2+100$. Which upon solving, we get, $x = \frac{119}{12}$ .

Back to $\triangle AED$, we see that $(ED)^2 = (24-x)^2 - 169$. So, we know $x$ and we can get $ED$ as well. The final touch, $$Area(BEDC) = Area(ABC) - Area(EDA) = \frac{24 \cdot 10}{2} - \frac{ED \cdot 13}{2}$$

Edit: Details of calculation: $(ED)^2 = (24 - x)^2 - 169 = (24 - \frac{119}{12})^2 - 169 = (\frac{169}{12})^2 - 169 = 169(\frac{169}{144}- 1)= \frac{169 \cdot 25 }{144}$ Therefore, $ED = \sqrt{\frac{169 \cdot 25 }{144}} = \frac{13 \cdot 5 }{12}$.

So, $$Area(BEDC) = \frac{24 \cdot 10}{2} - \frac{ED \cdot 13}{2} = 120 - \frac{13 \cdot 13 \cdot 5}{24} = 120- \frac{845}{24} = \frac{2035}{24} = 84.791666...$$