Triangle Properties #1

Since $\triangle ABC$ is a right triangle and $\angle ACB=30^{\circ}$ , we know that $AB:AC=1:2$ (because $\sin(30^{\circ})=\displaystyle{AB\over{AC}}=0.5$ ).

Since $\overline{AE}$ is drawn such that $\overline{AE} \perp \overline{BD}$ and $\overline{DE} = \overline{BE}$ , we can conclude that $\triangle ABD$ is isosceles ( $SAS$ postulate of similarity: $\overline{DE} = \overline{BE},\ \overline{AE}=\overline{AE},\ \angle AED = \angle AEB = 90^{\circ}$ , $\triangle AED$ is $\triangle AEB$ scaled by a factor of $1$ , and therefore is congruent to $\triangle AEB$ , then $\overline{AB}=\overline{AD}$ ).

Following, we can split $\overline{BD}$ in half (3.5), and use Pythagoras’ Theorem to obtain $\overline{AD}$ (and $\overline{AB}$ ).

$\overline{AB} = \overline{AD} = \sqrt{\overline{AE}^{\,2}+\overline{DE}^{\,2}}=\sqrt{6^2+3.5^2}$ .

Since $AB:AC=1:2$ , $AC=2\sqrt{6^2+3.5^2}$ .

Now, we need to find $\overline{BC}$ .

$\overline{BC}= \sqrt{(2\sqrt{6^2+3.5^2})^2-(\sqrt{6^2+3.5^2})^2}=\sqrt{4(6^2+3.5^2)-(6^2+3.5^2)}=\sqrt{3(6^2+3.5^2)}=\sqrt{144.75}$

The area of $\triangle ABC$ is: $\displaystyle{{AB\times BC}\over {2}}={{\sqrt{144.75} \times \sqrt{6^2+3.5^2}}\over 2}\approx \boxed{42}$

ABC triangle area is the half of an equilateral triangle area that is $\frac{\sqrt {3}}{2} \cdot \overline{AB}^2 = \frac{\sqrt {3}}{2} \cdot (\overline{AE}^2 + ( \frac{\overline{BD}}{2})^2) = \frac{\sqrt {3}}{2} \cdot (6^2 + ( \frac{7}{2})^2) \simeq 41,8$ .

The nearest whole number is $42$ .