A geometry problem by A Former Brilliant Member

The sum of interior angles of a triangle is $180^\circ$ . Since the ratio is $1:2:6$ , the total is $1+2+6=9$ , so the measure of the angles are $\dfrac{1}{9}(180)=20^\circ$ , $\dfrac{2}{9}(180)=40^\circ$ and $\dfrac{6}{9}(180)=120^\circ$ . We let $a$ , $b$ and $c$ be the sides of the triangle. So, $a+b+c=24$ . Then by sine law,

$\dfrac{a}{sin~120}=\dfrac{b}{sin~40}=\dfrac{c}{sin~20}$

Adding the lengths of the three sides, we get

$a+b+c=a+a\dfrac{sin~40}{sin~120}+a\dfrac{sin~20}{sin~120}=24$ $\implies$ $a=11.22986673$

It follows that,

$b=11.22986673\dfrac{sin~40}{sin~120}=8.335112528$

Finally, the area of the triangle is

$A=\dfrac{1}{2}ab~sin~20=\dfrac{1}{2}(11.22986673)(8.335112528)(sin~20)=$ $\boxed{16~ft^2}$

Let the angles of the triangle be $A$ , $B$ and $C$ , and their corresponding opposite sides be $a$ , $b$ and $c$ respectively. Further let $A : B:C = 1:2:6$ . Since $A+B+C=180^\circ$ , $\implies A=20^\circ$ , $B =40^\circ$ and $C=120^\circ$ .

By sine rule , we have:

$\dfrac a{\sin 20^\circ} = \dfrac b{\sin 40^\circ} = \dfrac c{\sin 120^\circ} = k \implies \begin{cases} a = k \sin 20^\circ \\ b = k \sin 40^\circ \\ c = k \sin 120^\circ \end{cases}$

$\implies a+b+c = k (\sin 20^\circ + \sin 40^\circ + \sin 120^\circ) = 24$ .

$\implies k = \dfrac {24}{\sin 20^\circ + \sin 40^\circ + \sin 120^\circ}$

Area of the triangle is given by:

$\begin{aligned} A & = \frac 12 ab \sin C \\ & = \frac 12 \cdot k \sin 20^\circ \cdot k \sin 40^\circ \cdot \sin 120^\circ \\ & = \frac {24^2 \sin 20^\circ \sin 40^\circ \sin 120^\circ}{2(\sin 20^\circ + \sin 40^\circ + \sin 120^\circ)^2} \\ & = 16.007 \\ & \approx \boxed{16} \end{aligned}$