euclidean-geometrygeometryplane-geometry
The sides of an acute-angled triangle measure
$3\sqrt2$, $\sqrt{26}$ and $\sqrt{20}$.
Calculate the area of the triangle (Answer:$9$)
My progress…
Is there any way other than Heron's formula since the accounts would be laborious or algebraic manipulation for the resolution?
$p=\frac{\sqrt{18}+\sqrt{20}+\sqrt{26}}{2}\implies S_{ABC} =\sqrt{p(p-\sqrt{20})(p-\sqrt{26})(p-\sqrt{18})}$
Here is an approach that avoids trigonometry. In $\triangle ADH$, $\frac{AH}{AD} = \frac{3}{5} \implies AD = R = \frac{5}{3} AH$.
As $AH = AC/2$, $R = \frac{5}{6} AC \tag1$
$H$ is the circumcenter of right triangles $\triangle AFC$ and $\triangle AEC$.
So, $FH = EH = AC/2$
$\angle AHF = 180^\circ - 2 \angle A, \angle EHC = 180^\circ - 2 \angle C$
That leads to $\angle EHF = 180^\circ - 2 \angle B$
As $\triangle EHF$ is isosceles and $M$ is the foot of perpendicular from $H$ to $FE$,
$\angle HFM = \angle HEM = \angle B$ and $FM = ME = 12$.
In $\triangle FHM$, $ \displaystyle \frac{FM}{FH} = \frac{4}{5}$
$FH = 15 = AC/2 \implies AC = 30$
Using $(1)$, $R = 25$
$15^\circ$ is not the correct answer. GeoGebra shows the same.
In $\triangle ABD$, traversal $EC$ intersects $BA$ and $BD$ internally and $AD$ externally. Applying Menelaus's theorem,
$ \displaystyle \frac{AC}{CD} \cdot \frac{DN}{NB} \cdot \frac{BE}{EA} = 1 \implies EA = 3 BE$
Or, $EM + MA = 3 BM - 3 EM$.
As $BM = MA$, $BM =2 EM \implies E$ is midpoint of $BM$.
As $CE$ is also the angle bisector, $BC = CM = \sqrt3$ and $CE \perp AB$
If $BE = x$, applying Pythagoras in $\triangle AEC$ and $\triangle BEC$,
$CE^2 = BC^2 - x^2 = AC^2 - (3x)^2 \implies 3 - x^2 = 9 - 9 x^2$
$ \displaystyle x = \frac{\sqrt3}{2}, AB = 2 \sqrt3$
As $BC = CM = MB$, $\triangle BCM$ is equilateral triangle and $\angle ABC = 60^\circ$.
Now given $AD:DC = BA:BC$, $BD$ is angle bisector of $\angle ABC$.
$\therefore \angle ABD = 30^\circ$
(As a side note, $\triangle ACB$ is right triangle with $\angle ACB = 90^\circ$)
Best Answer
Here is a method (generating two different proofs in fact) that does not use Heron's formula.
Having observed the following relationships:
$$20=4^2+2^2, \ \ \ \ 26=5^2+1^2,$$
we can "embed" the triangle into a coordinate grid in this way:
where we have, in the different right triangles, Pythagoras theorem giving:
$$AB=\sqrt{20}=\sqrt{5^2+1^2}, \ \ AC=3 \sqrt{2}=\sqrt{3^2+3^2}, \ \ BC=\sqrt{4^2+2^2}$$
It remains to subtract to the area of rectangle $DEBF$ the areas of the right triangles:
$$4 \times 5 - (5 \times 1/2 + 4 \times 2/2+3\times 3/2)=20-11=9.$$
In fact, with this figure, we do not need to do all these calculations. It is simpler to use Pick's theorem, saying that the area of a polygon whose vertices are on a square grid is given by:
$$A=\color{red}{i}+\frac{b}{2}-1$$
where $i$ is the number of interior points and $b$ the number of boundary points, giving in our case:
$$A=7+\frac{6}{2}-1=9$$
(Many thanks to A-B-C who has recalled me this method I had forgotten).