What's the value of the area of the triangle with sides $3\sqrt2$, $\sqrt{26}$, and $\sqrt{20}$?

Question

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?

$p=\frac{\sqrt{18}+\sqrt{20}+\sqrt{26}}{2}\implies S_{ABC} =\sqrt{p(p-\sqrt{20})(p-\sqrt{26})(p-\sqrt{18})}$

Answer 1

Heron's formula, with sides $a,\,b,\,c$ equal to square roots of rational numbers, is easier to use if you rewrite it as$$\tfrac14\sqrt{4b^2c^2-(b^2+c^2-a^2)^2},$$which is the form in which it's first obtained in a proof of Heron's formula from the cosine rule. In the case $a^2=18,\,b^2=26,\,c^2=20$, we get $\tfrac14\sqrt{2080-28^2}=\sqrt{130-49}=9$.

Answer 2

Use cosine law,

$c^2=a^2+b^2-2ab\cos{\gamma}$

$(\sqrt{26})^2=(3\sqrt2)^2+(\sqrt{20})^2-2\cdot3\sqrt2\sqrt{20}cos\gamma$

$\implies cos(\gamma) =\frac{1}{\sqrt{10}}\Rightarrow sin(\gamma)=\frac{3}{\sqrt{10}}$

$A=\frac12\cdot 3\sqrt2\cdot \sqrt{20}\cdot sin(\gamma)$

$\implies...A=9$

Answer 3

If $a = \sqrt{18}, b = \sqrt{20}, c = \sqrt{26}$, first note that $a^2 + b^2 \gt c^2$ where $c$ is the longest side. That means it is an acute-angled triangle and the orthocenter is inside the triangle. So if we drop a perp from vertex $A$, then using Pythagoras,

$AD^2 = 20 - x^2 = 26 - (\sqrt{18} - x)^2$

$ \implies x = \sqrt 2$ and we get $AD = \sqrt{18}$

So, $A = \frac 12 a \cdot AD = 9$

By the way, if you notice that $26 - 8 = 20 - 2$ and $\sqrt{8} + \sqrt{2} = a$, you can get to altitude $AD = \sqrt{18} ~$ a bit faster.

Answer 4

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).