areageometrytriangles
I have this triangle, of which I want to find an area. Is the given information enough? The length of the dotted line is 2, whereas, the length of dotted + line is sqrt(5). I can get the area of two small right angled triangle in top with 1/2 * base * height. But how do I get the area of lower trapezium?
Thanks a lot!
The problem evidently intends the base to be the third side that is not necessarily equal to the other two.
The height of the perpendicular, then, is $15$, so the area is indeed $$\frac12\cdot 30 \cdot 15 = 225$$
You have assumed that the base and one of the other sides constitute the pair of equal sides.
I think your interpretation isn't incorrect, but is not what the problem intended.
This is from an article I remember reading in an AMS journal.
Let $D$ be the point of intersection of the internal bisectors of $\angle A$, $\angle B$, and $\angle C$. Let $\alpha$, $\beta$, and $\gamma$ be the measures of the corresponding bisected angles. Note $$\alpha + \beta + \gamma = \frac 12 \pi.$$ Let $r$ be the common distance from the point $D$ to the sides $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$. Define $$s = \frac 12(a+b+c).$$ Note that the feet of the perpendiculars drawn from point $D$ to the sides $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$ create segments with the indicated lengths. Let $u,v,$ and $w$ be the respective lengths of the segments drawn from point $D$ to points $A$, $B$, and $C$. Then
\begin{align} (s-a)+ir = ue^{i \alpha}\\ (s-b)+ir = ve^{i \beta}\\ (s-c)+ir = we^{i \gamma}\\ \end{align}
So $(s-a+ir)(s-b+ir)(s-c+ir)=uvwe^{\frac 12 i\pi}=iuvw$. Hence the real part of $(s-a+ir)(s-b+ir)(s-c+ir)$ is equal to $0$.
We compute \begin{align} \Re[(s-a+ir)(s-b+ir)(s-c+ir)] &= 0 \\ (s-a)(s-b)(s-c)-r^2[(s-a)+(s-b)+(s-c)] &= 0 \\ sr^2 &= (s-a)(s-b)(s-c) \\ r &= \sqrt{\frac{(s-a)(s-b)(s-c)}{s}} \end{align}
Hence the area of $\triangle ABC$ is
$$ \frac 12r(a+b+c) = rs = \sqrt{s(s-a)(s-b)(s-c)} $$
Best Answer
You can use similar Triangles. Using Pythagoras the side length of the smaller triangle is $\sqrt{1^2 + 2^2} = \sqrt{5}$.
Knowing that the side length of the bigger triangle is the side length of the smaller triangle $+ \sqrt{5}$ we have that the side length of the larger triangle is $2\sqrt{5}$ so the ratio of smaller:bigger is 1:2.
Can you figure out the height of the bigger triangle now and consequently the area of the bigger triangle?