The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is $A = 1/2 × b × h$, where $b$ is the base and $h$ is the height of the given triangle. Here we have to find the side of triangle.
2011 AMC 10B-Problem 9 | Geometry
The area of $ \triangle EBD $ is one third of the area of $ \triangle ABC $. Segment is perpendicular to segment $ AB $. What is $ BD $?
![[asy] unitsize(10mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(5,0), C=(1.8,2.4), D=(5-4sqrt(3)/3,0), E=(5-4sqrt(3)/3,sqrt(3)); pair[] ps={A,B,C,D,E}; draw(A--B--C--cycle); draw(E--D); draw(rightanglemark(E,D,B)); dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NE); label("$3$",midpoint(A--C),NW); label("$4$",midpoint(C--B),NE); label("$5$",midpoint(A--B),SW); [/asy]](https://latex.artofproblemsolving.com/9/5/3/953e7cbe2b2ee84c7c3ff7f6c642569bc2483c4e.png)
(A) \(\frac{4}{3}\)
(B) \(\sqrt{5}\)
(C) \(\frac{9}{4}\)
(D) \(\frac{4 \sqrt{3}}{3}\)
(E) \(\frac{5}{2}\)
Challenges and Thrills in Pre College Mathematics
Notice that here \(\triangle A B C\) and \(\triangle B D E\) are similar.
Therefore \(D E=\frac{3}{4} B D\)
Now we have to find the area.
We know this is \(=\frac{1}{2} \times\) base\(\times\) height.
Now using this formula can you find the area of \(\triangle A B C\) and \(\triangle B D E\) ?
Now \(\triangle A B C=\frac{1}{2} \times 3 \times 4=6\) and \(\triangle B D E=\frac{1}{2} \times D E \times B D=\frac{1}{2} * \frac{3}{4} \times B D \times B D=\frac{3}{8} B D^2\)
Again we know the area of \(\triangle B D E\) is one third of the area of \(\triangle A B C\).
Therefore, \(\frac{3}{8} B D^2=6 \times \frac{1}{3}\)
\(9 * B D^2=48), or, (B D^2=\frac{48}{9}), or, (B D^2=\frac{16}{3}\)
so, the answer is= \(B D=\frac{4 \sqrt{3}}{3}\)
