Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 10B Problems/Problem 9"

m (→‎Solution 3 (Shortcut))
Line 29: Line 29:
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
  
−
== Solution ==
+
== Solution 1==
  
 
<math>\triangle ABC \sim \triangle EBD</math> by AA Similarity. Therefore <math>DE = \frac{3}{4} BD</math>. Find the areas of the triangles.
 
<math>\triangle ABC \sim \triangle EBD</math> by AA Similarity. Therefore <math>DE = \frac{3}{4} BD</math>. Find the areas of the triangles.

Revision as of 19:17, 1 May 2021

Problem

The area of $\triangle$$EBD$ is one third of the area of $\triangle$$ABC$. Segment $DE$ 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]

$\textbf{(A)}\ \frac{4}{3} \qquad\textbf{(B)}\ \sqrt{5} \qquad\textbf{(C)}\ \frac{9}{4} \qquad\textbf{(D)}\ \frac{4\sqrt{3}}{3} \qquad\textbf{(E)}\ \frac{5}{2}$

Solution 1

$\triangle ABC \sim \triangle EBD$ by AA Similarity. Therefore $DE = \frac{3}{4} BD$. Find the areas of the triangles. \[\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6\] \[\triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2\] The area of $\triangle EBD$ is one third of the area of $\triangle ABC$. \begin{align*} \frac{3}{8} BD^2 &= 6 \times \frac{1}{3}\\ 9BD^2 &= 48\\ BD^2 &= \frac{16}{3}\\ BD &= \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}} \end{align*}

Solution 2

$\triangle ABC \sim \triangle EBD$ by AA Similarity. Since the area of $\triangle EBD$ is $\frac{1}{3}$ of $\triangle ABC$ and the bases/heights are in the same ratio, we use the formula forarea of a triangle for these ratios. Thus, \[\frac{1}{3}[\triangle ABC] = [\triangle EBD]\] \[\frac{1}{3} \times \frac{1}{2} \times AC \times BC = \frac{1}{2} \times ED \times DB\] In order to scale the sides of ED and DB to make $\frac{1}{3}$ (since the ratios of sides are the same), we take the square root of $\frac{1}{3} = \frac{\sqrt(3)}{3}$ to scale each side by the same amount.

Thus $BD = 4 \times \frac{\sqrt(3)}{3}$ and the answer is $BD = \boxed{\textbf{(D)} \frac{4\sqrt{3}}{3}}$

Solution 3 (Shortcut)

The ratio of the areas of $\triangle$$EBD$ and $\triangle$$ABC$ is $1 : 3$, meaning the ratio of the sides is $1 : \sqrt{3}$. The only answer choice involving $\sqrt{3}$ is $\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}$.

-Solution by Joeya

See Also

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_9&oldid=153070"