Difference between revisions of "2011 AMC 10B Problems/Problem 9"
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[[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. | ||
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-Solution by Joeya | -Solution by Joeya | ||
+ | |||
+ | |||
+ | ==Remark (slightly more vigorous than Solution 3)== | ||
+ | The ratio of the areas is equal to twice the ratio of sides (in similar figures) because area is a second-degree property of similar figures. So like solution 3, the ratio of sides is <math>\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}=\sqrt{3}.</math> By similarity, <math>\frac{ED}{DB}=\frac{\sqrt{3}}{DB}=\frac{3}{4}</math>, so solving for DB, we get <math>\boxed{\textbf{(D) } \frac{4\sqrt{3}}{3}}</math>. | ||
+ | |||
+ | ~JH. L | ||
+ | |||
+ | ==Solution 4== | ||
+ | It is apparent that <math>\Delta ABC~\Delta EBD</math> by <math>AA</math> similarity (<math>\angle B=\angle B</math> and <math>\angle EDB=\angle ACB</math>). Thus, let the side length of <math>ED</math> equal <math>3x</math> and <math>DB=4x.</math> We can then see that <math>[EDB]=\dfrac{3x\cdot4x}2=6x^2</math>, and we are given that <math>[ABC]=3\cdot[EDB]</math>. Thus, <math>\dfrac{3\cdot4}2=3\cdot6x^2\implies6=18x^2\implies x=\dfrac{\sqrt{3}}3</math>. Since we let <math>BD=4x</math>, we know that <math>BD=\boxed{\textbf{D}~\dfrac{4\sqrt3}3}</math>. | ||
+ | ~Technodoggo | ||
== See Also== | == See Also== |
Latest revision as of 22:08, 31 August 2023
Contents
Problem
The area of is one third of the area of
. Segment
is perpendicular to segment
. What is
?
![[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]](http://latex.artofproblemsolving.com/9/5/3/953e7cbe2b2ee84c7c3ff7f6c642569bc2483c4e.png)
Solution 1
by AA Similarity. Therefore
. Find the areas of the triangles.
The area of
is one third of the area of
.
Solution 2
by AA Similarity. Since the area of
is
of
and the bases/heights are in the same ratio, we use the formula forarea of a triangle for these ratios. Thus,
In order to scale the sides of ED and DB to make
(since the ratios of sides are the same), we take the square root of
to scale each side by the same amount.
Thus and the answer is
Solution 3 (Shortcut)
The ratio of the areas of and
is
, meaning the ratio of the sides is
. The only answer choice involving
is
.
-Solution by Joeya
Remark (slightly more vigorous than Solution 3)
The ratio of the areas is equal to twice the ratio of sides (in similar figures) because area is a second-degree property of similar figures. So like solution 3, the ratio of sides is By similarity,
, so solving for DB, we get
.
~JH. L
Solution 4
It is apparent that by
similarity (
and
). Thus, let the side length of
equal
and
We can then see that
, and we are given that
. Thus,
. Since we let
, we know that
.
~Technodoggo
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.