Difference between revisions of "2011 AMC 10B Problems/Problem 9"
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<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. | ||
− | <cmath>\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6 | + | <cmath>\triangle ABC: 3 \times 4 \times \frac{1}{2} = 6</cmath> |
− | \triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2</cmath> | + | <cmath>\triangle EBD: BD \times \frac{3}{4} BD \times \frac{1}{2} = \frac{3}{8} BD ^2</cmath> |
The area of <math>\triangle EBD</math> is one third of the area of <math>\triangle ABC</math>. | The area of <math>\triangle EBD</math> is one third of the area of <math>\triangle ABC</math>. | ||
<cmath> | <cmath> |
Revision as of 13:27, 27 July 2018
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
by AA Similarity. Therefore
. Find the areas of the triangles.
The area of
is one third of the area of
.
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 |
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