Difference between revisions of "1995 AHSME Problems/Problem 8"
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<math>\triangle BDE</math> is similar to <math>\triangle BAC</math> by angle-angle similarity since <math>E=C = 90^\circ</math> and <math>B=B</math>, and thus <math>\frac{BD}{BA} = \frac{DE}{AC}</math>. | <math>\triangle BDE</math> is similar to <math>\triangle BAC</math> by angle-angle similarity since <math>E=C = 90^\circ</math> and <math>B=B</math>, and thus <math>\frac{BD}{BA} = \frac{DE}{AC}</math>. | ||
− | Solving the above for <math>BD</math>, we get <math>BD=\frac{BA\cdot DE | + | Solving the above for <math>BD</math>, we get <math>BD=\frac{BA\cdot DE}{AC} = 10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}</math>. |
==See also== | ==See also== | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:58, 5 July 2013
Problem
In , and . Points and are on and , respectively, and . If , then
Solution
is a right triangle with hypotenuse .
is similar to by angle-angle similarity since and , and thus .
Solving the above for , we get .
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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