Difference between revisions of "1986 AHSME Problems/Problem 11"
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Now <math>AM=MB=6.5</math>. Using Stewart's Theorem on <math>\triangle{ABH}</math>, letting <math>HM=x</math>: | Now <math>AM=MB=6.5</math>. Using Stewart's Theorem on <math>\triangle{ABH}</math>, letting <math>HM=x</math>: | ||
− | <math>13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5</math> (<math>man+dad=bmb+cnc</math>). | + | <math>13*6.5*6.5+13x^2=12*6.5*12+5*6.5*5</math> (remember that Stewart's Theorem is <math>man+dad=bmb+cnc</math>). |
Thus <math>x=6.5</math> or <math>x=-6.5</math> (reject this solution since <math>x</math> is positive). Thus <math>HM=6.5</math>. Select <math>\boxed{B}</math>. | Thus <math>x=6.5</math> or <math>x=-6.5</math> (reject this solution since <math>x</math> is positive). Thus <math>HM=6.5</math>. Select <math>\boxed{B}</math>. |
Latest revision as of 10:41, 1 December 2022
Contents
[hide]Problem
In and . Also, is the midpoint of side and is the foot of the altitude from to . The length of is
Solution
In a right triangle, the length of the hypotenuse is twice the length of the median which bisects it. If the hypotenuse is , the median must be .
Solution 2 (Self Torture)
Warning: this solution is very intensive in calculation. Please do NOT try this on the test!
Let's start by finding . By Heron's Formula, . Using the area formula , . Now using the Pythagorean Theorem, .
Now . Using Stewart's Theorem on , letting :
(remember that Stewart's Theorem is ).
Thus or (reject this solution since is positive). Thus . Select .
~hastapasta
P.S.: Although this is torturous, this is a good practice of Heron's formula and Stewart's theorem though.
See also
1986 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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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