Difference between revisions of "1964 AHSME Problems/Problem 29"
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We examine <math>\triangle RDF</math> and <math>\triangle SFR</math>. We are given <math>\angle RDF \cong \angle SFR</math>. Also note that <math>\frac{SF}{RD} = \frac{7.5}{6} = 1.25</math> and <math>\frac{FR}{DF} = \frac{5}{4} = 1.25</math>, so <math>\frac{SF}{RD} = \frac{FR}{DF}</math>. | We examine <math>\triangle RDF</math> and <math>\triangle SFR</math>. We are given <math>\angle RDF \cong \angle SFR</math>. Also note that <math>\frac{SF}{RD} = \frac{7.5}{6} = 1.25</math> and <math>\frac{FR}{DF} = \frac{5}{4} = 1.25</math>, so <math>\frac{SF}{RD} = \frac{FR}{DF}</math>. | ||
− | If two triangles have two pairs of sides that are proportional, and the included | + | If two triangles have two pairs of sides that are proportional, and the included angles are congruent, then the two triangles are similar by SAS congruence. Therefore, the third pair of sides must also be in the same proportion, so |
<math>\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25</math> | <math>\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25</math> |
Latest revision as of 23:35, 24 July 2019
Problem
In this figure , inches, inches, inches, inches. The length of , in inches, is:
Solution
We examine and . We are given . Also note that and , so .
If two triangles have two pairs of sides that are proportional, and the included angles are congruent, then the two triangles are similar by SAS congruence. Therefore, the third pair of sides must also be in the same proportion, so
, which is answer
See Also
1964 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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