Difference between revisions of "1954 AHSME Problems/Problem 8"
Katzrockso (talk | contribs) (Created page with "== Problem 8== The base of a triangle is twice as long as a side of a square and their areas are the same. Then the ratio of the altitude of the triangle to the side of the ...") |
|||
| Line 9: | Line 9: | ||
Let the base and altitude of the triangle be <math>b, h</math>, the common area <math>A</math>, and the side of the square <math>s</math>. Then <math>\frac{2sh}{2}=s^2\implies sh=s^2\implies s=h</math>, so <math>\fbox{C}</math> | Let the base and altitude of the triangle be <math>b, h</math>, the common area <math>A</math>, and the side of the square <math>s</math>. Then <math>\frac{2sh}{2}=s^2\implies sh=s^2\implies s=h</math>, so <math>\fbox{C}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1954|num-b=7|num-a=9}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 19:38, 17 February 2020
Problem 8
The base of a triangle is twice as long as a side of a square and their areas are the same. Then the ratio of the altitude of the triangle to the side of the square is:
Solution
Let the base and altitude of the triangle be 
, the common area 
, and the side of the square 
. Then 
, so 
See Also
| 1954 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
