Difference between revisions of "1956 AHSME Problems/Problem 38"
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− | The area of this triangle can be expressed in two ways; the first being <math>\frac{ab}{2}</math> (as this is a right triangle), and the second being <math>\frac{cx}{2}</math>. But by the Pythagorean Theorem, <math>c=\sqrt{a^2+b^2}</math>. Thus a second way of finding the area of the triangle is <math>\frac{x\sqrt{a^2+b^2}}{2}</math>. By setting them equal to each other we get <math>x=\frac{ab}{\sqrt{a^2+b^2}}</math>, and we can observe that the correct answer | + | The area of this triangle can be expressed in two ways; the first being <math>\frac{ab}{2}</math> (as this is a right triangle), and the second being <math>\frac{cx}{2}</math>. But by the [[Pythagorean Theorem]], <math>c=\sqrt{a^2+b^2}</math>. Thus a second way of finding the area of the triangle is <math>\frac{x\sqrt{a^2+b^2}}{2}</math>. By setting them equal to each other we get <math>x=\frac{ab}{\sqrt{a^2+b^2}}</math>, and we can observe that the correct answer is <math>\boxed{\textbf{(D) \ }}</math>. |
~anduran | ~anduran | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AHSME box|year=1956|num-b=37|num-a=39}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:AHSME]][[Category:AHSME Problems]] |
Latest revision as of 23:35, 2 January 2024
Problem 38
In a right triangle with sides and
, and hypotenuse
, the altitude drawn on the hypotenuse is
. Then:
Solution
The area of this triangle can be expressed in two ways; the first being (as this is a right triangle), and the second being
. But by the Pythagorean Theorem,
. Thus a second way of finding the area of the triangle is
. By setting them equal to each other we get
, and we can observe that the correct answer is
.
~anduran
See Also
1956 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 37 |
Followed by Problem 39 | |
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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