Difference between revisions of "1969 AHSME Problems/Problem 26"
(Created page with "== Problem == A parabolic arch has a height of <math>16</math> inches and a span of 40 inches. The height, in inches, of the arch at the point <math>5</math> inches from the cen...") |
Rockmanex3 (talk | contribs) (Solution to Problem 26) |
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== Problem == | == Problem == | ||
− | A parabolic arch has a height of <math>16</math> inches and a span of 40 inches. The height, in inches, of the arch at the point <math>5</math> inches from the center <math>M</math> is: | + | <asy> |
+ | draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); | ||
+ | draw((-1.7,0)--(0,0)--(1.7,0),dot); | ||
+ | draw((0,0)--(0,.98),dot); | ||
+ | MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); | ||
+ | </asy> | ||
+ | |||
+ | A parabolic arch has a height of <math>16</math> inches and a span of <math>40</math> inches. The height, in inches, of the arch at the point <math>5</math> inches from the center <math>M</math> is: | ||
<math>\text{(A) } 1\quad | <math>\text{(A) } 1\quad | ||
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== Solution == | == Solution == | ||
− | <math>\ | + | |
+ | Because the arch has a height of <math>16</math> inches, an equation that models the arch is <math>y = ax^2 + 16</math>, where <math>x</math> is the horizontal distance from the center and <math>y</math> is the height. The arch has a span of <math>40</math> inches, so the arch meets the ground <math>20</math> inches from the center. That means <math>0 = 400a + 16</math>, so <math>a = -\frac{1}{25}</math>. | ||
+ | |||
+ | Thus, the equation that models height based on distance from the center <math>y = -\frac{1}{25}x^2 + 16</math>, so the height of the arch <math>5</math> inches from the center is <math>-\frac{1}{25} \cdot 5^2 + 16 = \boxed{\textbf{(B) } 15}</math> inches. | ||
== See also == | == See also == |
Latest revision as of 19:41, 9 June 2018
Problem
A parabolic arch has a height of inches and a span of
inches. The height, in inches, of the arch at the point
inches from the center
is:
Solution
Because the arch has a height of inches, an equation that models the arch is
, where
is the horizontal distance from the center and
is the height. The arch has a span of
inches, so the arch meets the ground
inches from the center. That means
, so
.
Thus, the equation that models height based on distance from the center , so the height of the arch
inches from the center is
inches.
See also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 | ||
All AHSME Problems and Solutions |
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