Difference between revisions of "1969 AHSME Problems/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:
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<asy>
 +
draw(arc((0,-1),2,30,150),dashed+linewidth(.75));
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draw((-1.7,0)--(0,0)--(1.7,0),dot);
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draw((0,0)--(0,.98),dot);
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MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N);
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</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

Revision as of 22:40, 30 September 2014

Problem

[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 $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ inches from the center $M$ is:

$\text{(A) } 1\quad \text{(B) } 15\quad \text{(C) } 15\tfrac{1}{3}\quad \text{(D) } 15\tfrac{1}{2}\quad \text{(E) } 15\tfrac{3}{4}$

Solution

$\fbox{B}$

See also

1969 AHSC (ProblemsAnswer KeyResources)
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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