# Difference between revisions of "1972 AHSME Problems/Problem 27"

(Created page with "==Problem== If the area of <math>\triangle ABC</math> is <math>64</math> square units and the geometric mean (mean proportional) between sides <math>AB</math> and <math>AC</ma...") |
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==Solution== | ==Solution== | ||

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+ | Draw Diagram later | ||

+ | |||

+ | We can let <math>AB=s</math> and <math>AC=r</math>. We can also say that the area of a triangle is <math>\frac{1}{2}rs\sin A</math>, which we know, is <math>64</math>. We also know that the geometric mean of <math>r</math> and <math>s</math> is 12, so <math>\sqrt{rs}=12</math>. | ||

+ | <math>\newline</math> | ||

+ | Squaring both sides gives us that <math>rs=144</math>. We can substitute this value into the equation we had earlier. This gives us | ||

+ | <math>\newline</math> | ||

+ | <math>\frac{1}{2}\times144\times\sin A=64</math> | ||

+ | <math>\newline</math> | ||

+ | <math>72\sin A=64</math> | ||

+ | <math>\newline</math> | ||

+ | <math>\sin A=\frac{64}{72}=\frac{8}{9} \Rightarrow \boxed{\text{D}}</math> |

## Latest revision as of 13:27, 13 January 2018

## Problem

If the area of is square units and the geometric mean (mean proportional) between sides and is inches, then is equal to

## Solution

Draw Diagram later

We can let and . We can also say that the area of a triangle is , which we know, is . We also know that the geometric mean of and is 12, so . Squaring both sides gives us that . We can substitute this value into the equation we had earlier. This gives us