Difference between revisions of "1950 AHSME Problems/Problem 35"

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The inradius is equal to the area divided by semiperimeter. The area is <math>(10)(24)/2 = 120</math> because it's a right triangle. The semiperimeter is <math>30</math>. Therefore the inradius is <math>4</math> (B)
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==Problem==
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In triangle <math>ABC</math>, <math>AC=24</math> inches, <math>BC=10</math> inches, <math>AB=26</math> inches. The radius of the inscribed circle is:
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<math>\textbf{(A)}\ 26\text{ in} \qquad
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\textbf{(B)}\ 4\text{ in} \qquad
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\textbf{(C)}\ 13\text{ in} \qquad
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\textbf{(D)}\ 8\text{ in} \qquad
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\textbf{(E)}\ \text{None of these}</math>
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==Solution==
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The inradius is equal to the area divided by semiperimeter. The area is <math>\frac{(10)(24)}{2} = 120</math> because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is <math>30</math>. Therefore the inradius is <math>\boxed{\textbf{(B)}\ 4}</math>.
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==Solution 2==
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Since this is a right triangle, we have
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<cmath>\frac{a+b-c}{2}=\boxed{4}</cmath>
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- kante314
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==See Also==
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{{AHSME 50p box|year=1950|num-b=34|num-a=36}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Revision as of 02:04, 26 September 2021

Problem

In triangle $ABC$, $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:

$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$

Solution

The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$. Therefore the inradius is $\boxed{\textbf{(B)}\ 4}$.

Solution 2

Since this is a right triangle, we have \[\frac{a+b-c}{2}=\boxed{4}\]

- kante314

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 34
Followed by
Problem 36
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All AHSME Problems and Solutions

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