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

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==Solution==
 
==Solution==
 
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>.
 
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>.
 +
 +
==Solution 2==
 +
Since this is a right triangle, we have
 +
<cmath>\frac{a+b-c}{2}=\boxed{4}</cmath>
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 +
- kante314
  
 
==See Also==
 
==See Also==

Revision as of 01:04, 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 (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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