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{(p + b) - h} divided by 2

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{(p + b) - h} divided by 2

by sanju09 » Wed Feb 25, 2009 3:25 am
What is the radius of the incircle of the triangle whose sides measure 5, 12 and 13 units?

A. 2 units
B. 12 units
C. 6.5 units
D. 6 units
E. 7.5 units
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by DanaJ » Wed Feb 25, 2009 4:02 am
Again, this is a problem involving a right triangle. You will notice that:
12^2 = 144
5^2 = 25
13^2 = 169
After noticing this, you have that 13^2 = 12^2 + 5^2 or that the triangle in question is a right triangle with a 13 hypotenuse.
It's useful to know that the length of the radius of an incircle is equal to 2*(Area)/(Perimeter).

In this case, Area = 5*12/12 = 30 (since this is a right triangle)
Perimeter = 12 + 13 + 5 = 30.
So the radius will be 2*30/30 = 2.
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by bluementor » Wed Feb 25, 2009 5:30 am
The radius clearly has to be less than 5, the shortest edge of the triangle.

The only possible option is A.

-BM-
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by x2suresh » Wed Feb 25, 2009 9:37 am
bluementor wrote:The radius clearly has to be less than 5, the shortest edge of the triangle.

The only possible option is A.

-BM-
agreed.



Let's consider the general situation of a right triangle:

B
| \
| Z I is the center of the inscribed circle.
X I \ X,Y and Z are the points where the inscribed
| \ circle meets the sides, so XI = YI = ZI = r.
C---Y----------A

I write a = BC (length), b = AC and c = AB.

We know that area(ABC) = 0.5*a*b .........................[1].

Also area(ABC) = area(ABI) + area(CIA) + area(BCI)
= 0.5*r*a + 0.5*r*b + 0.5*r*c
= 0.5*r*(a+b+c) .........................[2].

From [1] = [2] we find the formula:

r = a*b/(a+b+c). = 2*Area /Perimeter


https://mathforum.org/library/drmath/view/55138.html
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by sanju09 » Thu Feb 26, 2009 2:47 am
How about the subject of this post? :)
The mind is everything. What you think you become. -Lord Buddha



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