{(p + b) - h} divided by 2 - The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice

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

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

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

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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Source: Beat The GMAT — Problem Solving | Wed Feb 25, 2009 3:25 am

DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members; Thanked: 712 times; GMAT Score:770

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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bluementor: Master | Next Rank: 500 Posts; Posts: 418; Joined: Wed Jun 11, 2008 5:29 am; Thanked: 65 times

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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x2suresh: Master | Next Rank: 500 Posts; Posts: 258; Joined: Thu Aug 07, 2008 5:32 am; Thanked: 16 times

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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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

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

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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