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radius of the circum circle

by sanju09 » Tue Apr 05, 2011 1:56 am
What is the radius of the circum circle of the triangle whose sides are 5, 12 and 13 respectively?
(A) 2
(B) 6
(C) 6 ½
(D) 7 ½
(E) 12
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by manpsingh87 » Tue Apr 05, 2011 2:07 am
sanju09 wrote:What is the radius of the circum circle of the triangle whose sides are 5, 12 and 13 respectively?
(A) 2
(B) 6
(C) 6 ½
(D) 7 ½
(E) 12
its a pythagorus triplet of sides 5,12,13 therefore its area is 1/2*5*12=30

circum radius= abc/4 (area of a triangle)=5*12*13/4*30=6(1/2) hence C
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by sanju09 » Tue Apr 05, 2011 2:25 am
manpsingh87 wrote:
sanju09 wrote:What is the radius of the circum circle of the triangle whose sides are 5, 12 and 13 respectively?
(A) 2
(B) 6
(C) 6 ½
(D) 7 ½
(E) 12
its a pythagorus triplet of sides 5,12,13 therefore its area is 1/2*5*12=30

circum radius= abc/4 (area of a triangle)=5*12*13/4*30=6(1/2) hence C
This is not an ideal approach that could be commonly exercised by the test takers mainly when they come from various other branch of studies than mathematics.

Beside, radius of the circum circle of a right triangle happens to be half of its hypotenuse. Please tell us why?
The mind is everything. What you think you become. -Lord Buddha



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by Anurag@Gurome » Tue Apr 05, 2011 2:39 am
sanju09 wrote:What is the radius of the circum circle of the triangle whose sides are 5, 12 and 13 respectively?
(A) 2
(B) 6
(C) 6 ½
(D) 7 ½
(E) 12
The center of the circum circle of a triangle is the point of intersection of the perpendicular bisectors of the triangle.
In this case the triangle with sides 5, 12, 13 is a right angled triangle, as (13)² = (12)² + (5)² holds true.
In case of a right angled triangle, is the diameter of circum circle, and its center is exactly at the mid point of the hypotenuse. (You can imagine this as the situation wherein the diameter subtends a right angle at any point on the circumference of circle).
Here hypotenuse = 13, so radius of circum circle = 13/2 = 6.5 units.

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The correct answer is C.
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by manpsingh87 » Tue Apr 05, 2011 2:40 am
sanju09 wrote:
manpsingh87 wrote:
sanju09 wrote:What is the radius of the circum circle of the triangle whose sides are 5, 12 and 13 respectively?
(A) 2
(B) 6
(C) 6 ½
(D) 7 ½
(E) 12
its a pythagorus triplet of sides 5,12,13 therefore its area is 1/2*5*12=30

circum radius= abc/4 (area of a triangle)=5*12*13/4*30=6(1/2) hence C
This is not an ideal approach that could be commonly exercised by the test takers mainly when they come from various other branch of studies than mathematics.

Beside, radius of the circum circle of a right triangle happens to be half of its hypotenuse. Please tell us why?
hi sanju its a theorem (may be thales ), according to which angle subtended by the diameter at circumference of the circle is right angle...!! i can prove the same thing for you but what's the purpose..???
i've remember few Pythagoras triplets which i found recurring in most of the question..1!!!

if you don't like this method than offcourse you can always follow the hero's formula to evaluate it.!!!
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by sanju09 » Tue Apr 05, 2011 2:58 am
manpsingh87 wrote:
sanju09 wrote:
manpsingh87 wrote:
sanju09 wrote:What is the radius of the circum circle of the triangle whose sides are 5, 12 and 13 respectively?
(A) 2
(B) 6
(C) 6 ½
(D) 7 ½
(E) 12
its a pythagorus triplet of sides 5,12,13 therefore its area is 1/2*5*12=30

circum radius= abc/4 (area of a triangle)=5*12*13/4*30=6(1/2) hence C
This is not an ideal approach that could be commonly exercised by the test takers mainly when they come from various other branch of studies than mathematics.

Beside, radius of the circum circle of a right triangle happens to be half of its hypotenuse. Please tell us why?
hi sanju its a theorem (may be thales ), according to which angle subtended by the diameter at circumference of the circle is right angle...!! i can prove the same thing for you but what's the purpose..???
i've remember few Pythagoras triplets which i found recurring in most of the question..1!!!

if you don't like than offcourse you can always follow the hero's formula to evaluate it.!!!
We really don't need to evaluate a thing using the Heron's formula in this problem in particular. Once we realize that the given is a right triangle, we must also realize that the radius of the circum circle of a right triangle happens to be half of its hypotenuse, and hypotenuse is 13 in our case whose half is (C) 6 ½, simple. Advance Trigonometric concepts like R = a b c/ (4 ∆) cannot be recommended to the common test takers for the same reason as I have cited before.
The mind is everything. What you think you become. -Lord Buddha



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by manpsingh87 » Tue Apr 05, 2011 3:09 am
sanju09 wrote:
manpsingh87 wrote:
sanju09 wrote:
manpsingh87 wrote:
sanju09 wrote:What is the radius of the circum circle of the triangle whose sides are 5, 12 and 13 respectively?
(A) 2
(B) 6
(C) 6 ½
(D) 7 ½
(E) 12
its a pythagorus triplet of sides 5,12,13 therefore its area is 1/2*5*12=30

circum radius= abc/4 (area of a triangle)=5*12*13/4*30=6(1/2) hence C
This is not an ideal approach that could be commonly exercised by the test takers mainly when they come from various other branch of studies than mathematics.

Beside, radius of the circum circle of a right triangle happens to be half of its hypotenuse. Please tell us why?
hi sanju its a theorem (may be thales ), according to which angle subtended by the diameter at circumference of the circle is right angle...!! i can prove the same thing for you but what's the purpose..???
i've remember few Pythagoras triplets which i found recurring in most of the question..1!!!

if you don't like than offcourse you can always follow the hero's formula to evaluate it.!!!
We really don't need to evaluate a thing using the Heron's formula in this problem in particular. Once we realize that the given is a right triangle, we must also realize that the radius of the circum circle of a right triangle happens to be half of its hypotenuse, and hypotenuse is 13 in our case whose half is (C) 6 ½, simple. Advance Trigonometric concepts like R = a b c/ (4 ∆) cannot be recommended to the common test takers for the same reason as I have cited before.
hmmm.. very interesting, the underlined part didn't came to my mind while solving it, any ways thanks for presenting a different view...!!!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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