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by Gurpinder » Mon Sep 20, 2010 7:09 am
Image


how in the world do you get the height of this triangle?
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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Source: — Problem Solving |

by selango » Mon Sep 20, 2010 7:20 am
A triangle inscribed in a circle having a diameter as one side is a right triangle.

ABC is a right angled triangle.Angle B is 90.

So AB^2=AC^2-BC^2=4-1=3

-->AB=sqrt(3)

Area of right angled triangle ABC=1/2*(Product of sides containing 90)=1/2*(BC*AB)

=1/2*(sqrt(3)*1)=sqrt(3)/2
--Anand--
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by Gurpinder » Mon Sep 20, 2010 7:35 am
selango wrote:A triangle inscribed in a circle having a diameter as one side is a right triangle.

Ahhhh...

but if its a right triangle and BC =1 and AC = 2, then should the area be 1?
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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by krazy800 » Mon Sep 20, 2010 8:07 am
Gurpinder wrote:
selango wrote:A triangle inscribed in a circle having a diameter as one side is a right triangle.

Ahhhh...

but if its a right triangle and BC =1 and AC = 2, then should the area be 1?
AC is the hypotenuse here..

so AB = (3)^0.5

Therefore area = 1/2 * AB * BC

= (3)^.5/2

HTH!!
Aiming High
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by Gurpinder » Mon Sep 20, 2010 8:11 am
krazy800 wrote:
AC is the hypotenuse here..
Thx
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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