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September 4th, 2012, 07:42 PM   #1
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Area of triangle ABC

[attachment=0:y2tsqyu1]Area of triangle ABC.JPG[/attachment:y2tsqyu1]
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 September 4th, 2012, 09:23 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,167 Thanks: 472 Math Focus: Calculus/ODEs Re: Area of triangle ABC Using coordinate geometry: Orient the coordinate axes such that D is the origin. Then we have A(-3,3), B(0,5) and C(5,0), and the area A of the triangle ABC is then: $A=\frac{1}{2}\|(5-3)(-3-5)-(0+3)(3-0)\|=\frac{25}{2}=12.5$
 September 5th, 2012, 03:19 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1618 AD and BC are parallel, so area of triangle ABC = area of triangle DBC = 25/2.
 September 5th, 2012, 04:40 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,610 Thanks: 845 Re: Area of triangle ABC Total area (including triangle ABH) = 25 + 9 + 3 = 37 "White area" = 5*5/2 + 3*8/2 = 24.5 "Yellow area" = 37 - 24.5 = 12.5 How come such an easy one, Albert; feeling guilty?
 September 5th, 2012, 05:17 PM #5 Senior Member   Joined: Sep 2009 From: Wisconsin, USA Posts: 227 Thanks: 0 Re: Area of triangle ABC Here is a vectored approach. BD+DC=BC. BH+HA=BA. Then the area is 1/2*|BA x BC|. You can do it from any of the points of the triangle and with different choices of direction.
September 5th, 2012, 06:09 PM   #6
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Re: Area of triangle ABC

Hello, Albert.Teng!

Quote:
 $\text{Find the area of }\Delta ABC.$ Code:  B G * - - - - - - - - * * | * | * | * | * | * | * | * | 5 A * - - - - * * | | * | * | 3 | | * * | | | * * | * - - - - * - - - - - - - - * F 3 D 5 C

$\text{Area }\Delta BDC \:=\:\frac{1}{2}(5)(5) \:=\:\frac{25}{2}$

$\text{Area trapezoid }ABDF \:=\:\frac{3}{2}(3\,+\,5) \:=\:12$

$\text{Area }\Delta AFC \:=\:\frac{1}{2}(8)(3) \:=\:12$

$\text{Area }\Delta ABC \;=\;\Delta BDC \,+\,\text{trap}\,ABDF \,-\,\Delta AFC$

[color=beige]. . . . . . . . . . . [/color]$=\;\;\frac{25}{2}\,+\,12\,-\,12 \;\;=\;\;12.5$

September 5th, 2012, 06:18 PM   #7
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Re: Area of triangle ABC

Quote:
 Denis wrote :How come such an easy one, Albert; feeling guilty?
I have posted so many tough problems, as you said may be a little bit

guilty.

Do you like easy ones or questions with challenging?

 September 5th, 2012, 07:29 PM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,610 Thanks: 845 Re: Area of triangle ABC NO easy ones please!
 September 6th, 2012, 12:49 AM #9 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Area of triangle ABC [color=#000000]Another solution (similar to soroban's).[/color] $\left(\triangle \textrm{AFC}\right)=\frac{1}{2}3(3+5)=12$ $\left(\textrm{ABCF}\right)=\left(\triangle \textrm{AHB}\right)+\left(\textrm{AHDF}\right)+\le ft(\triangle\textrm{BDC}\right)=\frac{1}{2}\cdot 3 \cdot 2+9+\frac{25}{2}=12+\frac{25}{2}$ $\left(\triangle \textrm{ABC}\right)=\left(ABCF\right)-\left(\triangle\textrm{AFC}\right)=12+\frac{25}{2}-12=\frac{25}{2}$
September 6th, 2012, 03:00 AM   #10
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Re: Area of triangle ABC

Quote:
 Originally Posted by Albert.Teng [attachment=0:1flowknc]Area of triangle ABC.JPG[/attachment:1flowknc]
Heres a hard way to do this easy problem...

By pythagoras, $AC= sqrt {73} \ BC = sqrt{50} \ AB = sqrt{13}$

By heron, $AREA \= \ \frac{1}{4}sqrt{(73 + 50 + 13)^2 \ - \ 2(73^2 + 50^2 + 13^2)} \ = \ \frac{1}{4}sqrt{2500} \ = \ \frac{50}{4} \ = \ \frac{25}{2}$

I like skipjack approach, he slides point A to point D along AD and yellow area never changes.

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