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Geometry prob #1

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Geometry prob #1

by papgust » Fri Sep 18, 2009 7:54 pm
"In triangle ABC, AD is the bisector of |A, AB=10 cm, AC=14 cm and area of triangle ABD = 140 sq cm. Find area of triangle ACD"

Can someone give a good easy solution to this prob?
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by papgust » Sun Sep 20, 2009 4:33 am
Hi ppl,

I'm sure that this problem is a cakewalk for many of you guys here.. Please respond and help me understand the problem.

Thank you in advance!
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by rohan_vus » Sun Sep 20, 2009 4:56 am
IMO - 196 is the area .Can explain if answer is correct
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by viju9162 » Sun Sep 20, 2009 6:05 am
Hi papgust,

Is the answer 140? My reasoning is as follows:

AD is the bisector of angle A. which means that it divides the angle into two equal parts.

Hence, BD opposite to angle A(one half) and DC opposite to angle A (one half) should be equal..

From Triangle ABD, Area = 1/2 * BD * x( X is height)..

140 = 1/2 * BD * x ; BD*x = 280..

Area of ADC = 1/2 * DC * x ( x is also the height for this triangle also) ..

And also, DC = BD, hence DC * x = 280.. Therefore, Area of ADC = 1/2*280 = 140..
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by papgust » Sun Sep 20, 2009 8:00 pm
Hi Rohan,

You are absolutely right. 196 is the correct answer. Could you please explain the method of solving?
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by papgust » Sun Sep 20, 2009 8:03 pm
viju9162 wrote: Hence, BD opposite to angle A(one half) and DC opposite to angle A (one half) should be equal..
Hi Viju,

I believe your hypothesis is wrong. BD and DC need not be the same if AD is the bisector of |A. This is the angle bisector formula --> If AD is the bisector of |BAC, then AB/AC=BD/DC.

Anyway the answer is not 140. 196 is the correct answer.
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by rohan_vus » Sun Sep 20, 2009 11:13 pm
Simple reasoning. Any angle bisector of any angle between 2 sides of a triangle divides the Area of the triangle into the ratio of sides .

Here's why its so..

Area of any triangle is 1/2 *(Product of any 2 sides of the triangle) * (Sin of Angle between those 2 sides)

Now coming to the question at concern.

Here area of ABD => 140 = 1/2*(AB * AD) *(Sin of anle g BAD) ---eqn (1)


Area of ACD = 1/2*(AC*AD) * (Sin of angle DAC) ---eqn(2)

angle DAC = angle BAD ---eqn(3) as angle A is bisected

Using eqn 1 aand 2 and 3, gives 196 as area of ACD.
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