Geometry prob #1

[papgust]

"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?

[papgust]

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!

[rohan_vus]

IMO - 196 is the area .Can explain if answer is correct

[viju9162]

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..

[papgust]

Hi Rohan,

You are absolutely right. 196 is the correct answer. Could you please explain the method of solving?

[papgust]

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.

[rohan_vus]

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.