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In triangle ABC angle bisectors AD, BE, and CF meet at I. If DI = 3, BD = 4, and BI = 5, then compute the area of triangle ABC.

 Nov 13, 2020
 #1
avatar+846 
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https://web2.0calc.com/questions/help-is-needed_5

 Nov 13, 2020
 #3
avatar+846 
+2

In triangle ABC angle bisectors AD, BE, CF intersect at a point I. If DI = 3, BD = 4, and BI = 5, then compute the area of triangle ABC.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Δ ABS is an isosceles triangle.

 

∠B = 2 [arctan(3 / 4)] = 73.73979529º

 

AD = tanB * BD = 13.71428571

 

Area = AD * BD = 54.85714286 u2

jugoslav  Nov 14, 2020
 #2
avatar+114221 
+1

Since DI =3, BD = 4  and BI = 5, then triangle IDB  is right with angle IDB = ADB = 90°

And since  ID = 3  and BI = 5.....then sin (DBI)  = 3/5

And cos DBI = 4/5

But since BE is an angle bisector, then sin (ABC)  = sin (2 * DBI)  =  2 sin(DBI)cos(DBI) = 2 ( 3/5) (4/5) = 24/25

 

And ADB is also a right triangle with BD  = 4

And since sin ABC = 24/25  then  triangle ADB  is a 7 - 24- 25  Pythagorean right triangle

Then sin ABD = 24/25

So sin of DAB  = 7/25

Using the Law of Sines

AB/ sin 90 = BD / sin (DAB)

AB/ 1  = 4 / (7/25)  =  100/7

And

AD  / sin (ABC)  = BD / sin (DAB)

AD = sin (ABC) * BD / sin (DAB)

AD = (24/25) * 4  / ( 7/25)  = 96/7

 

And since  angle ADB is right, then so is angle ADC

And since AD is an  angle bisector then  sin (BAD)  = sin (DAC)  = 7/25

So sin (ACD)   = 24/25

So we can find DC using the Law of Sines

DC / sin (DAC)  = AD / sin (ACD)

DC = sin (DAC) * AD / sin (ACD)

DC = (7/25) * ( 96/7) / (24/25)

DC = (7/25) * (96/7) * (25/24)

DC = (96/24)  = 4

So  BC =  BD + DC  = 4 + 4  = 8

 

And  the area of triangle ABC =  (1/2) (BC) ( AB) * sin ( ABC)  = (1/2) ( 8) ( 100/7) ( 24/25)  = 

(4) ( 4) (24/7)  = 

 

384/7  units^2

 

See the figure below to  get a feel  for this :

 

 

 

 

cool cool cool

 Nov 14, 2020

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