Thread: ..a very simple problem ..but different answers by different methods..

(see attached fig).
There is a triangle abc and an incircle to the triangle with radius 4cm.
The area of the triangle is given to be 84 cm^2
The base bc is divided by the radius into 6cm , 6cm
ab = ac
We need to find the value of these sides (ab or ac)
..
Methods of solving :
1)By using the pythagoras theorm, the answer comes to be appproximately 15 ( sqrrt(232) which is ~15
2)while by using area (1/2bh) the answer comes to be exactly 15..
what tis the reason for this difference?

Originally Posted by anshulbshah

2)while by using area (1/2bh) the answer comes to be exactly 15..
what tis the reason for this difference?

h = height
6h = 84 ; h = 14
sqrt(14^2 + 6^2) = sqrt(232)

HOW did you get sqrt(225) ?

Originally Posted by Wilmer

h = height
6h = 84 ; h = 14
sqrt(14^2 + 6^2) = sqrt(232)

HOW did you get sqrt(225) ?

well i got it by a diff method..
1)Join B and C to center (say o ) of the circle..
2)Radius if perpendicular to BC ( tangent to a circle)
therefore..

area of entire fig = thus area of remaining portion is 84 - 24 = 60
now join point of contact of AC and AB to O ..these will be radius and perpendicular to AB and AC respectively..
These will even be congruent ...as ab = ac & height / radius = 4cm
thus area of these =

the correctanswer to this problem is 15 because you must use the given in radius 4 to calculate the area .If you calculate that the altitude is 14 you would have a triangle with an in radius not equal to 4


bjh

Agree BJ. If area is 84 (as given) then height = 14: means radius = ~3.956

anshulbshah, check your post for accuracy, please.
If you want something similar to experiment with , use isosceles triangle 5-5-6:
will have height 4 and inradius 3/2.