point F is located on side AB of triangle ABC and point E is located on side AC. find radius of incircle BCEF. if known that
AF=9CM
AE=6CM
BF=12CM
CE=8CM
Dear Aditya, the problem is correctly formulated.
Only one trapezium BCEF has an incircle. In such case, the sums of 2 and 2 opposite sides are equal:
BF+CE=EF+BC
20=EF+BC (1)
BC:EF=AB:AF=AC:AE=7:3
hence 7EF=3BC (2)
(1) and(2) give BC=14, EF=6.
Try to finish the exercise or write, if there remains an obstacle.
Dear Aditya,
In order that your question be more clear, I think you should tell us who is M.
Sincerely, Octav
dear Panchatcharam Mariappan,
BC is unknown
BC=? .......( i don't know) did't mention in question.
As i think by looking question that in diagram,
but as i know "incircle" means circle inside the triangle. so may be i am wrong.... i am totally confused please solve this
BCEF is a trapezium. So, the circle should lie inside the trapezium
BCEF is atrapezium. For, AF/BF=9/12=3/4. AE/CE=6/8=3/4. Therefore, AF/BF=AE/CE. Therefore, EF is parallel to AB. Hence, BCEF is a trapezium.
For trapezium, the diameter of the incircle is height of the trapezium. Therefore, radius=height/2 of the trapezium. If BC is known, it is easy to find the radius.
thank you for answering question, but any other to find the solution its important .....
one more thing
as incircle means circle inside the triangle so it touch only three point but in it it touch 4 point so how is this incircle
InCircle means circle inside a polygon touching all sides of the polygon or each side of the polygon is a tangent to the circle.
We have to split the problem into three sub cases. Case1: Right-angled triangle, with hypotenues=AB, then radius of the circle = 8, if we change the hypotenus as BC, the result will change, similarly, if we change the hypotenuse as AC, the result will change to 12.
Case2: Isosceles triangle
If we assume, AB = BC, then radius of the circle=80*sqrt(6)/3. If we assume AC=BC, the result will change.
Case3: Scalene triangle
In this case, the answer will be in terms of angle, so exact value is not known to me.
Dear Aditya, the problem is correctly formulated.
Only one trapezium BCEF has an incircle. In such case, the sums of 2 and 2 opposite sides are equal:
BF+CE=EF+BC
20=EF+BC (1)
BC:EF=AB:AF=AC:AE=7:3
hence 7EF=3BC (2)
(1) and(2) give BC=14, EF=6.
Try to finish the exercise or write, if there remains an obstacle.
dear viera madam,
I tried but get confused ... we get BC and EF but ... radius of circle??? littile bit confused
is it BC is diameter of circle or it may be a chop?
You have to find the height of the trapezium which is the diameter of the circle
Dear Aditya,
1) draw the picture - you have an isosceles triangle ABC with a base AB=21 and sides AC=BC=14.
2) Compute the altitude (height) h from C to AB using Pythagorean theorem
142-(21/2)2=h2 (don't forget to compute the square root of the result)
3) Now compute the area of the triangle ABC
area=ABxh/2 (x denotes multiplication)
The same area using the height k from A to BC is area=BCxk/2.
You know the values of area and BC, you can compute k.
4) Denote d the height of the trapezium. As wrote prof. Panchatcharam Mariappan, d is the diameter of the circle.
Due to the proportionality that we have in this figure, k:d=7:4. Thus d=4k/7.
5) the radius r=d/2.
its very very thank you #Viera Cernanova
i got answer. it is r= 3.9685cm
but k:d = 7:4 you used but in line we use
BC:EF=AB:AF=AC:AE=7:3 ...... but using 7:4 i get answer
and also i use another method
by taking small triangle AEF
first taking area using base AF and then comparing with are of triangle AEF by base EF.
after this i subtract and i get height of trapezium ......
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