find the radius of a circle that is the tangent to all the three side of the triangle ABC given that the triangle hve lenght 7,24 and 25
help?
2 answers
http://mathcentral.uregina.ca/qq/database/qq.09.06/s/maria2.html
make a sketch
you will need the incentre of the triangle.
This consists of the intersection of the angle bisectors
label the triangle ABC, with the incentre O.
Sketch in the three radii, hitting sides AB, BC, and CA at P, Q, R respectively.
You should see that we have 3 pairs of congruent triangles, and
AP = AR = x
BP = BQ = y
CQ = CR = z
so x+y = 24 *
x+z = 25 **
y+z = 7 ***
*** - ** ----> z - y = 1
z = y+1
sub into ***
y + y+1 = 7
2y = 6
y = 3
z = 3+1 = 4
x+3 = 24
x = 21
We can find angle A by the cosine law:
7^2 = 24^2 + 25^2 - 2(24)(25)cosA
cosA = (576+625-49)/(1200)
A = appr 16.26°
Now in triangle AOP, angle P = 90°, angle OAP = 8.13°, AP = 21, OP = r
r/21 = tan 8.13
r = 21tan8.13° = 3
the radius is 3 units
you will need the incentre of the triangle.
This consists of the intersection of the angle bisectors
label the triangle ABC, with the incentre O.
Sketch in the three radii, hitting sides AB, BC, and CA at P, Q, R respectively.
You should see that we have 3 pairs of congruent triangles, and
AP = AR = x
BP = BQ = y
CQ = CR = z
so x+y = 24 *
x+z = 25 **
y+z = 7 ***
*** - ** ----> z - y = 1
z = y+1
sub into ***
y + y+1 = 7
2y = 6
y = 3
z = 3+1 = 4
x+3 = 24
x = 21
We can find angle A by the cosine law:
7^2 = 24^2 + 25^2 - 2(24)(25)cosA
cosA = (576+625-49)/(1200)
A = appr 16.26°
Now in triangle AOP, angle P = 90°, angle OAP = 8.13°, AP = 21, OP = r
r/21 = tan 8.13
r = 21tan8.13° = 3
the radius is 3 units
1 answer