In the following figure, ACB is a right-angled triangle. AD is the altitude. Circles are inscribed within the triangle ACD and triangle BCD. P and Q are the centers of the circles. The distance PQ is
The length of AB is 15 m and AC is 20 m
By Pythagoras theorem we get BC = 25 . Let BD = x;Triangle ABD is similar to triangle CBA => AD/15 = x/20 and also triangle ADC is similar to triangle ACB=> AD/20 = (25-x)/15. From the 2 equations, we get x = 9 and DC = 16
We know that AREA = (semi perimeter ) * inradius
For triangle ABD, Area = 1/2 x BD X AD = 1/2 x 12 x 9 = 54 and semi perimeter = (15 + 9 + 12)/2 = 18. On using the above equation we get, inradius, r = 3.
Similarly for triangle ADC we get inradius R = 4 .
PQ = R + r = 7 cm
Create a FREE account and get:
Hawk Learner
1Â month, 3Â weeks ago
is the inradius formula r = (a+b-c)/2 is only applicable in the right angled triangle or any triangle
JUNIOR SAMUEL
9Â months ago
thank you very much clear working out
Goiye Ata
9Â months, 1Â week ago
Thankyou for very clear working out.
Philimon BENARD
9Â months, 1Â week ago
Very clear and helpful.
AJAY
2Â years, 10Â months ago
wrong ans, final ans will be sq rt 50 for PQ. by right angle traingle formed by PQS.
Rupak
3Â years, 8Â months ago
thank you ..it was very helpful..