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Rahul Ahuja

· started a discussion

· 1 Months ago

Answer should be 7 only as per figure given (Question: 115785.0 )

Reason: As per figure, the two circles are incircles to the respective traingles. CD is common tangent to two circles so PQ must be s straight line because radius is always perpendicular to the tangent.

For calculating the inradius, the two other sides of right angled triangle should be added and hypotenuse should be subtracted and the result obtained must be divided by 2. So inradius= (a+b-c)/2.

Answer will be 7.

Question:
In the adjoining figure, ACB is a right angled triangle. CD is the altitude. Circles are inscribed within the triangles ACD, BCD. P and Q are the centres of the circles. The distance PQ is (approx) :  

Options:
A) 5
B) \(\sqrt[]{50}\)
C) 7
D) 8
Solution:
Ans: (b)


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