In the given figure, ACB is a right angled triangle. CD is the altitude. Circles are inscribed in the within the triangles ACD, BCD. P and Q are the centres of the circles. The distance PQ is.

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In the given figure, ACB is a right angled triangle. CD is the altitude. Circles are inscribed in the within the triangles ACD, BCD. P and Q are the centres of the circles. The distance PQ is.

Mathematics
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@Nnesha help plzzzz
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Is there any probability that you would be able to help me out with this question? @Nnesha
sorry i was on different tab wait a sec i'll see what i can do..
ok.
hmm not sure haven't done inscribed circle since 2014 so i need to review n this topic
Ohhhh. It's ok then. Still, thanks @Nnesha
there is no PQ in the pic i guess we have to draw that |dw:1442715126599:dw|
yeah! figure looks good! @Nnesha
lol
So, what next?
:p
idk LOL i guess we should find area of both right triangles tag some users they will help u
Yes You can find the lengths of CD, AD and DB from the given information and therefore the areas of the 2 triangles. There is a relation between the radii of the incircles of a triangle and its area. Knowing the radii of the 2 circles you can find PQ.
I cant recall that relation but its no doubt on the web somewhere
Yea r = A/s where A is the area of the triangle ans = semi perimeter of the triangle.
|dw:1442761194406:dw|
I've used x , y and z for the line segments AD, DB and CD so we get x + y = 25 15^2 + x^2 = z^2 20^2 + y^2 = z ^2 the last 2 are by the Pythagoras theorem
solve this system of equations
tnks :) I got them. @welshfella
yw

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