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In a right angled triangle, a semicircle is drawn such that its diameter lies on the hypotenuse and its center divides the hypotenuse into two segments of lengths 15 and 20.Find the length of the arc of the semicircle between the points at which the legs touch the semicircle.

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    $\begingroup$ There must be more given. There is nothing constraining the size of the semicircle, only where it's center lies. Making the semicircle small enough, it may not intersect the legs. Is the semicircle supposed to be tangent to the legs? $\endgroup$ – robjohn Aug 21 '15 at 17:28
  • $\begingroup$ the size of the semicircle is sufficiently large to intersect the legs $\endgroup$ – lokesh sangabattula Aug 21 '15 at 17:36
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    $\begingroup$ that still does not give sufficient information. If the semicircle intersects, but is not tangent to, the legs, then the semicircle intersects each leg at two points. Which are we supposed to use? $\endgroup$ – robjohn Aug 21 '15 at 17:40
  • $\begingroup$ okay let the semicircle touch both the legs $\endgroup$ – lokesh sangabattula Aug 21 '15 at 17:50
  • $\begingroup$ Is it like what I am understanding... A semi-circle is drawn inside a right angled triangle with hypotenuse length 35 in line with its diameter The center of touching semi-circle divides the hypotenuse into segments of length 15 and 20. What is length of arc between points of tangency forming $90^0?$... Do I understand it correctly? $\endgroup$ – Narasimham Aug 21 '15 at 18:47
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A Part of SemiCirc

Let the tangent length shown in sketch be T. The power of circle

$$ T^2 = (15-R) (15+R) \tag{1}$$

From similar triangles, (radius/hypotenuse) of right side right angled triangle:

$$ \frac{T}{R}= \frac{15}{20}= \frac{3}{4} \tag{2}$$

Solving

$$ R= 12, \; T = 9 \tag{3} $$

Arc Length is quarter circle $$ =\pi \, R/ 2 = \frac{ \pi \cdot 12}{2} = 6 \pi \tag{4}$$

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  • $\begingroup$ Is a sketch coming? $\endgroup$ – robjohn Aug 21 '15 at 19:49
  • $\begingroup$ Yes in 30 minutes $\endgroup$ – Narasimham Aug 21 '15 at 20:04
  • $\begingroup$ (+1) I was thinking of using the power of a point, but came up with a similar triangle approach first. $\endgroup$ – robjohn Aug 21 '15 at 20:28
  • $\begingroup$ Thanks, recently also I am under its power :) $\endgroup$ – Narasimham Aug 21 '15 at 20:31
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Here is the diagram:

enter image description here

The two smaller triangles are similar and the ratio of their legs is the ratio of their bases since the two $12$ length sides are also the radius of the circle tangent to the larger triangle.

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