AB is a line segment with P as midpoint. Three semicircles are drawn with AP,PB and AB as diameters. Another circle is drawn so as to touch all three semicircles. What’s the radius of this smaller circle?

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Solution

The correct option is B AB
Ans: b) $\frac{1}{6}$ AB.

Soln:

Let PB=r=PQ. QR=x. In triangle PRB, RB=2r-x. PR= r+x.

Since PRB is right angled, $(r + x)^{2} = (2 r - x)^{2} + r^{2}$ .

Solving we get x= $\frac{r}{3}$ = $\frac{A B}{6}$

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AB is a line segment with P as midpoint. Three semicircles are drawn with AP,PB and AB as diameters. Another circle is drawn so as to touch all three semicircles. What’s the radius of this smaller circle?

The correct option is B ABAns: b) 16AB. Soln: Let PB=r=PQ. QR=x. In triangle PRB, RB=2r-x. PR= r+x. Since PRB is right angled, (r+x)2=(2r−x)2+r2 . Solving we get x=r3=AB6

The correct option is B AB
Ans: b) $\frac{1}{6}$ AB.

Soln:

Let PB=r=PQ. QR=x. In triangle PRB, RB=2r-x. PR= r+x.

Since PRB is right angled, $(r + x)^{2} = (2 r - x)^{2} + r^{2}$ .

Solving we get x= $\frac{r}{3}$ = $\frac{A B}{6}$