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MATHMAA

# 16th KVJ Solutions5

Here you find 16th KVJ Solutions5 in detail explanation .

5Q) A semi-circle is drawn outwardly on chord AB of the circle with centre O and unit radius. The perpendicular from O to AB, meets the semi-circle on AB at C. Determine the measure of $$\angle AOB$$ and length AB so that OC has maximum length.

 Solution :Let  O be the center, OA be the radius of 1 unit, AB is the chord.Let length of AB =2$$\sqrt{a}$$. Let Perpendicular from O to AB meets at M on AB.

AM=MB=$$\sqrt{a}$$.

OM extended to meet semicircle at C.

We know that AM=MB=MC as these are length of radius of semicircle. From right triangle OAM , we get ,

OM= $$\sqrt{1^{2}-(\sqrt{a})^{2}}$$

OM = $$\sqrt{1-a}$$

OC= OM + MC = $$\sqrt{1-a} +\sqrt{a}$$

$$OC^{2} =(\sqrt{1-a} + \sqrt{a})^{2}$$

$$OC^{2}= 1-a + a + 2\sqrt{(1-a)a}$$

$$OC^{2} = 1 + 2 \sqrt{a-a^2}$$

OC is maximum then$$OC^{2}$$ also maximum.

$$OC^{2}$$ is maximum when a-a^2 is maximum.

We know that $$y = px^{2} +qx + r$$ has maximum when p<0 and   it occurs at x = -q/2p .

$$OC^{2}$$ is maximum since a^2 coefficient p=-1 < 0 , q=1 .

a= -q/2p = -1/-2 =1/2

OC is maximum when a = 1/2

AB= $$2\sqrt{a}$$ = 2$$\sqrt{1/2}$$ = $$\sqrt{2}$$

OA =1, AM=$$\frac{1}{\sqrt{2}}$$ , From Triangle OAM,

sin ($$\angle AOM$$ )= $$\frac{AM}{OA}$$

sin($$\angle AOM$$ =$$\frac{1}{\sqrt{2}}$$

$$\angle AOM$$ = $$45^{\circ}$$

$$\angle AOB =2 \angle AOM$$

$$\angle AOB = 90^{\circ}$$