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MATHMAA

# 16th KVJ Solutions8

Here you find the solutions of  16th KVJ Solutions8 in detail explanation

8Q) (a) Two parallel chords in a circle have lengths 10cm an 14cm and distance between them is 6 cm. If a chord parallel to these chords and midway between them is length √a, find the value of a.

Solution :

Let O be the center  and r be the radius of the circle.

Let AB and CD are two Parallel chords and distance between them is 6 cm .

Let AB =14 and CD =10 cm .

Let perpendicular from O to AB is Q and from O to CD be P, then PQ=6cm.

AQ =QB =7cm

CP=PD =5 cm

OA= OC =r cm

Let MN is a chord  of length $$\sqrt{a}$$, midway between AB and CD and intersect PQ at E.

MN = $$\sqrt{a}$$ , Then ME= EN =$$\frac{\sqrt{a}}{2}$$

PE = EQ =3 cm.

Let OQ = x cm then OP =6-x      (as PQ= 6cm).

From Triangle OAQ

$$7^{2} +x^{2} =r^{2}$$

x = $$\sqrt{r^{2} - 49}$$ ----------(1)

From Triangle OCP ,

$$OC^{2} =OP^{2} + PC^{2}$$

$$r^{2} = (6-x)^{2} + 5^{2}$$

$$r^{2} = (6 - \sqrt{r^{2}-49})^2 +25$$  (From (1))

$$r^{2} = 36 + r^{2} - 49 - 12\sqrt{r^{2}-49} +25$$

$$12 -12\sqrt{r^{2}-49} =0$$

$$r^{2} -49 = 1$$

$$r^{2} = 50$$

$$r = 5\sqrt{2}$$

$$x=\sqrt{r^2-49}$$

x=1

To find MN length.

MN is midway between AB and CD.

PE= EQ =3

EQ =x+ OE

3 = 1+OE

OE =2

From triangle OME , we get,

$$OM^{2} =OE^{2} +EM^{2}$$

$$r^{2} =2^{2} + (\frac{\sqrt{a}}{2})^{2}$$

50 = 4 + $$\frac{a}{4}$$

46 = $$\frac{a}{4}$$

a = 46 * 4 = 184

Therefore the value of a is 184.

(b) The line joining two points A(2, 0) and B(3, 1) is rotated about point A in anticlockwise direction through an angle of $$15^{\circ}$$. Find the equation of the line in the new position. IF B goes to C in new position, find the coordinates of C.

For Part (b) solution of  16th KVJ Solutions8 , Click on NEXT .