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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 :

Please see the fig below:

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 .

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