RD Chapter 16- Circles Ex-16.1 |
RD Chapter 16- Circles Ex-16.2 |
RD Chapter 16- Circles Ex-16.4 |
RD Chapter 16- Circles Ex-16.5 |
RD Chapter 16- Circles Ex-VSAQS |

**Answer
1** :
Let R, S and M be the position of Ishita, Isha and Nisha respectively.

Since OA is a perpendicular bisector on RS, so AR = AS = 24/2 =12 cm

Radii of circle = OR = OS = OM = 20 cm (Given)

In ΔOAR:

By Pythagoras theorem,

OA^{2}+AR^{2}=OR^{2}

OA^{2}+12^{2}=20^{2}

OA^{2 }= 400 – 144 =256

Or OA = 16 m …(1)

From figure, OABC is a kite since OA = OC and AB = BC. We knowthat, diagonals of a kite are perpendicular and the diagonal common to both theisosceles triangles is bisected by another diagonal.

So in ΔRSM, ∠RCS = 90^{0} and RC = CM …(2)

Now, Area of ΔORS = Area of ΔORS

=>1/2×OA×RS = 1/2 x RC x OS

=> OA ×RS = RC x OS

=> 16 x 24 = RC x 20

=> RC = 19.2

Since RC = CM (from (2), we have

RM = 2(19.2) = 38.4

So, the distance between Ishita and Nisha is 38.4 m.

**Answer
2** :

Since, AB = BC = CA. So, ABC is an equilateral triangle

Radius = OA = 40 m (Given)

We know, medians of equilateral triangle pass through thecircumcentre and intersect each other at the ratio 2 : 1.

Here AD is the median of equilateral triangle ABC, we can write:

OA/OD = 2/1

or 40/OD = 2/1

or OD = 20 m

Therefore, AD = OA + OD = (40 + 20) m = 60 m

Now, In ΔADC:

By Pythagoras theorem,

AC^{2} = AD^{2} + DC^{2}

AC^{2} = 60^{2} + (AC/2)^{ 2}

AC^{2 }= 3600 + AC^{2} / 4

3/4 AC^{2 }= 3600

AC^{2 }= 4800

or AC = 40√3 m

Therefore, length of string of each phone will be 40√3 m.

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