# Circle

## Exercise 10.4

Question 2: If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer:Given; circle with centre O in which chords AB and CD intersect at M.
To Prove: AM = DM and CM = BM

In ΔAOM and ΔDOM
OM = OM (common side)
So, ΔAOM ≈ Δ DOM
Hence, AM = DM proved
It is given that AB = CD
So, AB – AM = CD – DM
Or, CM = BM proved

Question 3: If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Answer: Let us use the figure in previous question to solve this problem.
To Prove: ∠AMO = ∠DMO
In ΔOMA and ΔOMD
OM = OM (common)
So, ΔOMA ≈ ΔOMD
Hence, ∠AMO = ∠DMO proved

Question 4: If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see the figure).

Answer:Given two concentric circles with centre O. A line intersects the bigger circle at A and B, and intersects the smaller circle at B and C.
To Prove: AB = CD

AD is the chord for bigger circle.
Hence, AM = MD ………..(1)
Similarly, BC is the chord for smaller circle.
Hence, BM = MC …………(2)
From equations (1) and (2)
AM – BM = MD – MC
Or, AB = CD proved

Question 5: Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Answer:Given a circle with centre O. Here; RS = SM = 6 m and radius OS = 5 m
Let us draw OS perpendicular RM and OL perpendicular SM

In ΔOMS; base = OS and height = MN
Area (OMS) = ½ xx MN xx OS = ½ xx MN xx 5 ………….(1)
If base = SM = 6 m then height = OL
Area = ½ xx OL xx SM = ½ xx OL xx 6 …………(2)

From equations (1) and (2)
(5/2)\ MN = 3OL
In ΔOLM; OL^2 = OM^2 - ML^2
= 5^2 - 3^2
= 25 – 9 = 16
Or, OL = 4
So, (5/2)\MN = 3 xx 4 = 12
Or, MN = 24/5 = 4.8
So, RM = 2 xx 4.8 = 9.6  m

Question 6: A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer: Given circle with centre O with radius = 20 m; in which chords AS = SD = AD = ?

In ΔADS; AM^2 = AD^2 - ((AD)/(2))^2
Or, AM^2 = AD^2 - (AD^2)/(4)
Or, AM^2 = (3AD^2)/(4)
Or, AM = (sqrt(3/4))AD
Or, AM = (sqrt(3/4))AD = OM + 20
Or, OM = (sqrt(3/4))AD - 20 …………(1)

In ΔODM; OM^2 = OD^2 - ((AD)/(2))^2
= 20^2 - (AD^2)/(4)
= 400 – (AD^2)/(4) ……………..(2)

From equations (1) and (2)
400 – (AD^2)/(4) = (sqrt3/4\AD - 20)^2
Or, 400 – (AD^2)/(4) = ¾ xx AD^2 + 400 – 20sqrt3 xx AD

Or, 400 – (AD^2)/(4) - ¾ xx AD^2 - 400 + 20sqrt3 xx AD = 0

Or, 20sqrt3\AD – AD^2 = 0
Or, 20sqrt3\AD = AD^2
Or, AD = 20sqrt3
This is the length of each wire