Question

A circular park of radius 20 m 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 1

LET A,B and C are the boys AND  AB, BC, AC are the strings of phones.

 O is the centre and COP is a line passing through centre and making a 90 degree angle with AB. Now AP = BP = x.

 And the line BO, bisects the angle ABC into two equal halves. So angle PBO is 30 degrees.

 Now  the right angled triangle PBO.

 In this Δ cos B = BP / BO

 i.e cos 30 = x / 20

 i.e x = 20 cos 30 = 17.32 m

 AS AP = BP = x,

 AB = 2x

 AB = 34.64 m

∴the length of each string is 34.64 m

Answer 2

Hello Mate!

Since arc[ AB = BC = CA )

So, chord[ AB = BC = CA ]

∆ABC is an equilateral ∆.

So, centroid G is also circumcentre

So, according to question length of radius i.e, AG = BG = CG = 20 cm

Now, AG = ⅔ AD

20 cm × 3/2 = AD

30 cm = AD

GD = 30 - 20 cm

GD = 10 cm

Now, since AD is perpendicular bisector of chord BC. So, BD = CD = ½ BC

In right ∆BGD,

BG² = GD² + BD²

20² = 10² + BD²

400 - 100 = BD²

√300 = BD or BD = 10√3 cm

Now, BD = ½ BC or 2BD = BC

2(10√2) = BC

20√2 = BC

Hence, AB = BC = AC = 20√2 cm.

Have great future ahead!