A circular park of radius 10 m is situated in a colony. Three students Ashok, Raman and
Karthika are standing at equal distances on its circumference each having a toy telephone in
his hands to talk each other about Honesty, Peace and Discipline.
(i) Find the length of the string of each phone.
(2) Find the altitude of the equilateral triangle
Answers
Length of the string of each phone = 10√3 m
Altitude of the equilateral triangle = 15 m
Given:
- A circular park of radius 10m is situated in a colony
- Three students Ashok, Raman and Karthika are standing at equal distances on its circumference each having a toy telephone in his hands to talk each other about
To Find:
- The length of the string of each phone
Solution:
Sitting at equal distance
=> Form Equilateral Triangle
Let say ABC is triangle (Ashok , Raman and Karthika are at A , B and C)
and O is center of circular park
then AO = BO = CO = Radius = 10 m
as Triangle is equilateral
Hence ∠AOB = ∠BOC = ∠COA = 360°/3 = 120°
Using Cosine Rule
AB² = AO² + BO² - 2AO. BO Cos∠AOB
=> AB² = 10² + 10² - 2* 10 * 10 Cos120
=> AB² = 200 - 200(-0.5)
=> AB² = 200 + 100
=> AB² = 300
=> AB = 10√3
Length of the string of each phone = 10√3 m
Altitude of the equilateral triangle = √3 ( side)/2
= √3 ( 10√3)/2
= 15 m
Answer:
See Let Us Make A Triangle According To The Question Keeping The Centre In Between.
As Given Ashok Radhika And Karthika Are There So Triangle Will Be ARK.
Now From The Centre Make A Radius Joining A
Let The Centre Be O.
Draw Perpendicular Line OD On Chord AK To Obtain A Triangle AOD. Again Draw Line From Point R To The Centre O.
So The Triangle Obtained Will Be ARD And AOD.
Now Prove=
Let AR Be 2x An We Know That A Perpendicular Chord Is Drawn So AD Will Be x,Let OD Be y,So RD Will Be 10+y.
By Applying Pythagoras Theorem In Triangle ARD H^2=B^2+P^22. So,
AR^2=AD^2+RD^2
By Substituting Value, 2x^2=x^2+(10+y)^2 , 4x^2=x^2+