Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5m drawn in a in a park. Reshma throws a ball to Salma, Salma to Mandeep, Mandeep to Reshma. If the distance between Reshma and Salma and between Salma and Mandeep is 6 m each, what is the distance between Reshma and Mandeep¿
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Answer:
Distance between Reshma and Mandeep is 9.6 m.
Step-by-step explanation:
Given :-
Distance between Reshma and salma and between salma and mandeep is 6 m.
Radius of circle is 5 m.
To find :-
Distance between Reshma and mandeep.
Solution :-
Let, Point where Reshma, salma and mandeep are standing be R, S and M respectively.
And Centre of circle be A.
Construction :- Draw perpendicular AB on RS.
The perpendicular from the center of a circle to a chord bisects the chord.
RS is cord. Then RB and BS are equal.
RS = RB + BS = 6 m
RB = BS = 3 m.
In ∆RBA,
By Pythagoras theorem :
Perpendicular² = Hypotenuse² - Base²
Perpendicular = AB
Base = RS = 3 m
Hypotenuse = AR = 5 m
(AB)² = (AR)² - (RS)²
AB² = (5)² - (3)²
AB² = 25 - 9
AB² = 16
AB = √16
AB = 4
Measure of AB is 4 m.
In ∆SAR,
We know,
Area of triangle = ½ × base × height
Base = RS
Height = AB
½ × 4 × 6
2 × 6
12
Area of ∆SAR is 12 m².
If we see triangle taking base AS and height be RO than area will be same.
So,
½ × RO × 5 = 12
RO × 5 = 12 × 2
RO × 5 = 24
RO = 24/5
RO = 4.8
Measure of RO is 4.8 m.
In Quadrilateral RSMA
RS = MS = 6 m [Given]
RS = AM = 5 m [Radius of circle]
We know,
Kite is only quadrilateral whose pair of adjacent sides are equal.
Thus,
RSMA is a kite.
And,
Diagonals of kite bisect each other at right angle
So, Diagonals are SA and RM,
Then, RO = MO
RO is 4.8 m.
Than,
MO = 4.8 m.
Now,
Distance between Reshma and Mandeep = RO + MO
= 4.8 + 4.8
= 9.6
Therefore,
Distance between Reshma and Mandeep is 9.6 m.