Q.☞Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m. Reshma throws to Salma, Salma to Mandip and Mandip to Reshma. If the distance between Reshma and Salma and Salma and Mandip is 6m each. What is the distance Reshma and Mandip?
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Answers
Answer:
Let O be the centre of the circle. A, B and C represent the positions of Reshma, Salma and Mandip.
AB = 6cm and BC = 6cm.
Radius OA = 5cm. (given)
Draw BM ⊥ AC
ABC is an isosceles triangle as AB = BC, M is mid-point of AC.
BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
In ΔOAM,
OA²=OM²+AM² ( by Pythagoras
OA²=OM²+AM² ( by Pythagorastheorem)
5²=x²+ y²—(i)
In ΔAMB,
AB²=BM² +AM²
AB²=BM² +AM²(by Pythagoras theorem)
6²= (5-x)²+y² — (ii)
Subtracting (i) from (ii),
36 – 25 = (5-x)² -x²
11 = (25+ x²– 2×5×x) - x²
11= 25+x²-10x - x²
11= 25-10x
10x = 14
x= 7/5
Substituting the value of x in (i), we get
y²+ 49/25 = 25
y² =25 – 49/25
y² =(625 – 49)/25
y²=576/25
y = 24/5
Thus,
AC = 2×AM
AC = 2×y
AC= 2×(24/5) m
= 48/5 m = 9.6 m
Hence, the Distance between Reshma and Mandip is 9.6 m.
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Solution :-
( 7 ¹ - 14 - ¹ ) ÷ 7 - ²
14 - ¹ × 7 - ¹
= - 14 × ( - 7 )
= 98 Ans
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