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Question 1. Amit, Deepak and Prabha have saved a good amount from their pocket money. They wish to donate it for a good cause. They sat on a round table of 20 cm radius, to decide the mode of donation. In the adjoining figure, Amit, Deepak and Prabha are sitting at A, S and D respectively, such that AS = SD = AD i.e., ∆ASD is an equilateral triangle.
(a) Find the distance between Amit, Deepak and Prabha.
(b) Which mathematical concept is used in the above problem?
(c) By donating the savings from the pocket money what values are depicted by Amit, Deepak and Prabha?
Sol. (a) Since, AS = SD = AD [Given]
â ∆ASD is an equilateral triangle.
⇒ Perpendicular AM will pass through O.
â SM = (1/2) (SD) = x cm (say) [∵ Perpendicular from the centre to a chord, bisects the chord]
Now, in right angled ∆ASM, we have AM^{2} + SM^{2 }=AS^{2}
⇒ AM^{2} =AS^{2} – SM^{2}
= (2 SM)^{2} – SM^{2} [∵ AS = SD and SM = (1/2) SD]
= 4SM^{2} – SM^{2} = 3 SM^{2} = 3x^{2}
⇒ AM = √3x
Now, OM = AM – OA = [ √3x – 20] cm
Also, in rt. ∆OBM, we have OS^{2} =SM^{2} + OM^{2}
20^{2} = x^{2} + OM^{2}
⇒ 400 = x^{2} + ( √3x – 20)^{2 }
⇒ 400 = x^{2} + 3x^{2} + 400 – 40 3x
⇒ 4x^{2 }– 40 3x =0
⇒ 4x (x – 10 √3= 0
⇒ x = 0 or x = 10√3
∵ x = 0 is not possible,
â x =10 3 cm
Now, SD = 2x cm = 2 × 10√3 cm = 20 √3 cm
⇒ Distance between Amit, Deepak and Prabha is 20√3 cm.
(b) Circles
(c)(i) Charity (ii) Habit of saving (iii) Taking right decision.
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