Question

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Answer 1

Answer:

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Answer 2

Answer:

mark as brainliest answer.

Step-by-step explanation:

1. OA = OB = AB = Radius

Δ AOB is an equilateral triangle

so ∠AOB = 60°

∠ACB = (1/2) ∠AOB = (1/2) * 60° = 30°

Arc AB = (60/360) * 2 π * Radius

= (1/6) 2π * Radius

Arc ACB = 2π * Radius - (1/6) 2π * Radius

=>Arc ACB =  2π * Radius ( 1 - 1/6)

=>Arc ACB =  2π * Radius ( 5/6)

=>Arc ACB =  (5/6) 2π * Radius

2.Given in right triangle PQR, QS =  SR

By Pythagoras theorem, we have

PR2 = PQ2 + QR2 → (1)

In right triangle PQS, we have

PS2 = PQ2 + QS2

PQ2 + (QR/2)2    [Since QS =  SR = 1/2 (QR)]

PQ2 + (QR2/4)

4PS2 =  4PQ2 + QR2

∴ QR2 = 4PS2 −  4PQ2 → (2)

Put (2) in (1), we get

PR2 = PQ2 + (4PS2 −  4PQ2)

∴ PR2 = 4PS2 −  3PQ2

3.Let the points C, D and E divide seg AB in four equal

Point D is the midpoint of seg AB. X co-ordinate of D = (x1 + x2)/2 = (-14 + 6)/2 = -8/2 = -4 y co-ordinate of D = (y1 + y2)/2 = (-10 -2)/2 = -12/2 = -6 ∴ Co-ordinate of D are (-4, -6). Point C is the midpoint of seg AF. ∴ By midpoint formula.