Question

Three girls A, B, and C are playing a game while standing on a circle of radius 5m drawn in a park. A throws a ball to B, B to C, and C to A. If the distance between A and B, B and C is 6m each, what is the distance between A and C?

Answer with a diagram

Answer 1

Step-by-step explanation:

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Answer 2

• The distance between A and C is equal to 9.6 m .

Given :-

• Radius of circle = 5 m
• AB = BC = 6 m

To Find :-

• AC = ?

Concept / Formula used :-

• Altitude of isosceles ∆ bisect the base . { Non equal side }
• Perpendicular from centre to the chord bisect the chord .
• pythagoras theorem :- (Perpendicular)² + (Base)² = (Hypotenuse)²
• Area of triangle = (1/2) × Base × perpendicular

Construction :-

• Join chord AC .
• ON ⟂ AC .
• OL ⟂ AB .

Solution :-

In ∆OAB we have,

→ OA = OB = 5 m { Radii of circle }

So, ∆OAB is an isosceles triangle .

since OL ⟂ AB,

→ AL = LB = 6/2 = 3 m { Altitude of isosceles ∆ bisect the base }

now in right angled ∆OLA we have,

→ OL² + AL² = OA² { By pythagoras theorem }

→ OL² = OA² - AL²

→ OL² = 5² - 3²

→ OL² = 25 - 9

→ OL² = 16

Square root both sides,

→ OL = √16

→ OL = 4 m --------- Equation (1)

then,

→ Area (∆OAB) = (1/2) × Base × perpendicular

→ Area (∆OAB) = (1/2) × AB × OL

putting value of Equation (1),

→ Area (∆OAB) = (1/2) × 6 × 4

→ Area (∆OAB) = 3 × 4

→ Area (∆OAB) = 12 m² ----------- Equation (2)

also, area of ∆OAB as base OB and perpendicular as AN,

→ Area (∆OAB) = (1/2) × OB × AN

from Equation (2),

→ 12 = (1/2) × 5 × AN

→ 12 × 2 = 5 × AN

→ 5 × AN = 24

dividing both sides by 5,

→ AN = (24/5)

→ AN = 4.8 m

therefore,

→ AC = AN + NC

→ AC = AN + AN { Since, Perpendicular from centre to the chord bisect the chord . N is mid point of AC . }

→ AC = 2•AN

→ AC = 2 × 4.8

→ AC = 9.6 m (Ans.)

Hence, the distance between A and C is equal to 9.6 m .

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