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AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to
(2019)
  • a)
    9.1
  • b)
    7.8
  • c)
    9.3
  • d)
    8.5
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
AB is a diameter of a circle of radius 5 cm. Let P and Q be two points...

Since AB  is a diameter, therefore ∠AQB and ∠APB are angles in the semi–circle and hence are right–angles.
i.e. ∠A PB = ∠AQB = 90°
Let AQ  = x, then AP = 2x
In right ΔAPB,
AP2 = AB2 – BP2 ⇒ AP2 = 102 – 62 = 82
⇒ AP = 8
⇒ 2x = 8
⇒ x = 4
In right ΔAQB
BQ2 = AB2 – AQ2
⇒ BQ2 = 102 – 42 = 84
⇒ BQ = √84 ≈ 9.1
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Most Upvoted Answer
AB is a diameter of a circle of radius 5 cm. Let P and Q be two points...
Given:
- AB is a diameter of a circle with radius 5 cm.
- PB = 6 cm.
- AP = 2 * AQ.

To find:
The length of QB, nearest to.

Solution:
Let's draw a diagram to better understand the given information.

Key points:
- The length of PB is given as 6 cm.
- AP is twice the length of AQ.

Approach:
- We will use the properties of a circle to solve this problem.
- First, we will find the length of AB using the formula 2 * radius.
- Then, we will find the length of AP by subtracting PB from AB.
- Finally, we will find the length of AQ by dividing AP by 3 (since AP is twice AQ).

Calculation:
- The length of AB = 2 * 5 = 10 cm.
- The length of AP = AB - PB = 10 - 6 = 4 cm.
- The length of AQ = AP / 3 = 4 / 3 = 1.33 cm.
- The length of QB = AB - AQ = 10 - 1.33 = 8.67 cm.

Answer:
The length of QB, nearest to, is 8.67 cm, which is closest to option A (9.1 cm).
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AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to(2019)a)9.1b)7.8c)9.3d)8.5Correct answer is option 'A'. Can you explain this answer?
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