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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?
Question Description
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