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 toOption~(a) 9.3(b) 7.8(c) 9.1(d) 8.5Correct answer is 'c'. Can you explain this answer?

Question

Answer

Since AB is a diameter, AQB and APB’ will right angles.

In right triangle APB,

Now, 2AQ = AP

=>AQ= 8/2=4

In right triangle AQB,

Answer

Given AB is a diameter of a circle with radius 5 cm, we need to find the length of QB.

Let's break down the information given in the question:

1. AB is a diameter of the circle: This means that AB is twice the length of the radius, which is 5 cm. So AB = 2 * 5 = 10 cm.

2. Length of PB is 6 cm: Let's denote the center of the circle as O. Since AB is a diameter, O lies on the line segment AB. Let's consider the point where PB intersects the circle and call it X. Since PB is a chord of the circle, it divides the diameter AB into two equal parts. Therefore, AX = BX = 10/2 - 6 = 5 - 6 = -1 cm. But distance can't be negative, so we take the absolute value of -1, which is 1 cm.

3. Length of AP is twice that of AQ: Let's consider the point where AP intersects the circle and call it Y. Since AP is a chord of the circle, it divides the diameter AB into two equal parts. Therefore, AY = YB = 10/2 = 5 cm. Given that AP is twice AQ, we can say that AQ = 5/3 cm and AP = 10/3 cm.

To find the length of QB, we can use the property of intersecting chords in a circle:

PB * QB = AB * YB

Substituting the values we know:

6 * QB = 10 * 5

6 * QB = 50

QB = 50/6

The length of QB is approximately 8.33 cm.

Therefore, the length of QB, nearest to the given options, is 8.5 cm, which corresponds to option (d).

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