Question

In a triangle ABC, If, AB = 6, BC = 8 and AC = 10. Further, if a perpendicular which is dropped from B, meets the side AC at D. A circle of radius BD (with center (b) is drawn. If the circle cuts AB and BC at P and Q respectively, the AP:QC will be equal to which of the following?

• A. 1:4
• B. 4:1
• C. 5:8
• D. 1:3
• E. 3:8

Hint:

In a right triangle, the altitude to the hypotenuse can be found using the area formula: $h = \frac{ab}{c}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
First, observe the lengths of the sides of \(\Delta ABC\): 6, 8, and 10. Since \(6^2 + 8^2 = 10^2\) (\(36 + 64 = 100\)), the triangle is a right-angled triangle with the right angle at vertex B.
Step 2: Key Formula or Approach:
1. Altitude to the hypotenuse in a right triangle: \( \text{Altitude} = \frac{\text{Base} \times \text{Perpendicular}}{\text{Hypotenuse}} \).
2. The radius of the circle is equal to this altitude BD. Points P and Q are on the sides AB and BC respectively, such that \(BP = BQ = \text{Radius}\).
Step 3: Detailed Explanation:
1. Calculate the length of the perpendicular BD (the altitude from B to hypotenuse AC):
\[ BD = \frac{AB \times BC}{AC} \]
\[ BD = \frac{6 \times 8}{10} = \frac{48}{10} = 4.8 \]
2. The radius of the circle with center B is \(BD = 4.8\).
3. The circle cuts side AB at P. Since B is the center, \(BP = \text{radius} = 4.8\).
Calculate AP:
\[ AP = AB - BP = 6 - 4.8 = 1.2 \]
4. The circle cuts side BC at Q. Similarly, \(BQ = \text{radius} = 4.8\).
Calculate QC:
\[ QC = BC - BQ = 8 - 4.8 = 3.2 \]
5. Determine the ratio AP : QC:
\[ AP : QC = 1.2 : 3.2 \]
\[ AP : QC = 12 : 32 \]
Dividing both by the common factor 4:
\[ AP : QC = 3 : 8 \].
Step 4: Final Answer:
The ratio AP:QC is equal to 3 : 8.