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: Identify Triangle Type:
$6^2 + 8^2 = 36+64 = 100 = 10^2$. So triangle ABC is right-angled at B (since AB and BC are legs, AC is hypotenus(e).
Step 2: Find BD (Altitude from B to A(c):
In a right triangle, area = $\frac{1}{2} \times AB \times BC = \frac{1}{2} \times 6 \times 8 = 24$. Also, area = $\frac{1}{2} \times AC \times BD = \frac{1}{2} \times 10 \times BD = 5 \times BD$. So $5 \times BD = 24 \implies BD = \frac{24}{5} = 4.8$.
Step 3: Understand the Circle:
Circle centered at B with radius $BD = 4.8$. It cuts AB at P and BC at Q. So BP = BQ = radius = 4.8. Given AB = 6, so AP = AB - BP = $6 - 4.8 = 1.2$. Given BC = 8, so QC = BC - BQ = $8 - 4.8 = 3.2$.
Step 4: Find the Ratio:
AP : QC = $1.2 : 3.2 = 12 : 32 = 3 : 8$? Wait, $1.2/3.2 = 12/32 = 3/8 = 0.375$. So ratio is 3:8. But the optionss have 4:1, 5:8, etc. 3:8 is not listed. Let's check: $1.2:3.2 = 12:32 = 3:8$. So AP:QC = 3:8. But the optionss don't have 3:8. Let's re-check: AB=6, BC=8, AC=10. BD = (6*8)/10 = 48/10 = 4.8. Yes. Then AP = 6 - 4.8 = 1.2; QC = 8 - 4.8 = 3.2. Ratio = 1.2:3.2 = 12:32 = 3:8. So it should be 3:8, which is the last options. The list shows 3:8 as an options? Yes, the last options is 3:8. So that is correct.
Step 5: Final Answer:
AP:QC = 3:8.