Question

If two chords AB and CD intersect externally at P and if AB = 4, PB = 5, PD = 3, then CD =

• A. 4.3
• B. 4
• C. 1.5
• D. 1.3

Answer 1

The correct option is D

1.3

According BPT

We have,

$\frac{PC}{CD}=\frac{PB}{BA}$

$\frac{PC}{CD}=\frac{5}{4}$

Since PC = PD - CD

$\frac{PD-CD}{CD}=\frac{5}{4}$

$4(PD-CD)=5CD$

$4PD-4CD=5CD$

$4PD=5CD+4CD$

$4PD=9CD$

$4\left(3\right)=9CD$ (Given PD = 3cm)

$12=9CD$

$CD=\frac{12}{9}$

$CD=\frac{4}{3}=1.3$