Question : In the figure, chords AB and CD of a circle intersect externally at P. If AB = 4 cm, CD = 11 cm and PD =15 cm, then the length of PB is:
Option 1: 10 cm
Option 2: 8 cm
Option 3: 14 cm
Option 4: 12 cm
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 10 cm
Solution : AB and CD are two chords of a circle intersecting externally at P AB = 4 cm, CD = 11 cm and PD = 15 cm If two chords AB and CD intersect at outer point P then, PA × PB = PC × PD Let the length of PA be $x$ PC = PD - CD = 15 - 11 = PC = 4 cm According to the question PA × PB = PC × PD ⇒ $x$ × ($x$ + 4) = 15 × 4 ⇒ $x$ = 6 PB = PA + AB = 6 + 4 = 10 $\therefore$ The length of PB is 10 cm. Hence, the correct answer is 10 cm.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : In the following figure, O is the centre of the circle. Its two chords AB and CD intersect each other at point P inside the circle. If AB =18 cm, PB = 6 cm, and CP = 4 cm, then find the measure of PD.
Question : In the figure, O is the centre of the circle. Its two chords AB and CD intersect each other at the point P within the circle. If
Question : In the given figure, the diameter AB and chord CD of a circle meet at P. PT is tangent to the circle at T. If CD = 8 cm, PD = 10 cm, and PB = 8 cm, find AB. 2 Views
Question : Two chords AB and CD of a circle meet inside the circle at point P. If AP = 12 cm, AB = 20 cm, and CP = 16 cm, find CD.
Question : In the following figure, two circles touch each other externally. The radius of the first circle with centre P is 25 cm. The radius of the second circle with centre Q is 4 cm. Find the length of their direct common tangent, AB. The figure is not to scale and is only for
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile