1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The figure below shows a circle centered at O and of radius 5 cm. AB and AC are two chords such that AB = AC = 6 cm. AP is perpendicular to BC. Find the length of the chord BC.

A

5.6 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

9.6 cm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

10.6 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

11.6 cm

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 9.6 cm In ΔAPC, ∵ AP is perpendicular to BC ⇒AC2=PC2+AP2⇒PC2=AC2−AP2 =(6)2−AP2 =36−AP2−−−(1) In ΔPOC, PC2=OC2−OP2 =25−(5−AP)2​ [∵ PO = AO - AP = 5 - AP] PC2=25−(25+AP2−10AP) PC2=10AP−AP2 ------- (2) Substitute (1) in (2) 36−AP2=10AP−AP2 ⟹AP=3.6 cm PC2=36−AP2 =36−12.96 =4.8 cm Since a perpendicular from the centre of a circle to a chord bisects the chord, BC=2×PC=2×4.8=9.6 cm.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Intersecting Chord Properties both Internally and Externally
MATHEMATICS
Watch in App
Join BYJU'S Learning Program