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 in cm.

• A. 9.6

Answer 1

The correct option is A 9.6

In $\Delta$APC,
$\because$ AP is perpendicular to BC

$\Rightarrow {AC}^2={PC}^2+{AP}^2\Rightarrow {PC}^2={AC}^2-{AP}^2$
$={\left(6\right)}^2-{AP}^2=36-{AP}^2---\left(1\right)$

In $\Delta$POC,

${PC}^2={OC}^2-{OP}^2$
$=25-{(5-AP)}^2$​
[$\because$ PO = AO - AP = 5 - AP]

${PC}^2=25-(25+{AP}^2-10AP)$

${PC}^2=10AP-{AP}^2$ ------- (2)

Substitute (1) in (2)

$36-{AP}^2=10AP-{AP}^2$

$⟹AP=3.6cm$

${PC}^2=36-{AP}^2=36-12.96=4.8cm$

Since a perpendicular from the centre of a circle to a chord bisects the chord,
$BC=2\times PC=2\times 4.8=9.6cm.$