Question

In the figure 6.19, C is the centre of the circle. seg QT is a diameter CT=13, CP=5, find the length of chord RS.

Answer 1

Solution:-

given by:- C is the centre of the circle.
CT = 13
CP = 5
THEN ,
QP = CT - CP = 13-5 = 8
in Triangle SPC
HERE , CS = 13 , CP = 5
$( {cs)}^{2} = {(ps)}^{2} + {(cp)}^{2} \\ ps = \sqrt{ {13}^{2} - {5}^{2} } \\ ps = \sqrt{169 - 25} \\ = \sqrt{144} = 12$
PS = 12 ,
THEN
RS = 2×PS

CHORD RS = 2×12 = 24

■I HOPE ITS HELP■

Answer 2

Given : C is the centre of the

circle .

CP = 5 , CT = CS = 13 ,

QT perpendicular bisector to RS.

In ∆CPS ,

<CPS = 90°

By Phythogarian theorem ,

PS² + CP² = CS²

5² + CP² = 13²

CP² = 13² - 5²

CP² = 169 - 25

CP² = 144

CP = √144 = 12

Therefore ,

RS = 2 × CP

=> RS = 2 × 12 = 24

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