Question

Please answer this question, and do not spam, if you know the correct answer then only answer this. correct answer wip be marked brainliest.​

Answer 1

Answer:

Recheck the numericals. And don't forget to mark me as brainliest.

Step-by-step explanation:

OA and OC are radii of the circle .

So, OA=OC=x

Since the perpendicular from the centre to a chord bisects the chord.

So, CP= 11/2 cm

AQ= 5/2 cm

OP=✓(OC^2-PC^2)

= ✓(x^2- 121/4 )

OQ=✓(OA^2 - AQ^2)

= ✓(x^2 - 25/4 )

Given,

PQ=3

=> OQ-OP= 3

=> ✓(x^2 - 25/4

$\sqrt{ {x}^{2} - \frac{25}{4} } - \sqrt{ {x}^{2} - \frac{121}{4} } = 3 \\ = > {x}^{2} - \frac{25}{4} + {x}^{2} - \frac{121}{4} - 2 \sqrt{( {x}^{2} - \frac{25}{4} )( {x}^{2} - \frac{121}{4} )} = 9 \\ = > 2 {x}^{2} - \frac{146}{4} - 9 = 2 \sqrt{( {x}^{2} - \frac{25}{4} )( {x}^{2} - \frac{121}{4} )} \\ = > (2 {x}^{2} - \frac{91}{2} ) = 2 \sqrt{( {x}^{4} - \frac{146}{4} {x}^{2} + \frac{3025}{16} ) } \\ = > 4 {x}^{4} + \frac{8281}{4} - 182 {x}^{2} = 4 {x}^{4} - 146 {x}^{2} + \frac{3025}{4} \\ = > \frac{8281}{4} - 182 {x}^{2} + 146 {x}^{2} - \frac{3025}{4} = 0 \\ = > - 36 {x}^{2} + \frac{5256}{4} =0 \\ = > 3 6{x}^{2} = \frac{5256}{4} \\ = > {x}^{2} = \frac{146}{4} \\ = > x = \frac{ \sqrt{146} }{2} \\ = > x = 6.04 \: \: \: \: \: \: \: \: \: \: \: (approx.)$

)- ✓(x^2- 121/4 )=3

=>