Question

hy--friends

plz give the solutions:-
1.) AB and AC are two chords of a circle of radius r such that AB= 2AC. If P and q are the distances of AB and AC from the centre then prove that:-
$4q {}^{2} = p {}^{2} + 3r {}^{2}$

Answer 1

Hi there !

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Given :

AB and AC are two chords of a circle of radius r such that AB= 2AC.

P and q are the perpendicular distances of AB and AC from the centre O, that is

OM = p and ON = q

And 'r' is the radius of the circle.

To prove :

$\mathsf {4q {}^{2} = p {}^{2} + 3r {}^{2}}$

Construction ; Join OA.

PROOF :

OM and ON are the perpendicular distances of AB and AC from the centre O.

AN = $\mathsf {\frac {AC} {2} }$

AM = $\mathsf {\frac {AB} {2} }$

[Because these are perpendicular from center to chord]

In right angled ΔONA,

$\mathsf{ON {} ^{2} + NA{} ^{2} = OA{} ^{2}}$

⇒ $\mathsf{q {} ^{2} + NA{} ^{2} = r{} ^{2}}$

⇒ $\mathsf{NA {} ^{2} = r{} ^{2} - q{} ^{2}} .... (i)$

In right angled ΔONA,

$\mathsf{OM {} ^{2} + AM{} ^{2} = OA{} ^{2}}$

⇒ $\mathsf{p {} ^{2} + AM{} ^{2} = r{} ^{2}}$

⇒ $\mathsf{AM {} ^{2} = r{} ^{2} - p{} ^{2}}....(ii)$

Since,

AM = $\mathsf {\frac {AB} {2} =\frac {2AC} {2} = AC = 2NA}$

Now, from (i) and (ii), we have ;

$\mathsf{r {}^{2} - p {}^{2} = AM {}^{2} = (2NA) {}^{2} = 4NA {}^{2} } \\ \\ \mathsf{r {}^{2} - p {}^{2} = 4[r {}^{2} - q {}^{2} ] } \\ \\ \mathsf{r {}^{2} - p {}^{2} = 4r {}^{2} - 4q {}^{2} } \\ \\ \mathsf{4 q {}^{2} = 3r {}^{2} + p {}^{2} }$

Hence, it is proved !

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Thanks for the question !

☺️❤️☺️

Answer 2

Given: AB and AC are two chords of a circle with center O, such that AB = 2AC

p and q are the perpendicular distances  of AB and AC from center O. i.e OM=p and ON=q

r is the radius of the circle.

TPT:

4q² =p² + 3r²


Proof: join OA.

OM and ON are perpendicular distances of AB and AC from center O.

AN=AC/2 and AM=AB/2

[perpendicular from center to chord intersect at mid-point of the chord]

In right angled triangle ONA,


ON² + NA² = OA²

q² + NA = r²

NA² = r² - q² .......... (1)


In the right angled triangle OMA,

OM² + AM² = OA²

p² + AM² = r²

AM² = r² - p² ......... (2)


Since,

AM = AB / 2 = 2AC / 2 = AC = 2NA


from (1) and (2)

r² - p² = AM² = (2NA)² = 4NA²

r² - p² = 4 [ r² - q²]

r² - p² = 4r² - 4q²

4q² = 3r² + p²

Hence, we got the required answer.

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Thanks !!.. ❤️