Question

i.
त्रिज्या r असलेल्या वर्तुळाच्या AB आणि AC या दोन जीवा आहेत. जर p आणि १ ही अनुक्रमे जीवा AB व
जीवा CD ची केंद्रापासून अंतर आहेत आणि जर AB = 2AC, तर
$4q {}^{2} = p {}^{2} + 3r {}^{2}$
सिद्ध करा

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Answer 1

Answer:

it is proved that 4q^2 = p^2 + 3r^2.

Step-by-step explanation:

Consider O as the centre of the circle with radius of r

So we get

OB = OC = r

Consider AC = x and AB = 2x

We know that OM ⊥ AB

So we get

OM = p

We know that ON ⊥ AC

So we get

ON = q

Consider △ OMB

Using Pythagoras' theorem

OB^2 = OM^2 + BM^2

We know that the perpendicular from the centre of the circle bisects the chord

So we get

r^2 = p^2 + ((1/2) AB)^2

It can be written as

r^2 = p^2 + ¼ × 4x^2

So we get

r^2 = p^2 + x^2 ……… (1)

Consider △ ONC

Using Pythagoras' theorem

OC^2 = ON^2 + CN^2

We know that the perpendicular from the centre of the circle bisects the chord

So we get

r^2 = q^2 + ((1/2) AC)^2

It can be written as

r^2 = q^2 + x^2/4

We get

q^2 = r^2 - x^2/4

Multiplying the equation by 4

4q^2 = 4r^2 - x^2

Substituting equation (1)

4q^2 = 4r^2 – (r^2 - p^2)

So we get

4q^2 = 3r^2 + p^2

Therefore, it is proved that 4q^2 = p^2 + 3r^2.