Question

75. ABand AC are two chords of a circle of
radius r such that AB = 2 AC. If p and q are
the distances of AB and AC from the centre,
then prove that 4q² = 4p² + 3r².​

Answer 1

Answer:

$\implies$

Step-by-step explanation:

$\rm{ let \: ac = a} \\ \rm{then \: AB = 2a} \\ \rm{ from \: centre \: 0 \: perpendicular \: is } \\ \rm {drawn \: to \: the \: cords \: ac \: and \: ab \: at \: m \: and \:n }$

$\rm{therefore \: am = mc = \frac{a}{2}} \\ \rm{and \: an \: = nb = a}$

$\rm{in \: triangle \: oma \: and \: tringle \: ona} \\ \rm{ \: \: \: \: \: by \: pythagras \: theorem} \\ \rm{ {ao}^{2} = {am}^{2} + {mo}^{2} } \\ \implies \rm{ {ao}^{2} = { \frac{a}{2} }^{2} + {q}^{2} \: \: \: \rm {i)}} \\ \implies \rm{ {ao}^{2} = {a}^{2} + {p}^{2} } \: \: \: \: \: \rm{(2)}$

from equation I and 2 we get,

$\large\rm{ { \frac{a}{2} }^{2} + {q}^{2} = {a}^{2} + 4 {p}^{2} }$

$\implies \rm \large{ {4q}^{2} = {4p}^{2} + {3r}^{2} } \\ \rm{because \: in \: triangle \: ona \: {r}^{2} = {a}^{2} + {p}^{2} }$

hence proved

Answer 2

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