Question

$<b><i>Heya Mates !!$♥♥♥♥


___________________


AB and AC are the two chords of the circle of radius r. If AB = 2AC and the perpendiculars drawn from the centre on these chords are of length 'a' and 'b', prove that :

4b^2 = a^2 + 3r^2


___________________

Class 9th

Chapter : Circles


_____________________

NO SPAM

ONLY PERFECT ANSWER !!



___________________

#BE BRAINLY

Answer 1

Refer the attachment for figure

We know that perpendicular from the centre bisects the chord.

So, perpendicular of length a will bisect AB

given that, AB = 2AC

=> 1/2 AB = AC

Similarly, perpendicular of length b will bisect AC

Half of AC = 1/2 AC

Now, For chord AC, apply Pythagoras Theorem :-

(1/2 AC)² + b² = r²

(r is the radius)

=> AC²/4 + b² = r²

=> AC²/4 = r² - b²

=> AC² = 4(r² - b²)

=> AC² = 4r² - 4b²

Now for Chord AB, apply Pythagoras Theorem


(1/2 AB²) + a² = r²

=> (AC)² + a² = r²

(since, 1/2 AB = AC)

=> AC² = r² - a²

But we have shown that AC² = 4r² - 4b².
Hence,

4r² - 4b² = r² - a²

=> - 4b² = r² - a² - 4r²

=> - 4b² = -3r² - a²

=> - (4b²) = -(3r² + a²)

Minus minus cancel

=> 4b² = 3r² + a²

Hence Proved ;)

Answer 2

here is your answer OK ☺☺☺☺☺☺☺☺


$\huge \red{thanks \: }$


given : AB and AC are two chord of the circle with centre O and radius R


such that AB = 2AC............. (1)

and length of the perpendicular from the OM and ON are a and b are respectively...

since the perpendicular from the centre to the chord bisect the chord .

AM = AB /2 and AN = AC /2

in the right angle triangle ONA, use Pythagoras theorem.....

${b}^{2} + {an}^{2} = {r}^{2}$
${an}^{2} = {r}^{2} - {b}^{2}$

Similarly in the right angle triangle OMA ,

${a}^{2} + {am}^{2} = {r}^{2}$
${am}^{2} = {r}^{2} - {b}^{2}$
${ \frac{ab}{2} }2 \: = {r}^{2} - {a}^{2}$
(because AM = ab/2= AC = 2AN)

(2AN )'2= r'2 - a'2
(4AN) = r'2 - a'2

from 3 and 4


4 (r'2 - b'2) = r'2 - a'2

-4b + 4 r'2 = r'2 - a'2

4b'2 = 3r'2 + a'2 ( answer OK)

$\huge \boxed{hope \: its \: help \: you}$