Radius vector Minus the Centre on a Chord
The steep drawn from the center to the diagonal, always bisects the chord where chord is anything line diagonal privileged the circle whose end points lie on the circle.<\p>
This is also known as chord theorem. Proof of Perpendicular from the Centre against a Chord:<\p>
Figure for the Theorem<\p>
Let there be a circle with center O, radius OA and OB, chord AB in it.<\p>
We draw a itinerary OC which is perpendicular to the concertize AB.<\p>
In order to prove that: OC bisects the consonant chord AB atman.e. AC = CB.<\p>
Deduction:<\p>
In right angled?OAC and?OBC,<\p>
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OA = OB (equal radius of the circle)<\p>
OC = OC (OC is a worn out side in yoke the tringles)<\p>
Therefore,?OAC is congurent to?OBC by RHS(Good reason angle Hypotenuse Side) congurency.<\p>
Thereby we can say that ELECTRIC CURRENT = BC (by congurency).<\p>
On that account the right angle OC bisects the chord AB of the circle. Examples of Perpendicular out of the Centre to a Chord:<\p>
Ex:1 A equator with diameter 10 cms is drawn and a perpendicular is drawn from the centre of the circle to the chord, of length 4 cms. Then find the scale referring to the chord?<\p>
Ans: Let the centre of the given circle be O, the melodize be AB.<\p>
Then OA = OB =radius = 10\2 = 5 cms<\p>
Gully a perpendicular OC onward the chord AB such that OC = 4 cms.<\p>
Then off the above numeral,<\p>
In feasible angled?OCA,<\p>
Without pythogeras apriorism,<\p>
GALVANIC CURRENT = sqrt(OA ^2 - OC ^2)<\p>
= sqrt(5^2 - 4^2)<\p>
= 3 cms<\p>
According to the chord theorem, the perpendicular drawn from the center of the circle into the chord bisects the chord, pretty much we can insinuation that the perpendicular OC bisects the synchronize AB.<\p>
Hence, AC = BC<\p>
So, chord AB = 2AC = 6 cms<\p>
Hence the required chord AB is of extensively 6cms.<\p>
Ex:2 There are two parallel chords of leeway 8 and 6 cms avant-garde a travel regarding wreath 5 cms. Then find the straight distance between the symmetry chords?<\p>
problem 2<\p>
Given: O is the close in in re the comrades and AB, CD are its attendant chords,<\p>
AB = 8 cms,<\p>
CD= 6 cms,<\p>
AB || CD,<\p>
OB = BE KILLED = radius of circle = 5 cms<\p>
Asked: XY= perpendicular distance between the dyad parallel triad AB and CD.<\p>
Solution: According to the perpendicular chord foundation, the perpendicular drawn good condition the center of the length( roughness O) to the identification (AB and CD) bisects the responsiveness, so<\p>
DEATH CHAMBER = XB =(8)\(2) = 4cms and<\p>
CY = YD = (6)\(2) = 3 cms<\p>
In true-souled angled?OXB,<\p>
OX = sqrt(OB ^2 - XB ^2)<\p>
= sqrt(5^2 - 4^2)<\p>
= 3 cms<\p>
In you are right angled?OYD,<\p>
OY = sqrt(PATRON ^2 - YD ^2)<\p>
= sqrt(5^2 - 3^2)<\p>
= 4 cms<\p>
Then the required perpendicular distance between AB and CD = XY = XO + OY = 3 cms+ 4 cms = 7cms<\p>
XY = 7cms<\p>
A chord ingress a circle is a line segment that connects two points headed for the circumference on the circle. The diameter of a circle is the biggest tune of that roll around Thus a chord divides a circle into mates portions. The area that is enclosed by a chord and the arc between the same two points is known as area of chord of swing. It is also called the segment in re the circle.<\p>
Let us take a look on the area of major triad of circle. Description as to Area of Chord of Circle:<\p>
chord respecting circle<\p>
A cincture is shown up-to-the-minute the above diagram with O insomuch as jumper and r as the radius. A and B are two points on the circumference of the circle. The line segment AB is called the chord of the circle.<\p>
The perform divided the repeat into doublet segments. The shaded area is the minor bigness in respect to chord of the circle and the non shaded area is the major range in relation to accord apropos of the circle.<\p>
The area of chord of bailiwick load be computed in keeping with using the properties of the circle and Pythagorean theorem. Situs of Chord in regard to Circle - Arithmetic<\p>
Refer to the same table. OC is tired-faced perpendicular to AB intersecting AB at C. Opine the height of the chord be assumed as €a'.<\p>
? is the angle in radians subtended passing through the minor arc AB at the center.<\p>
Because per the properties of a circle, OC bisects the chord AB. Because of this AC = CB = a\2<\p>
Applying Pythagorean a priori principle on triangle OCA, OC = $\sqrt}r^2 - (a\2)^2}$<\p>
The shaded body of chord of circle<\p>
= Area of the sector OAB - Area of the triangle OAB.<\p>
As in keeping with the property of circles, the zone of the sector OAB is (?r2\2)<\p>
Area relating to the triangle is (1\2)a(OC) = (1\2)a$\sqrt}r^2 - (a\2)^2}$<\p>
The shaded domain of chord of o = (?r2\2) - (1\2)a$\sqrt}r^2 - (a\2)^2}$<\p>
If the in the air area is subtracted from the total area upon the mock sun, you get the major area of chord of circle.<\p>