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Showing posts from November, 2017

Pythagoras’ Theorem

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Over 2000 years ago there was an amazing discovery about triangles,  Although it is often argued that knowledge of the theorem predates him,  the theorem is named after the  ancient Greek  mathematician  Pythagoras  ( c.  570–495 BC) as it is he who, by tradition, is credited with its first recorded  proof . Pythagoras studied right triangles, and the relationships between the legs and the hypotenuse of a right triangle, before deriving his theory. When a triangle has a right angle (90°) ... ... and squares are made on each of the three sides, ... ... then the biggest square has the  exact same area  as the other two squares put together! It is called "Pythagoras' Theorem" and can be written in one short equation: a 2  + b 2  = c 2 Note: c  is the  longest side  of the triangle a  and  b  are the other two sides Definition The longest side of the triangl...

To prove that "a straight line drawn from the centre of a circle to bisect a chord which is not a diameter is perpendicular to the chord"

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Before going to the topic remember these terms, Chord:- A line segment connecting two points on a curve.  Diameter:- A straight line going through the center of a circle connecting two points on the circumference. Perpendicular Line:- Lines that are at right angles ( 90° ) to each other. When a perpendicular line is drawn from the center of the circle to the chord. Then the perpendicular line bisects (divide into 2 halves) the chord. Lets take both of them (AOP & BOP) as right angle triangle and we can see that from point O the radius is same. Here the distance of OP is same in both of the triangles. On the other hand, the angle AOP and BOP are also same. Hence, by using the property of S.A.S the length AP and BP are same and thus, are divided equally (bisect) by the chord. For its derivation lets proceed forward, Given a circle with centre O and AC = BC To  Prove: OC ⊥ AB Construction: Join points O, A and O, B. Proof: In Δ’s OCA and OCB OA =...

To prove that "one and only one circle can pass through three non-collinear points"

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Non-collinear points :- Non-collinear means simply ‘does not lies on a common line’, so it is a property of a set of points (at least two) of a geometry. We join these three points with line. After joining these points when we draw a perpendicular bisector to each of these lines. When perpendicular bisectors are drawn they cut each other and from that point the length of each line is equal. This line is the radius of the circle,hence a circle drawn on these non-collinear points. When there are three non-collinear points (not lying on a single line), then only one circle can be drawn on which all these points lie. But, if we try to draw any other circle on these points then it would also be of the  same  radius. Its derivation is given below, Given : Three non collinear points P, Q and R To prove : There is one and only one circle passing through the points P, Q and R. Construction:  Join PQ and QR. Draw perpendicular bisectors AB of PQ ...