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I have got a project to show that the perpendicular bisector of the chord passes thru the centre.?

I have to make a model which i know how 2 make but wht shud i do about the presentation part

pls help me regarding the presentation. Pls pls pls

Relevance
• albert
Lv 5

Draw the chord. Observe that , unless it goes thru the center of the circle, that it splits the circle into a large part and a smaller part. Now construct a perpendicular bisector on the chord. Note

that it's length to the edge of the circle on the larger part is greater than the radius else it would be on the smaller part.

Also the distance to the edge in the smaller part is less than

the radius. So somewhere on the perpendicular bisector the

distance to the edge is equal to the radius. At any

point on the bisector draw a pair of lines to the end points of the chord. They make a pair of congruent triangles by S.A.S. therefore the lines are of equal length. At some point the lengths must be equal to the length of the radius. The only place that can be is at the center.

• Anonymous

Surveyors use this fact to draw an arc based on 3 located points. The 3 points define 2 chords. If the perpendicular bisectors of these chords are drawn, they will intersect at the radius point of the arc. So to prove this, draw 3 points that are not in a straight line. Connect the points with 2 lines. Divide each line in half and construct a perpendicular at each midpoint. These lines will intersect at the radius point, so set the compass point at the radius point and the pencil (or chalk) at one of the points. Then draw an arc between the three points. Eureka! The arc will pass through each of the points.

• raj
Lv 7

join the ends of the chord to the centre.

now the figure will be an isosceles triangle with the chord as its base.

now the perpendicular from the vertex to the base of an isosceles triangle will bisect the base

hence proved

aliter

like before join the ends of the chord to the centre.

draw the perpendicular from the centre to the chord

now you have two right triangles with a common side,the hypotenuses equal and one of the angles a right angle.so by RHS congruency the triangles are congruent and by cpct the third sides are equal

hence proved