C-08-Q033mediumsingle_mcq

Corollary 1 of Theorem 17 states that the perpendicular bisector of any chord—

ais parallel to the chord
bis the chord itself
cpasses through the centre
dis the radius

ব্যাখ্যা

Since the centre is equidistant from the two endpoints of a chord, it lies on the chord's perpendicular bisector. Hence that perpendicular bisector always passes through the centre.

এই অধ্যায়ের আরও প্রশ্ন

C-08-Q034Corollary 2 of Theorem 17 states that a straight line cannot intersect a circle in more than—
C-08-Q035The converse stated in the Work box of Theorem 17 says: the perpendicular from the centre of a circle to a cho
C-08-Q036If is the midpoint of chord and is the centre, then holds for—
C-08-Q037From the centre , a perpendicular dropped on a chord of length cm meets the chord at distance—
C-08-Q038The Work box challenges the student to prove that the perpendicular from the centre—
C-08-Q039Theorem 18 states that all equal chords of a circle are—
C-08-Q040In Theorem 18, and are drawn—
C-08-Q041Because the perpendicular from the centre bisects the chord, equals—
C-08-Q042In the proof of Theorem 18, which congruence rule is used to show ?
C-08-Q043After , we conclude—

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