C-08-Q051mediumsingle_mcq

From $AE = CF$ and the fact that perpendicular from centre bisects chord, we conclude—

a $\frac{1}{2}AB = \frac{1}{2}CD$ so $AB = CD$
b $AB = 2CD$
c $AB = \frac{1}{2}CD$
d $AB \perp CD$

ব্যাখ্যা

Since the perpendicular from the centre bisects each chord, $AE = \tfrac12 AB$ and $CF = \tfrac12 CD$ ; with $AE = CF$ it follows that $\tfrac12 AB = \tfrac12 CD$ , hence $AB = CD$ .

এই অধ্যায়ের আরও প্রশ্ন

C-08-Q052Corollary 3 (following Theorem 19) states that—
C-08-Q053Of two chords of a circle, the one closer to the centre is—
C-08-Q054The greatest chord of a circle has length—
C-08-Q055The straight line joining the midpoints of two parallel chords of a circle—
C-08-Q056If two chords and of a circle subtend equal angles with the radius through , then—
C-08-Q057If a circle passes through the vertices of a right-angled triangle, then the centre of the circle is the—
C-08-Q058In two concentric circles a chord of the outer circle meets the inner circle at and . Then—
C-08-Q059If two equal chords of a circle intersect, the segments of one chord—
C-08-Q060Two equal chords drawn from the two ends of a diameter on its opposite sides are—
C-08-Q061Of two chords of a circle, the bigger chord is—

Circle — সব প্রশ্ন দেখুন (231টি) →

From $AE = CF$ and the fact that perpendicular from centre bisects chord, we conclude—

\frac{1}{2}AB = \frac{1}{2}CD

AB = CD

AB = 2CD

AB = \frac{1}{2}CD

AB \perp CD

ব্যাখ্যা

Since the perpendicular from the centre bisects each chord, $AE = \tfrac12 AB$ and $CF = \tfrac12 CD$ ; with $AE = CF$ it follows that $\tfrac12 AB = \tfrac12 CD$ , hence $AB = CD$ .

From AE=CFAE = CFAE=CF and the fact that perpendicular from centre bisects chord, we conclude—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

From AE=CFAE = CFAE=CF and the fact that perpendicular from centre bisects chord, we conclude—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

From $AE = CF$ and the fact that perpendicular from centre bisects chord, we conclude—

From $AE = CF$ and the fact that perpendicular from centre bisects chord, we conclude—