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Theorem of Equal Chords Subtending Angles at the Center

Two chords ...

Question

Two chords $A B$ and $C D$ of a circle intersect at $E$ such that $A E = 2.4$ cm., $B E = 3.2$ cm and $C E = 1.6$ cm. The length of $D E$ is

A

1.6

cm

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B

3.2

cm

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C

4.8

cm

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D

6.4

cm

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Solution

The correct option is B $4.8$ cm
Applying the rule, $A E \times E B = C E \times E D$
$∴ 2 \cdot 4 \times 3 \cdot 2 = 1 \cdot 6 \times E D$
$∴ E D = 4 \cdot 8$ cm

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