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JAMB Mathematics 2011 Paper

Question : 25

Total: 46

Solution:

Concept:

The perpendicular from the centre of a circle to a chord bisects the chord.

Explanation:

Let the chord be

AB

and

O

be the centre of the circle.

The perpendicular from

O

meets

AB

N

, so

ON = 5

cm and radius

OA = 7

cm.

Since

ON

bisects

AB

, we have

AN = NB

Apply Pythagoras theorem in right-angled triangle

ONA

OA^2 = ON^2 + AN^2

7^2 = 5^2 + AN^2

49 = 25 + AN^2

AN^2 = 24

AN = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\text{ cm}

Therefore, full length of the chord is:

AB = AN + NB = 2\sqrt{6} + 2\sqrt{6} = 4\sqrt{6}\text{ cm}

Answer:

4\sqrt{6}

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