Question

5) O is the centre of the circle with radius 26cm. seg AB is the chord . Seg OM ⊥ ℎ . l(OM) = 10cm. Find the length of chord AB.

Answer 1

Step-by-step explanation:

(refer to the attachment)

•QUESTION•

r = 26 cm

OM= 10 cm

AB= ?

•ANSWER•

1. OM is perpendicular to AB.

so,

AM=BM

2. AB=AM+BM=BM+BM=2BM....(1)

3. in triangle OMB,

Angle OMB is a right angle

so,

OMB is a right angled triangle

4. By Pythagoras theorem,

$ob {}^{2} = om {}^{2} + bm {}^{2} \\ = > bm {}^{2} = ob {}^{2} - {om}^{2} \\ = > bm = \sqrt{ {ob}^{2} - {om}^{2} } = \sqrt{26 {}^{2} - 10 {}^{2} } = \sqrt{676 - 100} = \sqrt{576} = 24$

BM=24

5. from (1),

AB=2BM=2×24=48