Question

A circle of radius 25 units has a chord going through a point that is located 10 units from the
centre. What is the shortest possible length that chord could have ?
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Answer 1

Answer:

Given, Chord going through a point that is located 10 units from the

centre and radius = 25 units.

Thus,

OD = 10 units and AO = 25 units.

By Pythagoras theorem,

AD = √AO² - OD²

=> AD = √25² - 10²

=> AD = √525

The line from the centre of circle divides chord in two equal parts.

AB = 2 CD

=> 2√525

=> √4 * 525

=> √2100 units

Therefore, shortest possible length = √2100 units.

Hope this help

Answer 2

For length of shortest chord we move in direction ⊥ to the length of chord .

From ΔAOC AC=

$\sqrt{ {25}^{2} - {10}^{2} } = \sqrt{525} \: \: units$

As OA=OB and ∠OCB=∠OCA=90⁰ (OC common)

∴ ΔOAC≃ΔOCB

⇒AC=BC

⇒AB= length of chord =2AC=

$2 \sqrt{525}$

$= \sqrt{2100} \: units$