In the given figure, O is the centre of the circle and B is a point of contact. seg OE ⊥ seg AD, AB = 12, AC = 8, find (1) AD (2) DC (3) DE.
Chapter 3 – Circle – Text Book Solution
Practice Set 3.5 | Q 3 | Page 82
In the given figure, O is the centre of the circle and B is a point of contact. seg OE ⊥ seg AD, AB = 12, AC = 8, find
(1) AD (2) DC (3) DE.
(1) Using tangent secant segment theorem, we have
AB2 = AC × AD
⇒ 122 = 8 × AD
⇒ AD = \[\frac{144}{8}\] = 18 units
(2)DC = AD − AC = 18 − 8 = 10 units
(3) CD is the chord of the circle with centre O.
Also, OE ⊥ CD (seg OE ⊥ seg AD)
∴ DE = EC = \[\frac{DC}{2} = \frac{10}{2}\] units (Perpendicular drawn from the centre of a circle on its chord bisects the chord)
Explanation:-
(1) Using the tangent secant segment theorem, we find that AB<sup>2</sup> = AC × AD. This gives us 12<sup>2</sup> = 8 × AD, or AD = 144/8 = 18 units.
(2) We can then find DC by subtracting AC from AD: DC = AD – AC = 18 – 8 = 10 units.
(3) CD is a chord of the circle with center O. Also, OE is perpendicular to CD (since segment OE is perpendicular to segment AD). Therefore, by the perpendicular bisector theorem for chords, we know that DE = EC = DC/2 = 10/2 = 5 units.
Chapter 3 – Circle – Text Book Solution
Practice set 3.5 | Q 3 | Page 82
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