Question

AB is tangent to a circle with centre O. It touches the circle at the point P. If the radius of circle is 5 cm and OB = AB = 10 cm then find AP.

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Answer 1

$\Large\underline{\underline{\sf{\color{lime}{GIVEN:-}}}}$

• AB is tangent to a circle with center O. It touches the circle at the point P.
• Radius of the circle, r = 5 cm
• OB = AB = 10 cm

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$\Large\underline{\underline{\sf{\color{lime}{TO\ FIND:-}}}}$

The length of AP.

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$\Large\underline{\underline{\sf{\color{lime}{SOLUTION:-}}}}$

Since tangent is perpendicular to the radius, (OP ⊥ AB) so POB is a right-angles triangle.

Applying Pythagoras Theorem in POB :

OP² + PB² = OB²

→ (5)² + PB² = (10)²

→ PB² = 100 - 25

→ PB = √75

→ PB = 5√3

Taking √3 as 1.7 :

→ PB = 8.5 cm

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It is given that AB = 10 cm.

So, AP = AB - PB

→ AP = 10 - 8.5

→ AP = 1.5 cm

Therefore, the length of AP is 1.5 cm.

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♕ Peace out ♕

Answer 2

Step-by-step explanation:

given that AB = 10 cm.

So, AP = AB - PB

→ AP = 10 - 8.5

→ AP = 1.5 cm

Therefore, the length of AP is 1.5 cm.