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CBSE Class 10 Basic Math 2025 All Sets Solved Paper

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Question : 1 of 20
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PQ is tangent to the circle centred at O. If OQ = 3 cm, PQ = 5 cm, then OP is equal to
Solution:  
The tangent at any point of a circle is perpendicular to the radius through the point of Contact.
Given:
OQ(Radius)=3 cmO Q(\text{Radius}) = 3 \text{ cm}
PQ=5 cmP Q = 5 \text{ cm}
By Pythagoras theorem:-
OP2=OQ2+PQ2O P^2 = O Q^2 + P Q^2
OP2=(3 cm)2+(5 cm)2O P^2 = (3 \text{ cm})^2 + (5 \text{ cm})^2
OP2=9 cm2+25 cm2O P^2 = 9 \text{ cm}^2 + 25 \text{ cm}^2
OP2=34 cm2O P^2 = 34 \text{ cm}^2
∴OP=34 cm\therefore O P = \sqrt{34} \text{ cm}
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