From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm find the radius of the circle.
Answers
[tex]\large\bold{{Question :}}[/tex]
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm find the radius of the circle.
[tex]\large\bold\red{\underline{Solution :-}}[/tex]
Here, O is the center of the circle.
⟼ Given :
OQ = 25 cmPQ = 24 cm⟼ To Find : We have to find the radius OP.
Since QP is tangent, OP perpendicular to QP.
(Since, Tangent is Perpendicular to Radius ⠀⠀⠀⠀⠀⠀⠀at the point of contact)
So, ∠OPQ=90°
⟼ By Applying Pythagoras Theorem :
OP² + RQ² = OQ²
OP² + (24)² = (25)²
OP² = 625 - 576
OP² = 49
OP = √49
OP = 7 cm
Hence, The Radius is 7 cm
⠀⠀
⠀
-MissAbhi
Here after drawing the diagram for the question we came to knew that the length of line OQ is 25 cm and length of line PQ is 24 cm.
(Angle OP is of 90°)So line OQ is hypotenuse , line OP is perpendicular , line PQ is base.
Let us simply apply the concept of Pythagoras theorem to find out the length of line OP.
Base (PQ) = 24 cm Hypotenuse (OQ) = 25 cm Perpendicular (OP) = ?[tex]: \: \implies \: \sf{(Hypotenuse) {}^{2} \: = \: (Base) {}^{2} \: + \: (Perpendicular) {}^{2} } \\ \\ : \: \implies \: \sf{(OQ) {}^{2} \: = \: (OP) {}^{2} \: + \: (PQ) {}^{2} } \\ \\ : \: \implies \: \sf{(25) {}^{2} \: = \: (OP) {}^{2} \: + \: (24) {}^{2} } \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: (25) {}^{2} - \: (24) {}^{2}} \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: (25 \times 25) - \: (24 \times 24)} \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: (625) - \: (576)} \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: 625 - \: 576} \\ \\ : \: \implies \: \sf{(OP) {}^{2} \: = \: 49} \\ \\ : \: \implies \: \sf{OP \: = \: \sqrt{49} } \\ \\ : \: \implies \: \red{\bf{OP \: = \: 7}}[/tex]
★ Therefore,
Radius of the circle is of 7 cm.
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