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Answer :
Steps to construct the required figure:
Step 1: Draw a circle with center O and radius =6cm and draw a point P such that OP=10cm.
Step 2: Draw a perpendicular bisector of OP and mark a point M such that M is the mid- point of OP.
Step 3: With center as M and radius PM=MO, draw a circle which cuts the given circle at S and T.
Step 4: Now join PS and PT.
Thus, we have the required tangents as PS and PT and also the lengths of each tangent is 8cm.
JUSTIFICATION:
Join OS
Now in the triangle PSO,we have
\(∠PSO=90°\)
\(∴ PS= \sqrt{{OP}^2 -{OS}^2} \)
[by pythagoras’ theorem]
\(=\sqrt{{(10)}^2-{(6)}^2}=\sqrt{100-36} \)
\(=\sqrt{64}=8cm \)