Two chords AB and CD of a circle intersect each other at P outside the circle. If AB 5 cm, BP 3 cm and PD 2 cm, find CD.
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It is given that AB and CD of a circle intersect each other at P outside the circle. If AB=6cm,BP=2cm and PD=2.5cm
So we get
AP×BP=CP×DP
From the figure we know that CP=CD+DP
By substituting the values
8×2=(CD+2.5)×2.5 cm
consider x=CD
So we get
8×2=(x+2.5)×2.5
On further calculation
16=2.5x+6.25
It can be written as
2.5x=16−6.25
By subtraction
2.5x=9.75
By division
x=
2.5
9.75
So we get
x=3.9cm
Therefore, CD=3.9cm
Step-by-step explanation:
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