
In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is
A) 7.5
B) 15
C) 10
D) 9
Answer
568.8k+ views
Hint: Tangents From The Same External Point
The two tangent theorem states that if we draw two lines from the same point which lies outside a circle, such that both lines are tangent to the circle, then their lengths are the same.
We are going to use the first point to solve this problem as AP, AQ; BP, BD and CQ, CD are tangents from the same external point A, B and C respectively.
Complete step-by-step answer:
We know that: -
AP is a tangent to the circle and the point of tangency is P.
AQ is a tangent to the circle and the point of tangency is Q.
So, therefore AP= AQ as tangent segments to a circle from the same external point are congruent.
BP is a tangent to the circle and the point of tangency is P.
BD is a tangent to the circle and the point of tangency is D.
So, therefore BP= BD as tangent segments to a circle from the same external point are congruent.
CQ is a tangent to the circle and the point of tangency is Q.
CD is a tangent to the circle and the point of tangency is D.
So, therefore CQ= CD as tangent segments to a circle from the same external point are congruent.
Given values are AB = 5 cm, AC = 6 cm and BC = 4 cm
Lets calculate the value of\[AP = AB + BP\] (As shown in figure)
\[ \Rightarrow AP = 5 + BD\]( as AB= 5 cm and BP=BD)………………………..(1)
Lets calculate the value of\[AQ = AC + CQ\] (As shown in figure)
\[ \Rightarrow AQ = 6 + CD\]( as AC= 6 cm and CQ=CD)…………………..(2)
Let's add equation (1) and equation (2)
\[ \Rightarrow AP + AQ = 5 + BD + 6 + CD\]
\[ \Rightarrow AP + AQ = 11 + BD + CD\]
\[ \Rightarrow 2AP = 11 + BC\]( as AP=AQ AND BD+CD=BC)
\[ \Rightarrow 2AP = 11 + 4\](as BC= 4 cm)
\[ \Rightarrow AP = \dfrac{{15}}{2} = 7.5\]
Hence, the correct answer is option(A)
Note: The shape of the drawing in the problem statement – where the two tangent lines create a sort of triangular ‘clown’s hat’ also suggests we would be served by constructing some triangles here, where the two tangent lines are sides. So whenever a clown hat is present in a problem diagram you can use this theorem.
The two tangent theorem states that if we draw two lines from the same point which lies outside a circle, such that both lines are tangent to the circle, then their lengths are the same.
We are going to use the first point to solve this problem as AP, AQ; BP, BD and CQ, CD are tangents from the same external point A, B and C respectively.
Complete step-by-step answer:
We know that: -
AP is a tangent to the circle and the point of tangency is P.
AQ is a tangent to the circle and the point of tangency is Q.
So, therefore AP= AQ as tangent segments to a circle from the same external point are congruent.
BP is a tangent to the circle and the point of tangency is P.
BD is a tangent to the circle and the point of tangency is D.
So, therefore BP= BD as tangent segments to a circle from the same external point are congruent.
CQ is a tangent to the circle and the point of tangency is Q.
CD is a tangent to the circle and the point of tangency is D.
So, therefore CQ= CD as tangent segments to a circle from the same external point are congruent.
Given values are AB = 5 cm, AC = 6 cm and BC = 4 cm
Lets calculate the value of\[AP = AB + BP\] (As shown in figure)
\[ \Rightarrow AP = 5 + BD\]( as AB= 5 cm and BP=BD)………………………..(1)
Lets calculate the value of\[AQ = AC + CQ\] (As shown in figure)
\[ \Rightarrow AQ = 6 + CD\]( as AC= 6 cm and CQ=CD)…………………..(2)
Let's add equation (1) and equation (2)
\[ \Rightarrow AP + AQ = 5 + BD + 6 + CD\]
\[ \Rightarrow AP + AQ = 11 + BD + CD\]
\[ \Rightarrow 2AP = 11 + BC\]( as AP=AQ AND BD+CD=BC)
\[ \Rightarrow 2AP = 11 + 4\](as BC= 4 cm)
\[ \Rightarrow AP = \dfrac{{15}}{2} = 7.5\]
Hence, the correct answer is option(A)
Note: The shape of the drawing in the problem statement – where the two tangent lines create a sort of triangular ‘clown’s hat’ also suggests we would be served by constructing some triangles here, where the two tangent lines are sides. So whenever a clown hat is present in a problem diagram you can use this theorem.
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