
From an external point \[P\], tangents \[PA\]and \[PB\]are drawn to a circle with centre \[O\].If \[CD\] is the tangent to the circle at point \[E\]and \[PA = 14cm\], find the perimeter of the \[\vartriangle PCD\].
Answer
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Hint: In the given question, we need to draw the figure and apply the concept of tangents on the circle from the point external to the circle. The tangents so formed will be of equal length and we can find the perimeter of the triangle by using the formula of adding all its side \[Perimeter = a + b + c\].
Complete step-by-step answer:
Perimeter of a triangle is the sum of the length of all the three sides. If the three sides of triangle are \[a\],\[b\],\[c\]then the formula to find perimeter will be:
\[Perimeter = a + b + c\]
Let us draw a diagram to solve the sum:
Draw a circle with centre \[O\]. From an external point \[P\], draw two straight lines bisecting the circle at point \[A\]and \[B\] such that\[PA = 14cm\]. We are given that \[CD\] is the tangent to the circle at point \[E\]. Hence the diagram will be as follows:
Since \[PA\]and \[PB\]are tangents to a circle, using the property that two tangents from a point external to circle are equal, we get,
\[PA = PB = 14cm\]
Moreover \[C\]and \[D\]are also points external to the circle forming two tangents from each point. So:
\[DB = DE\] and \[CA = CE\] ……(1)
We can further conclude from above that:
\[CD = CE + ED = DB + CA\] …..(2)
Also, from the diagram we can conclude that:
\[PA = PC + CA\] and \[PB = PD + DB\] …..(3)
Now the perimeter of \[\vartriangle PCD\] can be found out as follows:
\[Perimeter = PC + CD + PD\]
Using equation (2), we can say that
\[Perimeter = PC + CA + DB + PD\]
Now using equation (3), we can conclude that:
\[Perimeter = PA + PB\]
Since \[PA = PB = 14cm\], we get,
\[Perimeter = 14 + 14\]
\[Perimeter = 28cm\]
Hence the perimeter of \[\vartriangle PCD\]is \[28cm\].
So, the correct answer is “Option B”.
Note: The diagram should be properly drawn and observed to arrive at the solution. One should not forget to write the unit of measurement after finding out the perimeter.
One shouldn’t get confused by applying the radius of circle as radius is perpendicular to tangent as a method to solve the question. We have to apply the concept of tangent being equal from the external point.
Complete step-by-step answer:
Perimeter of a triangle is the sum of the length of all the three sides. If the three sides of triangle are \[a\],\[b\],\[c\]then the formula to find perimeter will be:
\[Perimeter = a + b + c\]
Let us draw a diagram to solve the sum:
Draw a circle with centre \[O\]. From an external point \[P\], draw two straight lines bisecting the circle at point \[A\]and \[B\] such that\[PA = 14cm\]. We are given that \[CD\] is the tangent to the circle at point \[E\]. Hence the diagram will be as follows:
Since \[PA\]and \[PB\]are tangents to a circle, using the property that two tangents from a point external to circle are equal, we get,
\[PA = PB = 14cm\]
Moreover \[C\]and \[D\]are also points external to the circle forming two tangents from each point. So:
\[DB = DE\] and \[CA = CE\] ……(1)
We can further conclude from above that:
\[CD = CE + ED = DB + CA\] …..(2)
Also, from the diagram we can conclude that:
\[PA = PC + CA\] and \[PB = PD + DB\] …..(3)
Now the perimeter of \[\vartriangle PCD\] can be found out as follows:
\[Perimeter = PC + CD + PD\]
Using equation (2), we can say that
\[Perimeter = PC + CA + DB + PD\]
Now using equation (3), we can conclude that:
\[Perimeter = PA + PB\]
Since \[PA = PB = 14cm\], we get,
\[Perimeter = 14 + 14\]
\[Perimeter = 28cm\]
Hence the perimeter of \[\vartriangle PCD\]is \[28cm\].
So, the correct answer is “Option B”.
Note: The diagram should be properly drawn and observed to arrive at the solution. One should not forget to write the unit of measurement after finding out the perimeter.
One shouldn’t get confused by applying the radius of circle as radius is perpendicular to tangent as a method to solve the question. We have to apply the concept of tangent being equal from the external point.
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