Q. 1

In the picture, a triangle is formed by two mutually perpendicular tangents to a circle and a third tangent.



Prove that the perimeter of the triangle is equal to the diameter of the circle.

Answer :

Let us label the diagram and let the radius be 'r'.


To Prove : Perimeter of triangle PQR = diameter of circle


i.e. PQ + QR + PR = 2r


Construction: Join OA and OB



Proof:


In Quadrilateral OAPB


OAB + APB + OBP + AOB = 360°


Also,


OAB = 90° [OA AP, as tangent at any point on the circle is perpendicular to the radius through point of contact]


OBP = 90° [OB BP, as tangent at any point on the circle is perpendicular to the radius through point of contact]


APB = 90° [AP BP, Given]


90 + 90 + 90 + AOB = 360


⇒ ∠AOB = 90°


Also,


OA = OB [radii of same circle]


AP = BP [Tangents drawn from same point to a circle are equal]


And we know, if in a quadrilateral


i) All angles are 90° and


ii) Adjacent sides are equal


then the quadrilateral is square.


OAPB is a square


AP = BP = OA = OB = 'r' [radius] …[1]


Also, as tangents drawn from same points to a circle are equal


AQ = CQ [tangents from Q] …[2]


BR = CR [tangents from R] …[3]


Also,


Perimeter of triangle PQR


= PQ + QR + PR


= PQ + QC + CR + PR


= PQ + AQ + BR + PR [From 2 and 3]


= AP + BP


= r + r [From 1]


= 2r


Hence Proved.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses