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.

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:

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

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.

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