Question

3. Let PA and PB are two tangents 0/2
of a circle with center O. If length
of radius is 2.4 cm and OP = 4 cm,
then perimeter of the quadrilateral
AOBP is
O 12.8 cm
Ο Ο Ο
O 11.2 cm
9.6 cm
None of these​

Answer 1

Answer:

Given, PA and PB are the tangents from the point P.

∠APB=54

∘

Now, In quadrilateral AOBP

∠OAP=∠OBP=90

∘

(Angle between tangent and radius)

Sum of angles = 360

∠OAP+∠OBP+∠OAB+∠APB=360

90+90+54+∠AOB=360

∠AOB=126

Now, In △OAB

OA=OB (Radius of circle)

∠OAB=∠OBA (Isosceles triangle property)

Sum of angles = 180

∠OAB+∠OBA+∠AOB=180

2∠OAB+126=180

∠OAB=27

∘