Two mutually perpendicular tangents are drawn to a circle with radius $R\sqrt{2}$ units. The shortest distance between the two points of contact is :

1. R units

2. $\dfrac{1}{2}R$ units

3. $R\sqrt{2}$ units

4. 2R units

For the three circles with centers A, B and C and radii 5 cm, 2 cm and 6 cm respectively.

Assertion (A) : To find the perimeter of the triangle ABC, add the radii of given three circles.

Reason (R) : The required perimeter is the product of sum of radii by 2.

1. A is true, R is true

2. A is true, R is false

3. A is false, R is true

4. A is false, R is false

Chords AD and BC one produce meet at exterior point P.

Assertion(A): PD x AD = PC x BC.

Reason(R): In triangles PAB and PCD.

∠PAB = ∠PCD ⇒ ΔPAB ∼ ΔPCD

$\Rightarrow \dfrac{PD}{PB} = \dfrac{PC}{PA}$

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Two circles touch each other externally at point P. OA and OB are the tangent of the two circles (as shown) and OA = 10 cm.

Statement (1): OB = 10 cm.

Statement (2): On joining O and P, tangent OP = tangent OA and tangent OP = tangent OB

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.