# R S Aggarwal Class 10 Circles Solutions

## Circles Exercise Ex. 8A (By:- Ajay Raj)

### In PAO, A = 90

Hence, the length of the tangent = 15 cm.

### Solution 2

PA is the tangent to the circle with centre O and radius, such that PO = 25 cm, PA = 24 cm

In PAO, A = 90,

By Pythagoras theorem: Hence, the radius of the circle is 7 cm.

### Solution 5

Given AP is a tangent at A and OA is radius through A and PA and PB are the tangent segments to circle with centre O.

Therefore, OA is perpendicular to AP, similarly, OB is perpendicular to BP.  OAP = 90

And OBP = 90

So, OAP = OBP = 90  OBP + OAP = (90 + 90) = 180

Thus, the sum of opposite angles of quad. AOBP is 180 AOBP is a cyclic quadrilateral

### Solution 7

Given: From an external point P, tangent PA and PB are drawn to a circle with centre O. CD is the tangent to the circle at a point E and PA = 14cm.

Since the tangents from an external point are equal, we have

PA = PB,

Also, CA = CE and DB = DE

Perimeter of PCD = PC + CD + PD

=(PA - CA) + (CE + DE) +(PB - DB)

= (PA - CE) + (CE + DE) + (PB - DE)

= (PA + PB) = 2PA = (2 14) cm

= 28 cm

Hence, Perimeter of PCD = 28 cm

### Solution 8

A circle is inscribed in a triangle ABC touching AB, BC and CA at P, Q and R respectively. Also, AB = 10 cm, AR = 7cm, CR = 5cm

AR, AP are the tangents to the circle

AB = 10 cm BP = AB - AP = (10 - 7)= 3 cm

Also, BP and BQ are tangents to the circle

Further, CQ and CR are tangents to the circle

BC = BQ + CQ = (3 + 5) cm = 8 cm

Hence, BC = 8 cm

### Solution 9

Let the circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively

We know that the length of tangents drawn from an exterior point to a circle are equal AP = AS ----(1)    {tangents from A}

BP = BQ ---(2)     {tangents from B}

CR = CQ ---(3)    {tangents from C}

DR = DS----(4)    {tangents from D}

Adding (1), (2) and (3) we get AP + BP + CR + DR = AS + BQ + CQ + DS (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) AB + CD = AD + BC AD = (AB + CD) - BC = {(6 + 4) - 7} cm = 3 cm