# RS Aggarwal Solutions Class 10 Chapter 12 Circles

Here you can get solutions of RS Aggarwal Solutions Class 10 Chapter 12 Circles. These Solutions are part of RS Aggarwal Solutions Class 10. we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles download pdf.

## RS Aggarwal Solutions Class 10 Chapter 12 Circles

### Exercise 12

Question 1:

PA is the tangent to the circle with center O and radius AO = 8 cm. The point P is at a distance of 17 cm from O.
In ∆ PAO, A = 90◦,

By Pythagoras theorem:

Hence, the length of the tangent = 15 cm.

Question 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.

Question 3:
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

Question 4:
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

Question 5:
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
AP = AR = 7cm
AB = 10 cm
BP = AB – AP = (10 – 7) = 3 cm
Also, BP and BQ are tangents to the circle
BP = BQ = 3 cm
Further, CQ and CR are tangents to the circle
CQ = CR = 5cm
BC = BQ + CQ = (3 + 5) cm = 8 cm
Hence, BC = 8 cm

Question 6:
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

Question 7:
Given: Two tangent segments BC and BD are drawn to a circle with centre O such that ∠CBD = 120◦.
Join OB, OC and OD.

In triangle OBC,
∠OBC = ∠OBD = 60◦
∠OCB = 90◦ (BC is tangent to the circle)
Therefore, ∠BOC = 30◦
[latex]frac { { BC } }{ { OB } } =sin { 30^{ circ } } =frac { 1 }{ 2 } [/latex]
⇒ OB = 2BC

Question 8:
Given O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. PA = 10cm. Join OA, OB and OP.
Then, OB = 4 cm, OA= 6 cm and PA = 10 cm

In triangle OAP,

Hence, BP = 10.9 cm

Question 9:

Join OR and OS, then OR = OS
OR ⊥DR and OS⊥DS
∴ ORDS is a square
Tangents from an external point being equal, we have
BP = BQ
CQ = CR
DR = DS
∴ BQ = BP = 27 cm
⇒ BC – CQ = 27 cm
⇒ 38 – CQ = 27
⇒ CQ = 11 cm
⇒ CR = 11 cm
⇒ CD – DR = 11 cm
⇒ 25 – DR = 11 cm
⇒ DR = 14 cm
⇒ r = 14 cm