In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 12 Circles Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 12 Circles Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 12 Circles Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 112 |

Chapter Name | Circles |

Exercise | 12 |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 12 Circles**

**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

Hence, AD = 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◦

BCOB=sin30∘=12

⇒ 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

Hence, radius = 14 cm

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