## RS Aggarwal Solutions Class 10 Chapter 12 Circles

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given 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.

**Read More:**

- How To Construct A Tangent To A Circle From An External Point
- Number Of Tangents From A Point On A Circle
- Parts of a Circle
- Perimeter of A Circle
- Common Chord of Two Intersecting Circle
- Construction of a Circle
- The Area of A Circle
- Properties of Circles
- Sector of A Circle
- The Area of A Segment of A Circle
- The Area of A Sector of A Circle

**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◦

\(\frac { { BC } }{ { OB } } =\sin { 30^{ \circ } } =\frac { 1 }{ 2 } \)

⇒ 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

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles are helpful to complete your math homework.

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