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◦,
rs-aggarwal-class-10-solutions-circles-12-q1-1
By Pythagoras theorem:
rs-aggarwal-class-10-solutions-circles-12-q1-2
Hence, the length of the tangent = 15 cm.





Read More:





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◦,
RS Aggarwal Solutions Class 10 Chapter 12 Circles ex 12
By Pythagoras theorem:
RS Aggarwal Class 10 Solutions Chapter 12 Circles ex 12
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.
Solution of RS Aggarwal Class 10 Chapter 12 Circles ex 12
∴ ∠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.
RS Aggarwal Solutions Class 10 2017 Chapter 12 Circles ex 12
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.
Class 10 RS Aggarwal Solutions Chapter 12 Circles ex 12
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
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 12 Circles ex 12
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.
RS Aggarwal Class 10 Book pdf download Chapter 12 Circles ex 12
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
Class 10 Maths RS Aggarwal Solutions Chapter 12 Circles ex 12
In triangle OAP,
Maths RS Aggarwal Class 10 Solutions Chapter 12 Circles ex 12
Hence, BP = 10.9 cm





Question 9:
RS Aggarwal Solutions Class 10 2018 Chapter 12 Circles ex 12
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





Complete RS Aggarwal Solutions Class 10





If You have any query regarding this chapter, please comment on below section our team will answer you. We Tried our best to give complete solutions so you got good marks in your exam.





If these solutions have helped you, you can also share alarity.in to your friends.





Best of Luck For Your Future!!


Comments

Popular posts from this blog

KSEEB Solutions For Class 6 Social Science

KSEEB Solutions For Class 7 Hindi

KSEEB Solutions For Class 6 Science