RS Aggarwal Solutions Chapter 12 Circles Exercise 12A Class 10 Maths
Chapter Name | RS Aggarwal Chapter 12 Circles |
Book Name | RS Aggarwal Mathematics for Class 10 |
Other Exercises |
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Related Study | NCERT Solutions for Class 10 Maths |
Exercise 12A Solutions
1. Find the length of tangent drawn to a circle with radius 8 cm form a point 17 cm away from the center of the circle.
Solution
Let P be a point, such that
OP = 17 cm
Let OT be the radius, where
OT = 5 cm
Join TP, where TP is a tangent.
Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.
∴ OT ⊥ PT
In the right ∆OTP, we have:
OP2 = OT2 + TP2 [By Pythagoras’ theorem]
∴ The length of the tangent is 15 cm.
2. A point P is 25 cm away from the center of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
Solution
Draw a circle and let P be a point such that OP = 25 cm
Let TP be the tangent, so that TP = 24 cm
Join OT where OT is radius.
Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.
∴ OT ⊥ PT
In the right ∆OTP, we have:
OP2 = OT2 + TP2 [By Pythagoras’ theorem]
∴ The length of the tangent is 7 cm.
3. Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the larger circle which touches the smaller circle.
Solution
In right triangle AOP
AO2 = OP2 + PA2
⇒ (6.5)2 = (2.5)2 + PA2
⇒ PA2 = 36
⇒ PA = 6 cm
Since the perpendicular drawn from the center bisects the chord.
∴ PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle is 12 cm.
4. In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F Respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the length of AD, BE and CF.
We know that tangent segments to a circle from the same external point are congruent.
Now, we have
AD = AF, BD = BE and CE = CF
Now, we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm ...(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm ....(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒ 2(AD + BD + FC) = 30
⇒ AD + BD + FC = 15 cm ...(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm
5. In the given figure, a circle touches all the four of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
Given, AB = 6 cm, BC = 7 cm and CD = 4 cm
Tangents drawn from an external point are equal.
∴ AP = AS, BP = BQ, CR = CQ and DR = DS
Now, AB + CD (AP + BP) + (CR + DR)
⇒ AB + CD = (AS + BQ) + (CQ + DS)
⇒ AB + CD = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AD = (AB + CD) – BC
⇒ AD = (6 + 4) – 7
⇒ AD = 3 cm.
∴ The length of AD is 3 cm.
6. In the given figure, the chord AB of the larger of the two concentric circles, with center O, touches the smaller circle at C. Prove that AC = CB.
Construction: Join OA, OC and OB
We know that the radius and tangent are perpendicular at their point of contact
∴ ∠OCA = ∠OCB = 90°
Now, In ∆OCA and ∆OCB
∠OCA = ∠OCB = 90°
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
∆OCA ≅ ∆OCB
∴ CA = CB
7. From an external point P, tangents PA and PB are drawn to a circle with center O. If CD is the tangent to circle at a point E and PA = 14 cm, find the perimeter of ∆PCD.
Given, PA and PB are the tangents to a circle with center O and CD is a tangent at E and
PA = 14 cm
Tangents drawn from an external point are equal.
∴ PA = PB, 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 (∵ PA = PB)
= (2 × 14) cm
= 28 cm
∴ Perimeter of ∆PCD = 28 cm
8. A circle is inscribed in a ∆ABC touching AB, BC and AC at P, Q and R respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.
Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle
Tangents drawn to a circle from an external point are equal.
∴ AP = AR = 7 cm, CQ = CR = 5 cm
Now, BP = (AB – AP)
= (10 – 7)
= 3 cm
∴ BP = BQ = 3 cm
∴ BC = (BQ + QC)
⇒ BC = 3 + 5
⇒ BC = 8
∴ The length of BC is 8 cm.
9. In the given figure, PA and PB are the tangents segments to a circle with centre O. Show that the points A, O, B and P are concyclic.
Here, OA = OB
And OA ⊥ AP, OA ⊥ BP (Since tangents drawn from an external point are perpendicular to the radius at the point of contact)
∴ ∠OAP = 90°, ∠OBP = 90°
∴ ∠OAP + ∠OBP = 90° + 90° = 180°
∴ ∠AOB + ∠APB = 180° (Since, ∠OAP + ∠OBP + ∠AOB + ∠APB = 360°)
Sum of opposite angle of a quadrilateral is 180°.
Hence, A, O, B and P are concyclic.
10. In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that point of contact P bisects the base BC.
We know that tangent segments to a circle from the same external points are congruent
Now, we have
AR = AO, BR = BP and CP = CQ
Now, AB = AC
⇒ AR + RB = AQ + QC
⇒ AR + RB = AR + OC
⇒ RB = QC
⇒ BP = CP
Hence, P bisects BC at P.
11. In the given figure, O is the centre of the two concentric circles of radii 4 cm and 6 cm respectively. AP and PB are tangents to the outer and inner circle respectively. If PA = 10 cm, find the length of PB upto one place of the decimal.
Given, O is the center of two concentric circles of radii OA = 6 cm and OB = 4 cm.
PA and PB are the two tangents to the outer and inner circles respectively and PA = 10 cm.
Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.
∴ ∠OAP = ∠OBP = 90°
∴ From right-angled ∆OAP, OP2 = OA2 + PA2
∴ The length of PB is 10.9 cm.
12. In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BC and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ∆ABC = 54 cm2 then find the length of sides AB and AC.
Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F.
We know that tangent segments to a circle from me same external point are congruent
Now, we have
AE = AF, BD = BE = 6 cm and CD = CF = 9 cm
Now,
Area (∆ABC) = Area (∆BOC) + Area(∆AOB) + Area (∆AOC)
⇒ 54 = 1/2 × BC × OD + 1/2 × AB × OE + 1/2 × AC × OF
⇒ 108 = 15 × 3 + (6 + x) × 3 + (9 + x) × 3
⇒ 36 = 15 + 6 + x + 9 + x
⇒ 36 = 30 + 2x
⇒ 2x = 6
⇒ x = 3 cm
∴ AB = 6 + 3 = 9 cm and AC = 9 + 3 = 12 cm
13. PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
We know that the perpendicular drawn from the center to me chord bisects it.
∴ PR = RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pythagoras theorem, we have
PO2 = OR2 + PR2
⇒ 32 = OR2 + (2.4)2
⇒ OR2 = 3.24
⇒ OR = 1.8
Now, in right triangle TPR
By using Pythagoras theorem, we have
TP2 = TR2 + PR2
⇒ x2 = y2 + (2.4)2
⇒ x2 = y2 + 5.76 ...(1)
Again, In right triangle TPQ
By using Pythagoras theorem, we have
TO2 = TP2 + PO2
⇒ (y + 1.8)2 = x2 + 32
⇒ y2 + 3.6y + 3.24 = x2 + 9
⇒ y2 + 3.6y = x2 + 5.76 ...(2)
Solving (1) and (2), we get
x = 4 cm and y = 3.2 cm
∴ TP = 4 cm
14. Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.
Solution
Construction: Draw a line parallel to CD passing through O i.e., OP
We know that the radius and tangent are perpendicular at their point of contact.
∠OQC = ∠ORA = 90°
Now, ∠OQC + ∠POQ = 180° (co-interior angles)
⇒ ∠POQ = 180° - 90° = 90°
Similarly, Now, ∠ORA + ∠POR = 180° (co-interior angles)
⇒ ∠POQ = 180° - 90° = 90°
Now, ∠POR + POQ = 90° + 90° = 180°
Since, ∠POR and ∠POQ are linear pair angles whose sum is 180°
Hence, QR is a straight line passing through center O.
15. In the given figure, with center O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm then find the radius of the circle.
We know that tangent segments to a circle from the same external point are congruent
Now, we have
DS = DR, AR = AQ
Now AD = 23 cm
⇒ AR + RD = 23
⇒ AR = 23 – RD
⇒ AR = 23 – 5 [∴ DS = DR = 5]
⇒ AR = 18 cm
Again. AB = 29 cm
⇒ AQ + QB = 29
⇒ QB = 29 – AQ
⇒ QB = 29 – 18 [∵ AR = AQ = 18]
⇒ QB = 11 cm
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
16. In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA : AT = 2 : 1.
AB is the chord passing through the center
So, AB is the diameter
Since, angle in a semicircle is a right angle
∴ ∠APB = 90°
By using alternate segment theorem
We have ∠APB = ∠PAT = 30°
Now, in ∆APB
∠BAP + ∠APB + ∠BAP = 180° (Angle sum property of triangle)
⇒ ∠BAP = 180° - 90° - 30° = 60°
Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)
⇒ 60° = 30° + ∠PTA
⇒ ∠PTA = 60° - 30° = 30°
We know that sides opposite to equal angles are equal
∴ AP = AT
In right triangle ABP
sin ∠ABP = AP/BA
⇒ sin 30° = AT/BA
⇒ 1/2 = AT/BA °
∴ BA : AT = 2 : 1