RD Chapter 10 Circles Ex10.2 Solutions
Question - 31 : - In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS. [CBSE 2015, NCERT Exemplar]
Answer - 31 : -
In the given figure,
PQ and PR are tangents to the circle with centre O drawn from P
∠RPQ = 30°
Chord RS || PQ is drawn
To find ∠RQS
PQ = PR (tangents to the circle)
∠PRQ = ∠PQR But ∠RPQ = 30°
Question - 32 : - From an external point P, tangents PA = PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB. [CBSE 2016]
Answer - 32 : -
PA = PB [tangents drawn from external point are equal]
∠PBA = ∠PAB = 50° [angles equal to opposite sides]
∠APB = 180° – 50° – 50° = 80° [angle-sum property of a A]
In cyclic quad. OAPB
∠AOB + ∠APB = 180° [sum of opposite angles of a cyclic quadrilateral is 180°]
∠AOB + 80° = 180°
∠AOB = 180°- 80° = 100°
Question - 33 : - In the figure, two tangents AB and AC are drawn to a circle with centre O such that ∠BAC = 120°. Prove that OA = 2AB.
Answer - 33 : -
Given : In the figure, O is the centre of the circle.
AB and AC are the tangents to the circle from A such that
∠BAC = 120° .
To prove : OA = 2AB
Proof : In ∆OAB and ∆OAC
∠OBA = ∠OCA – 90° (OB and OC are radii)
OA = OA (common)
OB = OC (radii of the circle)
∆OAB ~ ∆OAC
∠OAB = ∠OAC = 60°
Now in right ∆OAB,
Question - 34 : - The lengths of three concecutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm, and 7 cm respectively. Determine the length of the fourth side.
Answer - 34 : -
In quadrilateral ABCD which is circumerscribing it
BC = 4 cm, CD = 5 cm and DA = 7 cm
We know that if a quad, is circumscribed in a circle, then
AB + CD = AD + BC
=> AB + 5 = 4 + 7
=> AB + 5 = 11
AB = 11 – 5 = 6
AB = 6 cm
Question - 35 : - The common tangents AB and CD to two circles with centres O and O’ intersect at E between their centres. Prove that the points O, E and O’ are collinear. [NCERT Exemplar]
Answer - 35 : -
Joint AO, OC and O’D, O’B
Now, in ∆EO’D and ∆EO’B
O’D = O’B [radius]
O’E = O’E [common side]
ED = EB
Question - 36 : - In the figure, common tangents PQ and RS to two circles intersect at A. Prove that PQ = RS.
Answer - 36 : -
Given : Two common tangents PQ and RS intersect each other at A.
To prove : PQ = RS
Proof: From A, AQ and AR are two tangents are drawn to the circle with centre O.
AP = AR ….(i)
Similarly AQ and AS are the tangents to the circle with centre C
AQ = AS ….(ii)
Adding (i) and (ii)
AP + AQ = AR + AS
=> PQ = RS
Hence proved.
Question - 37 : - Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle. [CBSE 2014]
Answer - 37 : -
Let R be the radius of outer circle and r be the radius if small circle of two concentric circle
AB is the chord of the outer circle and touches the smaller circle at P
Join OP, OA
Question - 38 : - AB and CD are common tangents to two circles of equal radii. Prove that AB = CD. [NCERT Exemplar]
Answer - 38 : -
Given : AB and CD are tangents to two circles of equal radii.
To prove :
Construction : Join OA, OC, O’B and O’D
Proof: Now, ∠OAB = 90°
[tangent at any point of a circle is perpendicular to radius through the point of contact]
Thus, AC is a straight line.
Also, ∠OAB + ∠OCD = 180°
AB || CD
Similarly, BD is a straight line and ∠O’BA = ∠O’DC = 90°
Also, AC = BD
[radii of two circles are equal] In quadrilateral ABCD,
∠A = ∠B = ∠C = ∠D = 90°
andAC = BD
ABCD is a rectangle
Hence, AB = CD
[opposite sides of rectangle are equal]
Question - 39 : - A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆PQR is 336 cm², find the sides PQ and PR. [CBSE 2014]
Answer - 39 : -
∆PQR is circumscribed by a circle with centre O and radius 8 cm
T is point of contact which divides the line segment OT into two parts such that
QT = 14 cm and TR = 16 cm
Area of ∆PQR = 336 cm²
Let PS = x cm
QT and QS are tangents to the circle from Q
QS = QT = 14 cm
Similarly RU and RT are tangents to the circle
RT = RU = 16 cm
Similarly PS and PU are tangents from P
PS = PU = x
Now PQ = x + 14 and PR = x + 16 and QR = 14 + 16 = 30 cm
Now area of ∆PQR = Area of ∆POQ + area of ∆QOR + area of ∆POR
Question - 40 : - In the figure, the tangent at a point C of a circle and a diameter AB when extended itersect at P. If ∠PCA = 110°, find ∠CBA. [NCERT Exemplar]
Answer - 40 : -
Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.
Join OC. Here, OC is radius.
Since, tangent at any point of a circle is perpendicular to the radius through point of contact circle.
OC ⊥ PC
Now, ∠PCA = 110° [given]
=> ∠PCO + ∠OCA = 110°
=> 90° + ∠OCA = 110°
=> ∠OCA = 20°
OC = OA = Radius of circle
∠OCA = ∠OAC = 20°
[since, two sides are equal, then their opposite angles are equal]
Since, PC is a tangent, so
∠BCP = ∠CAB = 20°
[angles in a alternate segment are equal]
In ∆PBC, ∠P + ∠C + ∠A= 180°
∠P = 180° – (∠C + ∠A)
∠P = 180° – (110° + 20°)
∠P = 180° – 130° = 50°
In ∆PBC,
∠BPC + ∠PCB + ∠PBC = 180°
[sum of all interior angles of any triangle is 180°]
=> 50° + 20° + ∠PBC = 180°
=> ∠PBC = 180° – 70°
∠PBC = 110°
Since, ∆PB is a straight line.
∠PBC + ∠CBA = 180°
∠CBA = 180° – 110° = 70°