Important Questions for CBSE Class 10 Maths Chapter 10 - Circles 4 Mark Question

CBSE Class 10 Maths Chapter-10 Circles – Free PDF Download

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CBSE Class 10 Maths Chapter-10 Circles Important Questions

CBSE Class 10 Maths Important Questions Chapter 10 – Circles

4 Mark Questions

1. In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that AOB = 

Ans. Given: In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.
To Prove: AOB = 
Construction: Join OC
Proof: OPA =  ……….(i)
OCA =  ……….(ii)
[Tangent at any point of a circle is  to the radius through the point of contact]
In right angled triangles OPA and OCA,
OA = OA [Common]
AP = AC [Tangents from an external point to a circle are equal]
 OPAOCA [RHS congruence criterion]
 OAP = OAC [By C.P.C.T.]
 OAC = PAB ……….(iii)
Similarly, OBQ = OBC
 OBC = QBA ……….(iv)
 XYX’Y’ and a transversal AB intersects them.
 PAB + QBA =  [Sum of the consecutive interior angles on the same side of the transversal is ]
 PAB + QBA =  ……….(v)
 OAC + OBC = 
[From eq. (iii) & (iv)]
In AOB,
OAC + OBC + AOB = 
[Angel sum property of a triangle]
  + AOB = 
[From eq. (v)]
 AOB = 

2. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Ans. Join OE and OF. Also join OA, OB and OC.

Since BD = 8 cm
 BE = 8 cm
[Tangents from an external point to a circle are equal]
Since CD = 6 cm
 CF = 6 cm
[Tangents from an external point to a circle are equal]
Let AE = AF = 
Since OD = OE = OF = 4 cm
[Radii of a circle are equal]
 Semi-perimeter of ABC
= 
=  cm
 Area of ABC = 
= 
= cm²
Now, Area of ABC = Area of OBC + Area of OCA + Area of OAB
 
= 
 
= 
 
= 
 = 
Squaring both sides,

 
 
 
 AB =  = 7 + 8 = 15 cm
And AC =  = 7 + 6 = 13 cm

3. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Ans. Given: ABCD is a quadrilateral circumscribing a circle whose centre is O.
To prove: (i) AOB + COD =  (ii) BOC + AOD = 
Construction: Join OP, OQ, OR and OS.
Proof: Since tangents from an external point to a circle are equal.
 AP = AS,
BP = BQ ……….(i)
CQ = CR
DR = DS
In OBP and OBQ,
OP = OQ [Radii of the same circle]
OB = OB [Common]
BP = BQ [From eq. (i)]
 OPBOBQ [By SSS congruence criterion]
  [By C.P.C.T.]
Similarly, 
Since, the sum of all the angles round a point is equal to 
 
 
 
 
 
 AOB + COD = 
Similarly, we can prove that
BOC + AOD = 

4. In the given figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that 

Ans. Join OC.
In and DAOC, we have
AP = AC [tangents from A to the circle are equal]
AO = AO
OP = OC [radius]
[By CPCT]


Similarly,
Now,

But in 

5. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Ans. Let the sides BC, CA, AB of touch the incircle at D, E, F respectively.
Join the centre O of the circle with A, B, C, D, E, F
Since, tangents to a circle from an external point are equal

Area of 
=
Area of 
area
area
Again, perimeter of 

Area of 

[By 4 and 5]


But is not possible

6. In the given figure, PT is tangent and PAB is a secant. If PT = 6 cm, AB = 5 cm. Find the length PA.

Ans. Join OT, OA, OP. Draw OM AB
Let radius of the circle = r
[OT is radius and PT is a tangent]
 [From right ]

Also from right 




[bisects AB]

[It cannot be -ve]

7. From a point P two tangents are drawn to a circle with centre O. If OP = diameter of the circle, show that  is equilateral.

Ans. Join OP.
Suppose OP meets the circle at Q. Join AQ.
We have
i.e., OP = diameter
OQ + PQ = diameter
PQ = Diameter – radius []
PQ = radius
Thus, OQ = PQ = radius
Thus, OP is the hypotenuse of right triangle
OAP and Q is the mid-point of OP
OA = AQ = OQ
[mid-point of hypotenuse of a right triangle is equidistant from the vertices]
is equilateral

So,
Also 
But 
Hence,is equilateral.