tell me the topic off 10.2 of 10th class
Answers
Answer:
Hiii
Explanation:
In Q 1 to 3, choose the correct option and give justification.
1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Ans. (A) OPQ =
[The tangent at any point of a circle is to the radius
through the point of contact]
In right triangle OPQ,
[By Pythagoras theorem]
= 625 – 576 = 49
OP = 7 cm
2. In figure, if TP and TQ are the two tangents to a circle with centre O so that POQ = then PTQ is equal to:
(A)
(B)
(C)
(D)
Ans. (B) POQ = , OPT = and OQT =
[The tangent at any point of a circle is to the radius through the point of contact]
In quadrilateral OPTQ,
POQ + OPT + OQT + PTQ =
[Angle sum property of quadrilateral]
+ PTQ =
+ PTQ =
PTQ =
3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of , then POA is equal to:
(A)
(B)
(C)
(D)
Ans. (A) OPQ =
[The tangent at any point of a circle is to the radius
through the point of contact]
OPA = BPA
[Centre lies on the bisector of the
angle between the two tangents]
In OPA,
OAP + OPA + POA =
[Angle sum property of a triangle]
+ POA =
+ POA =
POA =
4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Ans. Given: PQ is a diameter of a circle with centre O.
The lines AB and CD are the tangents at P and Q respectively.
To Prove: AB CD
Proof: Since AB is a tangent to the circle at P and OP is the radius through the point of contact.
OPA = ……….(i)
[The tangent at any point of a circle is to the radius through the point of contact]
CD is a tangent to the circle at Q and OQ is the radius through the point of contact.
OQD = ……….(ii)
[The tangent at any point of a circle is to the radius through the point of contact]
From eq. (i) and (ii), OPA = OQD
But these form a pair of equal alternate angles also,
AB CD
5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Ans. We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact and the radius essentially passes through the centre of the circle, therefore the perpendicular at the point of contact to the tangent to a circle passes through the centre.
6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Ans. We know that the tangent at any point of a circle is to the radius through the point of contact.
OPA =
[By Pythagoras theorem]
= 9
OP = 3 cm
7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Ans. Let O be the common centre of the two concentric circles.
Let AB be a chord of the larger circle which touches the smaller circle at P.
Join OP and OA.
Then, OPA =
[The tangent at any point of a circle is to the radius through the point of contact
OA2 = OP2 + AP2
[By Pythagoras theorem]
= 16
AP = 4 cm
Since the perpendicular from the centre of a circle to a chord bisects the chord, therefore
AP = BP = 4 cm
AB = AP + BP
= AP + AP = 2AP
= = 8 cm