Courses

# Short Answer Type Questions(Part- 1)- Circles Class 10 Notes | EduRev

## Class 10 : Short Answer Type Questions(Part- 1)- Circles Class 10 Notes | EduRev

The document Short Answer Type Questions(Part- 1)- Circles Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

SHORT ANSWER TYPE QUESTIONS

Q1. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Sol.

Let NM be a chord of a circle with centre C.
Let the tangents at M and N meet at O.
âˆµ OM is a tangent at M

âˆµ âˆ OMC = 90Â° ...(1)
Similarly âˆ ONC = 90Â° ...(2)
Since, CM = CN [Radii of the same circle]
âˆµ In D CMN, âˆ 1= âˆ 2
From (1) and (2), we have
âˆ OMC â€“âˆ 1 = âˆ ONC â€“âˆ 2
â‡’ âˆ OML = âˆ ONL

Thus, tangents make equal angles with the chord.

Q2. Two concentric circles have a common centre O. The chord AB to the bigger circle touches the smaller circle at P. If OP = 3 cm and AB = 8 cm then find the radius of the bigger circle.

Sol. âˆµ AB touches the smaller circle at P.

âˆ´ OP âŠ¥ AB â‡’ âˆ OPA = 90Â°
Now, AB is a chord of the bigger circle.
Since, the perpendicular from the centre to a chord, bisects the chord,
âˆ´ P is the mid-point of AB

â‡’
In right âˆ† APO, we have
AO= OP2 + AP2
â‡’ AO2 = 32 + 42
â‡’ AO2 = 9 + 16 = 25 = 52

â‡’

Thus, the radius of the bigger circle is 5 cm.

Q3. In the given figure, O is the centre of the circle and PQ is a tangent to it. If its circumference is 12Ï€ cm, then find the length of the tangent.

Sol. âˆµ Circumference of the circle = 12Ï€ cm

âˆ´ 2Ï€ r =12Ï€
[âˆµ r is the radius of the circle]

â‡’
â‡’ Radius of the circle = 6 cm = OQ

Since a tangent to circle is perpendicular to the radius through the point of contact,
âˆ´ âˆ OQP = 90Â°
Now, in rt Î” OQP, we have:
OQ2 + QP2 = OP2
â‡’ 62 + QP2 =102
â‡’ QP2 = 102 âˆ’ 62 = (10 âˆ’ 6) (10 + 6) = 4 Ã— 16 = 64 = 82

â‡’

Thus, the length of the tangent is 8 cm.

Q4. Given two concentric circles of radii 10 cm and 6 cm. Find the length of the chord of the larger circle which touches the other circle.

Sol. The chord AB touches the inner circle at P.

âˆ´ AB is tangent to the inner circle.
â‡’ OP âŠ¥ AB

[âˆµ O is the centre and OP is radius through the point of contact P]
âˆ´ âˆ OPB = 90Â°.
Now, in right âˆ† OPB, we have:
OP2 + PB2 = OB2
â‡’ 62 + PB2 = 102
â‡’ PB2 = 102 âˆ’ 62
= (10 âˆ’ 6) Ã— (10 + 6)
â‡’ PB2 = 4 Ã— 16
â‡’ PB2 = 64 = 82

â‡’

âˆµ The radius perpendicular to a chord bisects the chord.
âˆ´ P is the mid-point of AB
âˆ´ AB = 2 Ã— PB = 2 Ã— 8 = 16 cm.

Q5. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that âˆ PTQ = 2 âˆ OPQ.

Sol. âˆµ Tangent to a circle is perpendicular to the radius through the point of contact.
In quadrilateral. OPTQ,
âˆ OPT + âˆ OQT + âˆ POQ + âˆ PTQ = 360Â°
or 90Â° + 90Â° + âˆ POQ + âˆ PTQ = 360Â°
â‡’ âˆ POQ + âˆ PTQ = 360Â° âˆ’ 90Â° âˆ’ 90Â° = 180Â°   ...(1)
In Î” OPQ, âˆ 1 + âˆ 2 + âˆ POQ = 180Â°   ...(2)

Since OP =OQ   [Radii of the same circle]
â‡’ âˆ 1 = âˆ 2    [Angles opposite to equal sides]
âˆ´  âˆ OPT = 90Â° = âˆ OQT
âˆ´ From (2), we have
âˆ 1 + âˆ 1 + âˆ POQ = 180Â°
â‡’ 2 âˆ 1 + âˆ POQ = 180Â° ...(3)
From (1) and (3), we have
2 âˆ 1 + âˆ POQ = âˆ POQ + âˆ PTQ
â‡’ 2 âˆ 1= âˆ PTQ
â‡’ 2 âˆ OPQ = âˆ  PTQ.

Q6. In the figure, the incircle of âˆ† ABC touches the sides BC, CA and AB at D, E and F respectively. If AB = AC, prove that BD = CD.

Sol. Since the lengths of tangents drawn from an external point to a circle are equal,
âˆ´ We have
AF = AE
BF = BD
CD = CE
Adding them, we get
(AF + BF) + CD = (AE + CE) + BD
â‡’ AB + CD = AC + BD
But AB = AC  (Given)
âˆ´ CD = BD.

Q7. A circle is touching the side BC of a D ABC at P and touching AB and AC produced at Q and R. Prove that:

Sol. Since, the two tangents drawn to a circle from an external point are equal.
âˆµ  AQ = AR ...(1)
Similarly, BQ = BP ...(2)
and CR = CP ...(3)

Now, Perimeter of Î” ABC
= AB + BC + AC
= AB + (BP + PC) + AC
= AB + (BQ + CR) + AC
[From (2) and (3)]
= (AB + BQ) + (CR + AC)
= AQ + AR
= AQ + AQ [From (1)] = 2AQ
â‡’

Q8. In the given figure, the radii of two concentric circles are 13 cm and 8 cm. AB is diameter of the bigger circle. BD is the tangent to the smaller circle touching it at D. Find the length AD.

Sol. Join AE and OD
âˆµ âˆ ODB = 90Â°
[BE is a tangent at D and OD is a radius] and âˆ AEB = 90Â° [AB is diameter so âˆ AEB is an angle in semicircle so âˆ AEB = 90Â°]
âˆ´ OD â•‘ Î‘Î• and âˆ†BEA are similar by AA-Similarity

So

Again in rt  Î” AED,

Q9. In two concentric circles, a chord of the larger circle touches the smaller circle. If the length of this chord is 8 cm and the diameter of the smaller circle is 6 cm, then find the diameter of the larger circle.

Sol. Let the common centre be O. Let AB be the chord of the larger circle.

âˆ´ AB = 8 cm

And CD is the diameter of the smaller circle i.e., CD = 6 cm

â‡’
Join OA. D is the point of contact.
âˆ´ OD âŠ¥ AB
â‡’ D is the mid point of AB
â‡’ AD = 4 cm Now, in right Î”ADO, we have:
AO2 = AD2 + OD2
= 42 + 32 = 16 + 9 = 25 = 52
â‡’ AO = 5 cm
â‡’ 2AO = 2(5 cm) = 10 cm

âˆ´ The diameter of the bigger circle is 10 cm.

Q10. In the following figure, PA and PB are two tangents drawn to a circle with centre O, from an external point P such that PA = 5 cm and âˆ APB = 60Â°. Find the length of chord AB.

Sol. Since the tangents to a circle from an external point are equal,
âˆ´ PA = PB = 5 cm
In Î”PAB, we have
âˆ PAB = âˆ PBA               [âˆµ PA = PB]
âˆ´âˆ PAB + âˆ PBA + âˆ APB = 180Â°
â‡’âˆ PAB + âˆ PAB + 60Â° = 180Â°
â‡’ 2 âˆ PAB + 60Â° = 180Â°
â‡’ 2 âˆ PAB = 180Â°âˆ’ 60Â°
= 120Â°
â‡’ âˆ PAB = 60Â°
â‡’ Each angle of âˆ†PAB is 60Â°.
â‡’ Î”PAB is an equilateral triangle. âˆ´
PA = PB
= AB = 5 cm
Thus, AB = 5 cm

Q11. In the following figure, AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm.

The tangents at A and B intersect at P. Find the length PA.

Sol.

Join OB.
Let PA = x cm and PR = y cm Since, OP is perpendicular bisector of AB

âˆ´ AR = BR = 9.6/2 = 4.8 cm

Now, in rt âˆ†OAR, we have:
OA2 = OR2 + AR2
[By Pythagoras theorem]
â‡’ OR2 = OA2 âˆ’ AR2
= 62 âˆ’ (4.8)2 = (6 âˆ’ 4.8) Ã— (6 + 4.8) = 1.2 Ã— 10.8
â‡’ = 12.96
OR = 3.6 cm.
Again, in right Î”OAP,
OP2 = AP2 + OA2
OP2 = (AR2 + PR2) + OA2
[âˆµ AP2 = AR2 + PR2]
â‡’ (y + 3.6)2 = (4.8)2 + y2 + 62
â‡’y2 + 12.96 + 7.2 y = 23.04 + y2 + 36
â‡’ 7.2 y = 46.08

â‡’
â‡’ PR = 6.4 cm
Now, AP2 = AP2 + PR2
= (4.8)2 + (6.4)2 = 23.04 + 40.96 = 64

â‡’

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;