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CBSE Solutions for Class 10 English

GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

The number of tangents that can be drawn from an external point to a circle is ____[CBSE 2011, 12]

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જવાબ : 2


In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR is equal to _____   [CBSE 2014]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127066.png

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જવાબ : 5 cm


In a circle of radius 7 cm, tangent PT is drawn from a point P, such that PT = 24 cm. If O is the centre of the circle, then OP = ?
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/2(427).png

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જવાબ : 25 cm


Which of the following pair of lines in a circle cannot be parallel?      [CBSE 2011]

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જવાબ : two diameters


The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm)  is ____  [CBSE 2014]

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જવાબ : 10√2 cm


In the given figure, PT is a tangent to a circle with centre O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127069.png

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જવાબ : 8 cm


In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then the radius of the circle is ______       [CBSE 2011, 12]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127070-I.png

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જવાબ : 10 cm


PQ is a tangent to a circle with centre O at the point P. If OPQ is an isosceles triangle, then OQP is equal to ______         [CBSE 2014]

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જવાબ : 45


In the given figure, AB and AC are tangents to a circle with centre O such that BAC = 40 .Then BOC is equal to _____      [CBSE 2011, 14]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127072.png

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જવાબ : 140


In the given figure, O is the centre of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of AB is _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127074.png

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જવાબ : 16 cm


If a chord AB subtends an angle of 60 at the centre of a circle, then the angle between the tangents to the circle drawn from A and B is to _____      [CBSE 2013C]

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જવાબ : 120 https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127073.png


In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm. If OA = 17 cm, then the length of AC (in cm) is ________

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127075.png

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જવાબ : 15 cm


In the given figure, O is the centre of a circle. AOC is its diameter, such that ACB = 50°. If AT is the tangent to the circle at the point A, then BAT = ?
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/13%2Cq(24).png

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જવાબ : 50°


In the given figure, O is the centre of the circle, PQ is a chord and PT is the tangent at P. If POQ = 70 , then TPQ is equal to _____   [CBSE 2011]

 https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127076.png

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જવાબ : 35


In the given figure, AT is a tangent to the circle with centre O, such that OT = 4 cm and OTA = 30°. Then, AT = ?
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/14%2Cq(19).png

 

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જવાબ : 2√3cm


If PA and PB are two tangents to a circle with centre O, such that AOB = 110°, find APB.
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/5%2Cq(60).png

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જવાબ : 70°


In the given figure, the length of BC is _____     [CBSE 2012, '14]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127077(1).png

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જવાબ : 10 cm


In the given figure, AOD = 135 then BOC is equal to _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127078.png

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જવાબ : 45


If is the centre of a circle and PT is the tangent to the circle. If PQ is a chord, such that QPT = 50° then POQ = ?

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જવાબ : 100°


In the given figure, PA and PB are two tangents to the circle with centre O. If  APB = 60 then OAB is ____      [CBSE 2011]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127079-I.png

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જવાબ : 30


If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is ____

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જવાબ : 3√3cm


In the given figure, PQ and PR are tangents to a circle with centre A. If QPA = 27 then QAR equals ____   [CBSE 2012]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127080.png

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જવાબ : 126


In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If  PA PB
then the length of each tangent ______         [CBSE 2013]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127081-I.png

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જવાબ : 4 cm


If PA and PB are two tangents to a circle with centre O, such that APB = 80°, then AOP = ?
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/6%2Cq(39).png

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જવાબ : 50°


In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If  APQ = 58  then the measue of  PQB is ____  [CBSE 2014]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127082.png

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જવાબ : 32


In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If  PAO = 30  then CPB + ACP is equal to _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127083(1).png

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જવાબ : 90


Question 27:

In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If  ∠PAB =  67, then the measure of ∠AQB is _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127084.png

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જવાબ : 44


Question 28:

In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of ∠ACB is ____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127085-I.png

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જવાબ : 90


O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quadrilateral PQOR is _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/12%2Cq(25).png

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જવાબ : 60 cm2


In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR, such that BQR = 70°. Then, AQB = ?
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/7%2Cq(40).png

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જવાબ : 40°


The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is

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જવાબ : √125 cm https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127086.png


In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If PBO = 30 then PTA = ?
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127087.png

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જવાબ : 30


In the given figure, a circle touches the side DF of EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of EDF is ____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127088.png

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જવાબ : Perimeter of EDF = 18 cm


To draw a pair of tangents to a circle, which are inclined to each other at angle of 450 , we have to draw the tangents at the end points of those two radii, the angle between which is ______    [CBSE 2011]

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જવાબ : AOB = 1350


In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively, and is a point on the circle, such that SQL = 50° and SRM = 60°. Find QSR.
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/16%2Cq(18).png

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જવાબ : 70°


In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T are of lengths 12 cm and 9 cm respectively. If the area of PQR = 189 cm2 then the length of side PQ is _____ [CBSE 2011]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127090-I.png

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જવાબ : 22.5 cm


In the given figure, QR is a common tangent to the given circle, touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm then the length of QR is _____     [CBSE 2014]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127091.png

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જવાબ : 7.6 cm


In the given figure, quadrilateral ABCD is circumscribed, touching the circle at PQR and S. If AP = 5 cm, BC = 7 cm and CS = 3 cm, AB = ?
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/19%2Cq(17).png

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જવાબ : 9 cm


In the given figure, O is the centre of the circle AB is a chord and AT is the tangent at A. If ∠AOB = 100 then ∠BAT is equal to _____   [CBSE 2011]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127092.png

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જવાબ : 50


In the given figure, quadrilateral ABCD is circumscribed, touching the circle at PQR and S. If AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm, then the perimeter of quadrilateral ABCD is _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/20%2Cq(16).png

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જવાબ :  36 cm


In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle is ____    [CBSE 2014]

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જવાબ : 2 cm


In the given figure, ΔABC is right-angled at B, such that BC = 6 cm and AB = 8 cm. A circle with centre O has been inscribed in the triangle. OP ⊥ ABOQ ⊥ BC and OR ⊥ AC.
If OP = OQ = OR = x cm, then x = ?
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/22%2Cq(8).png

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જવાબ : 2 cm
 


In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides Ab, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD CD then the length of CD is _____    [CBSE 2013]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127094-I.png

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જવાબ : 21 cm


Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then the length of AD is _____          [CBSE 2012]

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જવાબ : 3 cm


In the given figure, PA and PB are tangents to the given circle, such that PA = 5 cm and APB = 60°. The length of chord AB is _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/25%2Cq(8).png

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જવાબ : 5 cm


In the given figure, DE and DF are two tangents drawn from an external point D to a circle with centre A. If DE = 5 cm. and  DE DF then the radius of the circle is ____         [CBSE 2013]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127096-I.png

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જવાબ : 5 cm


In the given figure, three circles with centres AB, C, respectively, touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm, the radius of the circle with centre A is _____
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/23%2Cq(9).png

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જવાબ : 2 cm


In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is _____ [CBSE 2012]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127097-I.png

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જવાબ : 7.5 cm


In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA = 12 cm then PB is equal  to
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/24(74).png

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જવાબ : 4√10 cm


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   [CBSE 2015]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127051.png

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જવાબ : AB is the diameter
∴∠APB = 90 (angle in a semi circle is a right angle)
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
sinABP= AP/BA
sin30°=AT/BA 1/2=AT/BA
BA : AT = 2 : 1


A point P is at a distance of 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent drawn from P to the circle.    [CBSE 2017]

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/1_P485.png
OPB is a right angled triangle, with OBP=90°  ……..{ the tangent is perpendicular to the radius
By using pythagoras theorem in OPB, we get
OB2+PB2=OP2

(20)2+PB2=(29)2

400+PB2=841

PB2=841−400=441

PB=√441 =21 
So, length of the tangent from point P is 21 cm.


A point P is 25 cm away from the centre of a circle and the length of tangent drawn from to the circle is 24 cm. Find the radius of the circle.

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/pic(8).png
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.

OTPT In the right OTP, we have:

OP2=OT2+TP2   [By Pythagoras' theorem:]

OT=√ [OP2−TP2]= √(252−242) =√(625−576)  =√49 =7 cm

The length of the radius is 7 cm.


Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the chord of the larger circle which touches the smaller circle       [CBSE 2011]

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127043.png
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 centre bisect
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.


In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF       [CBSE 2013]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127044.png

 

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જવાબ : we have
AD = AF, BD = BE and CE = CF
Now, AD + BD = 12 cm          .....(A)
AF + FC = 10 cm 

AD + FC = 10 cm                    .....(B)
BE + EC = 8 cm 

BD + FC = 8 cm                   .....(C)
Adding all these we get
AD + BD +
AD + FC + BD + FC = 30
2(AD + BD + FC) = 30
AD + BD + FC = 15 cm           .....(D)
Solving (A) and (D), we get
FC = 3 cm
Solving (B) and (D), we get
BD = 5 cm
Solving (C) and (D), we get
and AD = 7 cm

AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm


From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ΔPCD.
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/4(280).png

 

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જવાબ : Given, PA ​and PB are the tangents to a circle with centre 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.


In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127045-I.png

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127045-II.png

We know that
OCA = OCB = 90 (the radius and tangent are perperpendular at their point of contact)
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


A circle is inscribed in ΔABC, touching AB, BC and AC at PQ and R, respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/5(198).png

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જવાબ : 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 = 8

The length of BC is 8 cm.


In the given figure, PA and PB are the tangents to a circle with centre O. Show that the points A, O, B, P are concyclic.
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/6%2Cq(38).png

 

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જવાબ : Here, OA=OB And  OAAP,  OABP

∴∠OAP=90ᵒ, OBP=90ᵒ

∴∠OAP+OBP= 90ᵒ +90ᵒ =180ᵒ

∴∠AOB+APB=180ᵒ      (Since,OAP+OBP+AOB+APB=360ᵒ)

Sum of opposite angle of a quadrilateral is 180°.

Hence, A,O,B and P are concyclic.


In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/6(183).png

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/pic(35).png
Given, AB=6 cm, BC=7 cm and CD=4 cm.

AP=AS, BP=BQ, CR=CQ  and DR=DS ……..( Tangents drawn from an external point are equal.)

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.


In the given figure, an isosceles triangle ABC with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC   [CBSE 2012]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127046.png

 

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જવાબ : AR = AQ, BR = BP and CP = CQ ……………..( tangent segments to a circle from the same external point are congruent.)
Now, AB = AC
AR + RB = AQ + QC
AR + RB = AR + QC
RB = QC
BP = CP
Hence, P bisects BC at P.


In the given figure, common tangents AB and CD to the two circle with centres O1 and O2 intersect at E. Prove that AB = CD   [CBSE 2014]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127055.png

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જવાબ : EA = EC for the circle having centre O1 (tangent segments to a circle from the same external point are congruent)

Similarly,
ED = EB for the circle having centre O1
Now, Adding ED on both sides in EA = EC, we get
EA + ED = EC + ED
EA + EB = EC + ED
AB = CD


In the given figure, O is the centre of a circle PT and PQ are tangents to the circle from an external point P. If TPQ = 70 then TRQ   [CBSE 2015]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127054-I.png

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127054-II.png

OTP = OQP = 90∘  (radius and tangent are perperpendular)
Now, In quadrilateral OQPT
QOT + OTP + OQP + TPQ = 360            [Angle sum property of a quadrilateral]
QOT + 90 + 90 + 70 = 360 
250 + QOT = 360 
QOT = 110 
∴∠TRQ= ½ (QOT) =55° (angle subtended by an arc at the centre is double the angle subtended by the arc)


In the given figure, PA and PB are two tangents to the circle with centre O. If  APB = 50 then what is the measure of OAB is      [CBSE 2015]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127053-I(1).png

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127053-II(1).png

OBP = OAP = 90∘  (radius and tangent are perperpendular)
Now, In quadrilateral AOBP
AOB + OBP + APB + OAP = 360            [Angle sum property of a quadrilateral]
AOB + 90 + 50+ 90 = 360 
230+ BOC = 360 
AOB = 130


Now, In isoceles triangle AOB
AOB + OAB + OBA = 180            [Angle sum property of a triangle]
130 +  2OAB = 1800                          [∵∠OAB = OBA ]
OAB = 25


In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of AD            [CBSE 2011]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127052.png

 

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જવાબ : AB + CD = AD + BC  (In quadrilateral circumscribes a circle, sum of opposites sides is equal to the sum of other opposite sides.)
6 + 8 = AD + 9
AD = 5 cm


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             [CBSE 2015]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127051.png

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જવાબ : AB is the chord passing through the centre
So, AB is the diameter
∴∠APB = 90ᵒ (angle in a semi circle is a right angle)


By using alternate segment theorem
We have APB = PAT = 30ᵒ 
Now, in APB
BAP + APB + BAP = 180ᵒ      (Angle sum property of triangle)
BAP = 180ᵒ − 90ᵒ − 30ᵒ = 60ᵒ
Now, BAP = APT + PTA        (Exterior angle property)
60ᵒ = 30ᵒ + PTA
PTA = 60ᵒ − 30ᵒ = 30ᵒ
We know that sides opposite to equal angles are equal.
AP = AT
In right triangle ABP
sinABP=AP/BA

sin30°=AT/BA

12=AT/BA

sinABP=AP/BA

sin30°=AT/BA

12=AT/BA
BA : AT = 2 : 1


In the given figure, a circle with centre 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.           [CBSE 2008, 13]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127050.png

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જવાબ : DS = DR, AR = AQ  (tangent segments to a circle from the same external point are congruent)
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
Hence, the radius of the circle is 11 cm.


Prove that the line joining the points of contact of two parallel tangents of a circle passes through its centre.      [CBSE 2014]

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127049.png
Suppose CD and AB are two parallel tangents of a circle with centre O
OQC = ORA = 90ᵒ  (radius and tangent are perperpendular)
Now, OQC + POQ = 180       (co-interior angles)
POQ = 180 − 90 = 90
Similarly, Now, ORA + POR = 180       (co-interior angles)
POR = 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 centre O.


PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP       [CBSE 2013C]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127048-I(1).png

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127048-II(1).png

Let TR = y and TP = x

 PR = RQ (perpendicular from the centre to the chord bisects it.)
Now, PR + RQ = 4.8
PR + PR = 4.8
PR = 2.4
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
PO
2 = OR2 + PR2
32 = OR2 + (2.4)2
OR2 = 3.24
OR = 1.8
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
TP
2 = TR2 + PR2
 x2 = y2 + (2.4)2
 x2 = y2 + 5.76          .....(A)
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
TO2 = TP2 + PO2

(y + 1.8)2 = x2 + 32
 y2 + 3.6y + 3.24 = x2 + 9
 y2 + 3.6x2 + 5.76          .....(B)
Solving (A) and (B), we get
x = 4 cm and y = 3.2 cm

TP = 4 cm


In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD 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 lengths of sides of AB and AC.    [CBSE 2011, '15]
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127047-I.png

 

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જવાબ : https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127047-II.png
We know that tangent segments to a circle from the 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= ½ ×BC×OD+  ½ ×AB×OE + ½ ×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 = 9cm and AC = 9 + 3 = 12 cm


In the given figure, O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circles, respectively. If PA = 10 cm, find the length of PB up to one decimal place.
https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/8(151).png

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જવાબ : Given-  O is the centre of two concentric circles of radii OA=6 cm and OB=4 cm.

OAP=OBP=90ᵒ

From right angled  OAP,  OP2=OA2+PA2

=>OP=√(OA2+PA2 )

=>OP=√(62+102)

=>OP=√136 cm.

From right-angled OAP,  OP2 =OB2 +PB2

 =>PB=√(OP2−OB2)  

 =>PB=√(136−16)

 =>PB=√120 cm

=>PB=10.9 cm.

The length of PB is 10.9 cm.


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 https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/2(427).png              

1

TP

A

7cm

2

TO

B

25cm

3

OP

C

24cm

4

D

90°

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જવાબ :

1-C, 2-A, 3-B, 4-D

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127069.png

1

T

A

6cm

2

TP

B

90°

3

TO

C

10cm

4

OP

D

8cm

 

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જવાબ :

1-B, 2-D, 3-A, 4-C

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127072.png

1

∠A

A

OC

2

O

B

40°

3

AB

C

50°

4

OB

D

AC

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જવાબ :

1-B, 2-C, 3-D, 4-A

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127073.png

1

O

A

AC

2

C

B

OA

3

OB

C

30°

4

BC

D

60°

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જવાબ :

1-D, 2-C, 3-B, 4-A

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127075.png

1

OB

A

90°

2

AB

B

<90°

3

∠B

C

AC

4

∠A

D

OC

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જવાબ :

1-D, 2-C, 3-A, 4-B

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/5%2Cq(60).png

1

OA

A

90°

2

P

B

110°

3

A

C

70°

4

∠O

D

OB

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જવાબ :

1-D, 2-C, 3-A, 4-B

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127077(1).png

1

AB

A

10cm

2

AE

B

7cm

3

CE

C

4cm

4

BC

D

7cm

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જવાબ :

1-D, 2-C, 3-B, 4-A

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127079-I.png

1

∠P

A

90°

2

∠AOB

B

60°

3

∠A

C

30°

4

AP

D

BP

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જવાબ :

1-B, 2-C, 3-D, 4-A

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127083(1).png

1

∠A

A

90°

2

∠AOP

B

30°

3

∠OPA

C

OP

4

OD

D

60°

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જવાબ :

1-B, 2-D, 3-A, 4-C

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/127084.png

1

∠BAP

A

113°

2

∠BAQ

B

90°

3

∠OAQ

C

67°

4

∠OAB

D

23°

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જવાબ :

1-C, 2-A, 3-B, 4-D

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Chapter 10 : Circles

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