{"id":3680,"date":"2021-08-10T23:30:30","date_gmt":"2021-08-10T18:00:30","guid":{"rendered":"https:\/\/cbsencertanswers.com\/?p=3680"},"modified":"2022-03-03T17:30:13","modified_gmt":"2022-03-03T12:00:13","slug":"circles-chapter-10-extra-questions-and-solutions-for-class-10-cbse-mathematics","status":"publish","type":"post","link":"https:\/\/cbsencertanswers.com\/2021\/08\/circles-chapter-10-extra-questions-and-solutions-for-class-10-cbse-mathematics.html","title":{"rendered":"Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics"},"content":{"rendered":"\r\n

You are going to go through Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics. This post presents to the students a clear conception of<\/em><\/em>\u00a0how to move with the basics of Extra\u00a0Questions and answers. The\u00a0expert\u00a0prepared The Extra\u00a0Questions\u00a0and And\u00a0Answers. https:\/\/cbsencertanswers.com\/is very much to make\u00a0things way simpler and easier for the students. Especially those who are appearing for the board exams. We took every care to make sure that the\u00a0effort\u00a0serves the\u00a0purpose. So, let us find out Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/em>.\u00a0On this page, you can find Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/em><\/em>.<\/p>\r\n\r\n\r\n\r\n

\"Circles<\/figure>\r\n\r\n\r\n\r\n

1)A point\u00a0P<\/em>\u00a0is at a distance of 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent drawn from\u00a0P<\/em>\u00a0to the circle?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

Considering\u00a0 the figure, We know that the tangent is perpendicular to the radius of a circle.\u00a0
So, OPB is a right angled triangle, with\u00a0\u2220OBP=90\u00b0
By using pythagoras theorem in\u00a0\u25b3OPB, we get
\u21d2 OB2<\/sup>+PB2<\/sup>=OP2<\/sup>
\u21d2 (20)2<\/sup>+PB2<\/sup>=(29)2<\/sup><\/p>\r\n\r\n\r\n\r\n

\u21d2 400+PB2<\/sup>=841<\/p>\r\n\r\n\r\n\r\n

\u21d2 PB2<\/sup>=841-400=441\u21d2PB=\u221a441=21\u00a0<\/p>\r\n\r\n\r\n\r\n

So, length of the tangent from point\u00a0P\u00a0is 21 cm(PB) .<\/p>\r\n\r\n\r\n\r\n

2) A point\u00a0P<\/em>\u00a0is 25 cm away from the centre of a circle and the length of tangent drawn from\u00a0P\u00a0<\/em>to the circle is 24 cm. Find the radius of the circle?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

A point\u00a0P<\/em>\u00a0is 25 cm away from the centre of a circle and the length of tangent drawn from\u00a0P\u00a0<\/em>to the circle is 24 cm.So, TP here is the tangent , OT is the radius So, OTP is a right angled triangle, with\u00a0\u2220OTP=90\u00b0
By using pythagoras theorem in\u00a0\u25b3OTP, we get<\/p>\r\n\r\n\r\n\r\n

\u21d2 OT2<\/sup>+TP2<\/sup>=OP2<\/sup>
\u21d2 (OT)2<\/sup>+(24)2<\/sup>=(25)2<\/sup><\/p>\r\n\r\n\r\n\r\n

\u21d2 OT2<\/sup>+576=625<\/p>\r\n\r\n\r\n\r\n

\u21d2 OT2<\/sup>=625 – 576=49\u21d2OT=\u221a49=7 cm.<\/p>\r\n\r\n\r\n\r\n

YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/h3>\r\n\r\n\r\n\r\n

Hence, the radius OT is 7 cm.<\/p>\r\n\r\n\r\n\r\n

3) 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?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

\u00a0Solution.<\/h2>\r\n\r\n\r\n\r\n

In this given figure, we know that the radius and tangent are perperpendular at their point of contact
In right\u00a0 triangle \u00a0\u25b3AOP,
\u00a0\u00a0\u00a0\u00a0 AO2<\/sup>\u00a0= OP2<\/sup>\u00a0+ PA2<\/sup>
\u21d2 (6.5)2<\/sup>\u00a0= (2.5)2<\/sup>\u00a0+ PA2<\/sup>
\u21d2 PA2<\/sup>\u00a0= 36
\u21d2 PA = 6 cm
Since, the perpendicular drawn from the centre bisect the chord.
\u2234 PA = PB = 6 cm
Now, AB = AP + PB = 6 + 6 = 12 cm
Hence, the length of the chord of the larger circle (AB) \u00a0is 12 cm.<\/p>\r\n\r\n\r\n\r\n

4) 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<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Construction:\u00a0 Join OA, OC and OB
We know that the radius and tangent are perpendicular at their point of contact
\u2234 \u2220OCA = \u2220OCB = 90\u2218<\/sup>\u00a0
Now, In \u25b3OCA and \u25b3OCB
\u2220OCA = \u2220OCB = 90\u2218<\/sup>
OA = OB\u00a0\u00a0\u00a0\u00a0 (Radii of the larger circle)
OC = OC\u00a0\u00a0\u00a0\u00a0 (Common)<\/p>\r\n\r\n\r\n\r\n

\u25b3OCA \u2245 \u25b3OCB\u00a0 (By RHS congruency
\u2234 CA = CB ( Proved).<\/p>\r\n\r\n\r\n\r\n

YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/h3>\r\n\r\n\r\n\r\n

5) 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\u00a0?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that, tangent segments to a circle from the same external point are congruent.
Now, we have
AR = AQ, BR = BP and CP = CQ
Now, AB = AC as \u25b3 ABC is isosceles triangle
\u21d2 AR + RB = AQ + QC
\u21d2 AR + RB = AR + QC
\u21d2 RB = QC
\u21d2 BP = CP
Hence, P bisects BC at P.<\/p>\r\n\r\n\r\n\r\n

6) 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,\u00a0\u2220B = 90\u2218<\/sup>\u00a0and DS = 5 cm then find the radius of the circle?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
DS = DR, AR = AQ
Now, AD = 23 cm
\u21d2 AR + RD = 23
\u21d2 AR = 23 \u2212 RD
\u21d2 AR = 23 \u2212 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [\u2235 DS = DR = 5]
\u21d2 AR = 18 cm
Again, AB = 29 cm
\u21d2 AQ + QB = 29
\u21d2 QB = 29 \u2212 AQ
\u21d2 QB = 29 \u2212 18 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [\u2235 AR = AQ = 18]
\u21d2 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.<\/p>\r\n\r\n\r\n\r\n

YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/h3>\r\n\r\n\r\n\r\n

7) 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?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides.
\u2234 AB + CD = AD + BC
\u00a0\u00a0 \u21d26 + 8 = AD + 9
\u00a0\u00a0 \u21d2 AD = 5 cm<\/p>\r\n\r\n\r\n\r\n

Hence, the length of AD is 5 cm.<\/p>\r\n\r\n\r\n\r\n

8) 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 \u2220TPQ = 70\u2218<\/sup>\u00a0then \u2220TRQ\u00a0?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that the radius and tangent are perperpendular at their point of contact
\u2235\u2220OTP = \u2220OQP = 90\u2218<\/sup>
Now, In quadrilateral OQPT
\u2220QOT + \u2220OTP + \u2220OQP + \u2220TPQ = 360\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [Angle sum property of a quadrilateral]
\u21d2 \u2220QOT + 90\u2218<\/sup>\u00a0+ 90\u2218<\/sup>\u00a0+ 70\u2218<\/sup>\u00a0= 360\u2218<\/sup>\u00a0
\u21d2 250\u2218<\/sup>\u00a0+ \u2220QOT = 360\u2218<\/sup>\u00a0
\u21d2 \u2220QOT = 110\u2218<\/sup>\u00a0
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.<\/p>\r\n\r\n\r\n\r\n

\u2234 , \u2220TRQ = \u00bd (\u2220QOT) = 55\u2218<\/sup>\u00a0.<\/p>\r\n\r\n\r\n\r\n

9) In the given figure, AB and AC are tangents to a circle with centre O such that \u2220BAC = 40\u2218<\/sup>\u00a0.Then find the value of \u2220BOC?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that the radius and tangent are perperpendular at their point of contact
\u2235\u2220OBA = \u2220OCA = 90\u2218<\/sup>
Now, In quadrilateral ABOC
\u2220BAC + \u2220OCA + \u2220OBA + \u2220BOC = 360\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [Angle sum property of a quadrilateral]
\u21d2 40\u2218<\/sup>\u00a0+ 90\u2218<\/sup>\u00a0+ 90\u2218<\/sup>\u00a0+ \u2220BOC = 360\u2218<\/sup>\u00a0
\u21d2 220\u2218<\/sup>\u00a0+ \u2220BOC = 360\u2218<\/sup>\u00a0
\u21d2 \u2220BOC = 140\u2218<\/sup>\u00a0<\/p>\r\n\r\n\r\n\r\n

Then the value of \u2220BOC is 140\u2218<\/sup>\u00a0.<\/p>\r\n\r\n\r\n\r\n

YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/h3>\r\n\r\n\r\n\r\n

10) In the given figure, \u2220AOD = 135\u2218<\/sup>\u00a0then find the angle \u2220BOC\u00a0?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that the sum of angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180 degrees.
\u2234\u2220AOD + \u2220BOC =\u00a0180\u2218<\/sup>
\u21d2\u2220BOC =\u00a0180\u2218<\/sup>\u00a0\u2212\u00a0135\u2218<\/sup>\u00a0= 45\u2218<\/sup>\u00a0<\/p>\r\n\r\n\r\n\r\n

11) In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If\u00a0 \u2220PAO = 30\u2218<\/sup>\u00a0 then find \u00a0the value of the sum of angles \u2220CPB + \u2220ACP ?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that a chord passing through the centre is the diameter of the circle.
\u2235\u2220DPC = 90\u2218<\/sup>\u00a0\u00a0\u00a0 (Angle in a semi circle \u00a0is 90\u2218<\/sup>)
\u00a0Now, In \u25b3CDP
\u2220CDP + \u2220DCP + \u2220DPC = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [Angle sum property of a triangle]
\u21d2 \u2220CDP + \u2220DCP +\u00a0 90\u2218<\/sup>\u00a0= 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\u21d2 \u2220CDP + \u2220DCP = 90\u2218<\/sup>
By using alternate segment theorem
We have \u2220CDP = \u2220CPB
\u2234\u2220CPB + \u2220ACP = 90\u2218<\/sup><\/p>\r\n\r\n\r\n\r\n

12) In the given figure, two circles touch each other at C and AB is a tangent to both the circles. Then find the value of \u2220ACB ?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have NA = NC and NC = NB. We also know that angle opposite to equal sides are equal.
\u2234\u00a0\u2220NAC = \u2220NCA and\u00a0\u2220NBC = \u2220NCB

Now, \u2220ANC + \u2220BNC = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0[Linear pair angles]
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u21d2 \u2220NBC + \u2220NCB + \u2220NAC + \u2220NCA= 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0 [Exterior angle property]
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u21d2 2\u2220NCB + 2\u2220NCA= 180\u2218<\/sup>\u00a0\u00a0
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u21d2 2(\u2220NCB + \u2220NCA) = 180\u2218<\/sup>
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u21d2 \u2220ACB = 90\u2218<\/sup>

<\/p>\r\n\r\n\r\n\r\n

YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/h3>\r\n\r\n\r\n\r\n

13) 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 \u2220PBO = 30\u2218<\/sup>\u00a0then \u00a0the value of \u2220PTA ?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that a chord passing through the centre is the diameter of the circle.
\u2235\u2220BPA = 90\u2218<\/sup>\u00a0\u00a0\u00a0 (Angle in a semi circle is 90\u2218<\/sup>)
By using alternate segment theorem
We have \u2220APT = \u2220ABP = 30\u2218<\/sup>
Now, In \u25b3ABP
\u2220PBA + \u2220BPA + \u2220BAP = 1800<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [Angle sum property of a triangle]
\u21d2 30\u2218<\/sup>\u00a0+\u00a0 900<\/sup>\u00a0+ \u2220BAP = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\u21d2 \u2220BAP = 60\u2218<\/sup>\u00a0
Now, \u2220BAP =\u00a0\u2220APT +\u00a0\u2220PTA
\u21d2 60\u2218<\/sup>\u00a0= 30\u2218<\/sup>\u00a0+ \u00a0\u2220PTA
\u21d2 \u2220PTA = 30\u2218<\/sup><\/p>\r\n\r\n\r\n\r\n

14) \u00a0In the given figure, a circle touches the side DF of \u25b3EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of \u25b3EDF is ?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
EK = EM = 9 cm
Now, EK + EM = 18 cm
\u21d2 ED + DK + EF + FM = 18 cm
\u21d2 ED + DH + EF + HF = 18 cm\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (\u2235DK = DH and FM = FH)
\u21d2 ED + DF + EF = 18 cm.<\/p>\r\n\r\n\r\n\r\n

Hence, the perimeter of \u25b3EDF = 18 cm.<\/p>\r\n\r\n\r\n\r\n

15) 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 ?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
PT = PQ = 3.8 cm and PT = PR = 3.8 cm
\u2234 QR = QP + PR = 3.8 + 3.8 = 7.6 cm.<\/p>\r\n\r\n\r\n\r\n

Hence, the length of QR is 7.6 cm.<\/p>\r\n\r\n\r\n\r\n

16) In the given figure, O is the centre of the circle AB is a chord and AT is the tangent at A. If \u2220AOB = 100\u2218<\/sup>\u00a0then \u2220BAT is equal to\u00a0?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

Given: AO and BO are the radius of the circle
Since, AO = BO
\u2234 \u25b3AOB is an isosceles triangle.
Now, in \u25b3AOB
\u2220AOB + \u2220OBA + \u2220OAB = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Angle sum property of triangle)
\u21d2 100\u2218<\/sup>\u00a0 + \u2220OAB + \u2220OAB = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0 (\u2220OBA = \u2220OAB)
\u21d2 2\u2220OAB = 80\u2218<\/sup>
\u21d2 \u2220OAB = 40\u2218<\/sup>
We know that the radius and tangent are perpendicular at their point of contact
\u2235 \u2220OAT = 90\u2218<\/sup>\u00a0
\u21d2 \u2220OAB\u00a0 + \u2220BAT = 90\u2218<\/sup>\u00a0
\u21d2 \u2220BAT = 90\u2218<\/sup>\u00a0\u2212\u00a040\u2218<\/sup>\u00a0= 50\u2218<\/sup>\u00a0<\/p>\r\n\r\n\r\n\r\n

YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/h3>\r\n\r\n\r\n\r\n

Then the value of \u2220BAT is 50\u2218<\/sup>\u00a0.<\/p>\r\n\r\n\r\n\r\n

17) 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 find the length of AP ?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that tangent segments to a circle from the same external point are congruent.
Therefore, we have
AP = AQ
BP = BD
CQ = CD
Now, AB + BC + AC = 5 + 4 + 6 = 15
\u21d2AB + BD + DC + AC = 15 cm
\u21d2AB + BP + CQ + AC = 15 cm
\u21d2AP + AQ= 15 cm
\u21d22AP = 15 cm
\u21d2AP = 7.5 cm.<\/p>\r\n\r\n\r\n\r\n

Then the length of AP is 7.5 cm.<\/p>\r\n\r\n\r\n\r\n

18) In the given figure, PA and PB are two tangents from an external point P to a circle with centre O. If \u2220PBA = 65\u2218<\/sup>\u00a0, find \u2220OABand \u2220APB?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that tangents drawn from the external point are congruent.
\u2234 PA = PB

<\/p>\r\n\r\n\r\n\r\n

Now, In isoceles triangle APB
\u2220APB + \u2220PBA + \u2220PAB = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0[Angle sum property of a triangle]
\u21d2 \u2220APB + 65\u2218<\/sup>\u00a0+ 65\u2218<\/sup>\u00a0= 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [\u2235\u2220PBA = \u2220PAB = 65\u2218<\/sup>\u00a0]
\u21d2 \u2220APB = 50\u2218<\/sup>

<\/p>\r\n\r\n\r\n\r\n

We know that the radius and tangent are perperpendular at their point of contact
\u2234\u2220OBP = \u2220OAP = 90\u2218<\/sup>

<\/p>\r\n\r\n\r\n\r\n

Now, In quadrilateral AOBP
\u2220AOB + \u2220OBP + \u2220APB + \u2220OAP = 360\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [Angle sum property of a quadrilateral]
\u21d2 \u2220AOB + 90\u2218<\/sup>\u00a0+ 50\u2218<\/sup>\u00a0+ 90\u2218<\/sup>\u00a0= 360\u2218<\/sup>\u00a0
\u21d2 230\u2218<\/sup>\u00a0+ \u2220BOC = 360\u2218<\/sup>\u00a0
\u21d2 \u2220AOB = 130\u2218<\/sup>\u00a0

<\/p>\r\n\r\n\r\n\r\n

Now, In isoceles triangle AOB
\u2220AOB + \u2220OAB + \u2220OBA = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 [Angle sum property of a triangle]
\u21d2 130\u2218<\/sup>\u00a0+\u00a0 2\u2220OAB = 1800<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [\u2235\u2220OAB = \u2220OBA ]
\u21d2 \u2220OAB = 25\u2218<\/sup><\/p>\r\n\r\n\r\n\r\n

19) 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\u00a0 DE \u22a5 DF then find\u00a0 the radius of the circle ?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Construction: Join AF and AE.
We know that the radius and tangent are perpendicular at their point of contact
\u2235\u2220AED = \u2220AFD = 90\u2218<\/sup>
Since, in quadrilateral AEDF all the angles are right angles
\u2234 AEDF is a rectangle
Now, we know that the pair of opposite sides are equal in rectangle
\u2234 AF = DE = 5 cm
Therefore, the radius of the circle is 5 cm.<\/p>\r\n\r\n\r\n\r\n

20) In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If\u00a0 \u2220PAB =\u00a0 67\u2218<\/sup>, then find \u00a0the measure of \u2220AQB ?<\/h2>\r\n\r\n\r\n\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Solution.<\/h2>\r\n\r\n\r\n\r\n

We know that a chord passing through the centre is the diameter of the circle.
\u2235\u2220BAC = 90\u2218<\/sup>\u00a0\u00a0\u00a0 (Angle in a semi circle is 90\u2218<\/sup>)
By using alternate segment theorem
We have \u2220PAB = \u2220ACB = 67\u2218<\/sup>

<\/p>\r\n\r\n\r\n\r\n

Now, In \u25b3ABC
\u2220ABC + \u2220ACB + \u2220BAC = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [Angle sum property of a triangle]
\u21d2 \u2220ABC + 67\u2218<\/sup>\u00a0<\/sup>+\u00a0 90\u2218<\/sup>\u00a0= 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\u21d2 \u2220ABC= 23\u2218<\/sup>

<\/p>\r\n\r\n\r\n\r\n

Now, \u2220BAQ = 180\u2218<\/sup>\u00a0\u2212 \u2220PAB\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [Linear pair angles]
= 180\u2218<\/sup>\u00a0\u2212 67\u2218<\/sup>
= 113\u2218<\/sup>

<\/p>\r\n\r\n\r\n\r\n

Now, In \u25b3ABQ
\u2220ABQ + \u2220AQB + \u2220BAQ = 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [Angle sum property of a triangle]
\u21d2 23\u2218<\/sup>\u00a0+ \u2220AQB + 113\u2218<\/sup>\u00a0= 180\u2218<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\u21d2 \u2220AQB = 44\u2218<\/sup><\/p>\r\n\r\n\r\n\r\n

Then the value of \u2220AQB is 44\u2218<\/sup><\/p>\r\n\r\n\r\n\r\n

YOU ARE READING: Circles Chapter 10 Extra Questions and Solutions For Class 10 CBSE Mathematics<\/h3>\r\n\r\n