{"id":545740,"date":"2021-10-05T05:57:23","date_gmt":"2021-10-05T05:57:23","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=545740"},"modified":"2021-10-05T10:06:38","modified_gmt":"2021-10-05T10:06:38","slug":"rd-sharma-solutions-for-class-9-maths-chapter-16-circles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-9-maths-chapter-16-circles\/","title":{"rendered":"RD Sharma Solutions for Class 9 Maths Chapter 16\u2013Circles"},"content":{"rendered":"\n

Class 9: Maths Chapter 16 solutions. Complete Class 9 Maths Chapter 16 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 16\u2013Circles<\/h2>\n\n\n\n

RD Sharma 9th Maths Chapter 16, Class 9 Maths Chapter 16 solutions<\/p>\n\n\n\n

Exercise 16.1 Page No: 16.5<\/h4>\n\n\n\n

Question 1: Fill in the blanks:<\/strong><\/p>\n\n\n\n

(i) All points lying inside\/outside a circle are called ______ points\/_______ points.<\/strong><\/p>\n\n\n\n

(ii) Circles having the same centre and different radii are called _____ circles.<\/strong><\/p>\n\n\n\n

(iii) A point whose distance from the center of a circle is greater than its radius lies in _________ of the circle.<\/strong><\/p>\n\n\n\n

(iv) A continuous piece of a circle is _______ of the circle.<\/strong><\/p>\n\n\n\n

(v) The longest chord of a circle is a ____________ of the circle.<\/strong><\/p>\n\n\n\n

(vi) An arc is a __________ when its ends are the ends of a diameter.<\/strong><\/p>\n\n\n\n

(vii) Segment of a circle is a region between an arc and _______ of the circle.<\/strong><\/p>\n\n\n\n

(viii) A circle divides the plane, on which it lies, in _________ parts.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> Interior\/Exterior<\/p>\n\n\n\n

(ii)<\/strong> Concentric<\/p>\n\n\n\n

(iii)<\/strong> The Exterior<\/p>\n\n\n\n

(iv)<\/strong> Arc<\/p>\n\n\n\n

(v)<\/strong> Diameter<\/p>\n\n\n\n

(vi)<\/strong> Semi-circle<\/p>\n\n\n\n

(vii)<\/strong> Center<\/p>\n\n\n\n

(viii)<\/strong> Three<\/p>\n\n\n\n

Question 2: Write the truth value (T\/F) of the following with suitable reasons:<\/strong><\/p>\n\n\n\n

(i) A circle is a plane figure.<\/strong><\/p>\n\n\n\n

(ii) Line segment joining the center to any point on the circle is a radius of the circle,<\/strong><\/p>\n\n\n\n

(iii) If a circle is divided into three equal arcs each is a major arc.<\/strong><\/p>\n\n\n\n

(iv) A circle has only finite number of equal chords.<\/strong><\/p>\n\n\n\n

(v) A chord of a circle, which is twice as long as its radius is the diameter of the circle.<\/strong><\/p>\n\n\n\n

(vi) Sector is the region between the chord and its corresponding arc.<\/strong><\/p>\n\n\n\n

(vii) The degree measure of an arc is the complement of the central angle containing the arc.<\/strong><\/p>\n\n\n\n

(viii) The degree measure of a semi-circle is 1800<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i)<\/strong> T<\/p>\n\n\n\n

(ii)<\/strong> T<\/p>\n\n\n\n

(iii)<\/strong> T<\/p>\n\n\n\n

(iv)<\/strong> F<\/p>\n\n\n\n

(v) <\/strong>T<\/p>\n\n\n\n

(vi)<\/strong> T<\/p>\n\n\n\n

(vii)<\/strong> F<\/p>\n\n\n\n

(viii)<\/strong> T<\/p>\n\n\n\n

\n\n\n\n

Exercise 16.2 Page No: 16.24<\/h4>\n\n\n\n

Question 1: The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Radius of circle (OA) = 8 cm (Given)<\/p>\n\n\n\n

Chord (AB) = 12cm (Given)<\/p>\n\n\n\n

Draw a perpendicular OC on AB.<\/p>\n\n\n\n

We know, perpendicular from centre to chord bisects the chord<\/p>\n\n\n\n

Which implies, AC = BC = 12\/2 = 6 cm<\/p>\n\n\n\n

In right \u0394OCA:<\/p>\n\n\n\n

Using Pythagoras theorem,<\/p>\n\n\n\n

OA2<\/sup> = AC2<\/sup> + OC2<\/sup><\/p>\n\n\n\n

64 = 36 + OC2<\/sup><\/p>\n\n\n\n

OC2<\/sup> = 64 \u2013 36 = 28<\/p>\n\n\n\n

or OC = \u221a28 = 5.291 (approx.)<\/p>\n\n\n\n

The distance of the chord from the centre is 5.291 cm.<\/p>\n\n\n\n

Question 2: Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Distance of the chord from the centre = OC = 5 cm (Given)<\/p>\n\n\n\n

Radius of the circle = OA = 10 cm (Given)<\/p>\n\n\n\n

In \u0394OCA:<\/p>\n\n\n\n

Using Pythagoras theorem,<\/p>\n\n\n\n

OA2<\/sup> = AC2<\/sup> + OC2<\/sup><\/p>\n\n\n\n

100 = AC2 <\/sup>+ 25<\/p>\n\n\n\n

AC2<\/sup> = 100 \u2013 25 = 75<\/p>\n\n\n\n

AC = \u221a75 = 8.66<\/p>\n\n\n\n

As, perpendicular from the centre to chord bisects the chord.<\/p>\n\n\n\n

Therefore, AC = BC = 8.66 cm<\/p>\n\n\n\n

=> AB = AC + BC = 8.66 + 8.66 = 17.32<\/p>\n\n\n\n

Answer: AB = 17.32 cm<\/p>\n\n\n\n

Question 3: Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Distance of the chord from the centre = OC = 4 cm (Given)<\/p>\n\n\n\n

Radius of the circle = OA = 6 cm (Given)<\/p>\n\n\n\n

In \u0394OCA:<\/p>\n\n\n\n

Using Pythagoras theorem,<\/p>\n\n\n\n

OA2<\/sup> = AC2<\/sup> + OC2<\/sup><\/p>\n\n\n\n

36 = AC2 <\/sup>+ 16<\/p>\n\n\n\n

AC2<\/sup> = 36 \u2013 16 = 20<\/p>\n\n\n\n

AC = \u221a20 = 4.47<\/p>\n\n\n\n

Or AC = 4.47cm<\/p>\n\n\n\n

As, perpendicular from the centre to chord bisects the chord.<\/p>\n\n\n\n

Therefore, AC = BC = 4.47 cm<\/p>\n\n\n\n

=> AB = AC + BC = 4.47 + 4.47 = 8.94<\/p>\n\n\n\n

Answer: AB = 8.94 cm<\/p>\n\n\n\n

Question 4: Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Given: AB = 5 cm, CD = 11 cm, PQ = 3 cm<\/p>\n\n\n\n

Draw perpendiculars OP on CD and OQ on AB<\/p>\n\n\n\n

Let OP = x cm and OC = OA = r cm<\/p>\n\n\n\n

We know, perpendicular from centre to chord bisects it.<\/p>\n\n\n\n

Since OP\u22a5CD, we have<\/p>\n\n\n\n

CP = PD = 11\/2 cm<\/p>\n\n\n\n

And OQ\u22a5AB<\/p>\n\n\n\n

AQ = BQ = 5\/2 cm<\/p>\n\n\n\n

In \u0394OCP:<\/p>\n\n\n\n

By Pythagoras theorem,<\/p>\n\n\n\n

OC2<\/sup> = OP2<\/sup> + CP2<\/sup><\/p>\n\n\n\n

r2 <\/sup>= x2 <\/sup>+ (11\/2) 2 <\/sup>\u2026..(1)<\/p>\n\n\n\n

In \u0394OQA:<\/p>\n\n\n\n

By Pythagoras theorem,<\/p>\n\n\n\n

OA2<\/sup>=OQ2<\/sup>+AQ2<\/sup><\/p>\n\n\n\n

r2<\/sup>= (x+3) 2 <\/sup>+ (5\/2) 2<\/sup> \u2026..(2)<\/p>\n\n\n\n

From equations (1) and (2), we get<\/p>\n\n\n\n

(x+3) 2 <\/sup>+ (5\/2) 2<\/sup> = x2 <\/sup>+ (11\/2) 2<\/sup><\/p>\n\n\n\n

Solve above equation and find the value of x.<\/p>\n\n\n\n

x2<\/sup> + 6x + 9 + 25\/4 = x2<\/sup> + 121\/4<\/p>\n\n\n\n

(using identity, (a+b) 2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab )<\/p>\n\n\n\n

6x = 121\/4 \u2013 25\/4 \u2212 9<\/p>\n\n\n\n

6x = 15<\/p>\n\n\n\n

or x = 15\/6 = 5\/2<\/p>\n\n\n\n

Substitute the value of x in equation (1), and find the length of radius,<\/p>\n\n\n\n

r2 <\/sup>= (5\/2)2 <\/sup>+ (11\/2) 2<\/sup><\/p>\n\n\n\n

= 25\/4 + 121\/4<\/p>\n\n\n\n

= 146\/4<\/p>\n\n\n\n

or r = \u221a146\/4 cm<\/p>\n\n\n\n

Question 5: Give a method to find the centre of a given circle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Steps of Construction:<\/p>\n\n\n\n

Step 1: Consider three points A, B and C on a circle.<\/p>\n\n\n\n

Step 2: Join AB and BC.<\/p>\n\n\n\n

Step 3: Draw perpendicular bisectors of chord AB and BC which intersect each other at a point, say O.<\/p>\n\n\n\n

Step 4: This point O is a centre of the circle, because we know that, the Perpendicular bisectors of chord always pass through the centre.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 6: Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

From figure, Let C is the mid-point of chord AB.<\/p>\n\n\n\n

To prove: D is the mid-point of arc AB.<\/p>\n\n\n\n

Now, In \u0394OAC and \u0394OBC<\/p>\n\n\n\n

OA = OB [Radius of circle]<\/p>\n\n\n\n

OC = OC [Common]<\/p>\n\n\n\n

AC = BC [C is the mid-point of chord AB (given)]<\/p>\n\n\n\n

So, by SSS condition: \u0394OAC \u2245 \u0394OBC<\/p>\n\n\n\n

So, \u2220AOC = \u2220BOC (BY CPCT)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Therefore, D is the mid-point of arc AB. Hence Proved.<\/p>\n\n\n\n

Question 7: Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Form figure: PQ is a diameter of circle which bisects the chord AB at C. (Given)<\/p>\n\n\n\n

To Prove: PQ bisects \u2220AOB<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

In \u0394BOC and \u0394AOC<\/p>\n\n\n\n

OA = OB [Radius]<\/p>\n\n\n\n

OC = OC [Common side]<\/p>\n\n\n\n

AC = BC [Given]<\/p>\n\n\n\n

Then, by SSS condition: \u0394AOC \u2245 \u0394BOC<\/p>\n\n\n\n

So, \u2220AOC = \u2220BOC [By c.p.c.t.]<\/p>\n\n\n\n

Therefore, PQ bisects \u2220AOB. Hence proved.<\/p>\n\n\n\n

\n\n\n\n

Exercise 16.3 Page No: 16.40<\/h4>\n\n\n\n

Question 1: Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let R, S and M be the position of Ishita, Isha and Nisha respectively.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Since OA is a perpendicular bisector on RS, so AR = AS = 24\/2 = 12 cm<\/p>\n\n\n\n

Radii of circle = OR = OS = OM = 20 cm (Given)<\/p>\n\n\n\n

In \u0394OAR:<\/p>\n\n\n\n

By Pythagoras theorem,<\/p>\n\n\n\n

OA2<\/sup>+AR2<\/sup>=OR2<\/sup><\/p>\n\n\n\n

OA2<\/sup>+122<\/sup>=202<\/sup><\/p>\n\n\n\n

OA2 <\/sup>= 400 \u2013 144 = 256<\/p>\n\n\n\n

Or OA = 16 m \u2026(1)<\/p>\n\n\n\n

From figure, OABC is a kite since OA = OC and AB = BC. We know that, diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.<\/p>\n\n\n\n

So in \u0394RSM, \u2220RCS = 900<\/sup> and RC = CM \u2026(2)<\/p>\n\n\n\n

Now, Area of \u0394ORS = Area of \u0394ORS<\/p>\n\n\n\n

=>1\/2\u00d7OA\u00d7RS = 1\/2 x RC x OS<\/p>\n\n\n\n

=> OA \u00d7RS = RC x OS<\/p>\n\n\n\n

=> 16 x 24 = RC x 20<\/p>\n\n\n\n

=> RC = 19.2<\/p>\n\n\n\n

Since RC = CM (from (2), we have<\/p>\n\n\n\n

RM = 2(19.2) = 38.4<\/p>\n\n\n\n

So, the distance between Ishita and Nisha is 38.4 m.<\/p>\n\n\n\n

Question 2: A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Since, AB = BC = CA. So, ABC is an equilateral triangle<\/p>\n\n\n\n

Radius = OA = 40 m (Given)<\/p>\n\n\n\n

We know, medians of equilateral triangle pass through the circumcentre and intersect each other at the ratio 2 : 1.<\/p>\n\n\n\n

Here AD is the median of equilateral triangle ABC, we can write:<\/p>\n\n\n\n

OA\/OD = 2\/1<\/p>\n\n\n\n

or 40\/OD = 2\/1<\/p>\n\n\n\n

or OD = 20 m<\/p>\n\n\n\n

Therefore, AD = OA + OD = (40 + 20) m = 60 m<\/p>\n\n\n\n

Now, In \u0394ADC:<\/p>\n\n\n\n

By Pythagoras theorem,<\/p>\n\n\n\n

AC2<\/sup> = AD2<\/sup> + DC2<\/sup><\/p>\n\n\n\n

AC2<\/sup> = 602<\/sup> + (AC\/2) 2<\/sup><\/p>\n\n\n\n

AC2 <\/sup>= 3600 + AC2<\/sup> \/ 4<\/p>\n\n\n\n

3\/4 AC2 <\/sup>= 3600<\/p>\n\n\n\n

AC2 <\/sup>= 4800<\/p>\n\n\n\n

or AC = 40\u221a3 m<\/p>\n\n\n\n

Therefore, length of string of each phone will be 40\u221a3 m.<\/p>\n\n\n\n

\n\n\n\n

Exercise 16.4 Page No: 16.60<\/h3>\n\n\n\n

Question 1: In figure, O is the centre of the circle. If \u2220APB = 500<\/sup>, find \u2220AOB and \u2220OAB.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220APB = 500<\/sup> (Given)<\/p>\n\n\n\n

By degree measure theorem: \u2220AOB = 2\u2220APB<\/p>\n\n\n\n

\u2220AOB = 2 \u00d7 500<\/sup> = 1000<\/sup><\/p>\n\n\n\n

Again, OA = OB [Radius of circle]<\/p>\n\n\n\n

Then \u2220OAB = \u2220OBA [Angles opposite to equal sides]<\/p>\n\n\n\n

Let \u2220OAB = m<\/p>\n\n\n\n

In \u0394OAB,<\/p>\n\n\n\n

By angle sum property: \u2220OAB+\u2220OBA+\u2220AOB=1800<\/sup><\/p>\n\n\n\n

=> m + m + 1000<\/sup> = 1800<\/sup><\/p>\n\n\n\n

=>2m = 1800<\/sup> \u2013 1000<\/sup> = 800<\/sup><\/p>\n\n\n\n

=>m = 800<\/sup>\/2 = 400<\/sup><\/p>\n\n\n\n

\u2220OAB = \u2220OBA = 400<\/sup><\/p>\n\n\n\n

Question 2: In figure, it is given that O is the centre of the circle and \u2220AOC = 1500<\/sup>. Find \u2220ABC.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220AOC = 1500<\/sup> (Given)<\/p>\n\n\n\n

By degree measure theorem: \u2220ABC = (reflex\u2220AOC)\/2 \u2026(1)<\/p>\n\n\n\n

We know, \u2220AOC + reflex(\u2220AOC) = 3600<\/sup> [Complex angle]<\/p>\n\n\n\n

1500<\/sup> + reflex\u2220AOC = 3600<\/sup><\/p>\n\n\n\n

or reflex \u2220AOC = 3600<\/sup>\u22121500<\/sup> = 2100<\/sup><\/p>\n\n\n\n

From (1) => \u2220ABC = 210 o<\/sup> \/2 = 105o<\/sup><\/p>\n\n\n\n

Question 3: In figure, O is the centre of the circle. Find \u2220BAC.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: \u2220AOB = 800<\/sup> and \u2220AOC = 1100<\/sup><\/p>\n\n\n\n

Therefore, \u2220AOB+\u2220AOC+\u2220BOC=3600<\/sup> [Completeangle]<\/p>\n\n\n\n

Substitute given values,<\/p>\n\n\n\n

800<\/sup> + 1000<\/sup> + \u2220BOC = 3600<\/sup><\/p>\n\n\n\n

\u2220BOC = 3600<\/sup> \u2013 800<\/sup> \u2013 1100<\/sup> = 1700<\/sup><\/p>\n\n\n\n

or \u2220BOC = 1700<\/sup><\/p>\n\n\n\n

Now, by degree measure theorem<\/p>\n\n\n\n

\u2220BOC = 2\u2220BAC<\/p>\n\n\n\n

1700<\/sup> = 2\u2220BAC<\/p>\n\n\n\n

Or \u2220BAC = 1700<\/sup>\/2 = 850<\/sup><\/p>\n\n\n\n

Question 4: If O is the centre of the circle, find the value of x in each of the following figures.<\/strong><\/p>\n\n\n\n

(i)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220AOC = 1350<\/sup> (Given)<\/p>\n\n\n\n

From figure, \u2220AOC + \u2220BOC = 1800<\/sup> [Linear pair of angles]<\/p>\n\n\n\n

1350<\/sup> +\u2220BOC = 1800<\/sup><\/p>\n\n\n\n

or \u2220BOC=1800<\/sup>\u22121350<\/sup><\/p>\n\n\n\n

or \u2220BOC=450<\/sup><\/p>\n\n\n\n

Again, by degree measure theorem<\/p>\n\n\n\n

\u2220BOC = 2\u2220CPB<\/p>\n\n\n\n

450<\/sup> = 2x<\/p>\n\n\n\n

x = 450<\/sup>\/2<\/p>\n\n\n\n

(ii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220ABC=400<\/sup> (given)<\/p>\n\n\n\n

\u2220ACB = 900<\/sup> [Angle in semicircle]<\/p>\n\n\n\n

In \u0394ABC,<\/p>\n\n\n\n

\u2220CAB+\u2220ACB+\u2220ABC=1800<\/sup> [angle sum property]<\/p>\n\n\n\n

\u2220CAB+900<\/sup>+400<\/sup>=1800<\/sup><\/p>\n\n\n\n

\u2220CAB=1800<\/sup>\u2212900<\/sup>\u2212400<\/sup><\/p>\n\n\n\n

\u2220CAB=500<\/sup><\/p>\n\n\n\n

Now, \u2220CDB = \u2220CAB [Angle is on same segment]<\/p>\n\n\n\n

This implies, x = 500<\/sup><\/p>\n\n\n\n

(iii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220AOC = 1200<\/sup> (given)<\/p>\n\n\n\n

By degree measure theorem: \u2220AOC = 2\u2220APC<\/p>\n\n\n\n

1200 <\/sup>= 2\u2220APC<\/p>\n\n\n\n

\u2220APC = 1200<\/sup>\/2 = 600<\/sup><\/p>\n\n\n\n

Again, \u2220APC + \u2220ABC = 1800<\/sup> [Sum of opposite angles of cyclic quadrilaterals = 180 o<\/sup> ]<\/p>\n\n\n\n

600<\/sup> + \u2220ABC=1800<\/sup><\/p>\n\n\n\n

\u2220ABC=1800<\/sup>\u2212600<\/sup><\/p>\n\n\n\n

\u2220ABC = 1200<\/sup><\/p>\n\n\n\n

\u2220ABC + \u2220DBC = 1800<\/sup> [Linear pair of angles]<\/p>\n\n\n\n

1200 <\/sup>+ x = 1800<\/sup><\/p>\n\n\n\n

x = 1800<\/sup>\u22121200<\/sup>=600<\/sup><\/p>\n\n\n\n

The value of x is 600<\/sup><\/p>\n\n\n\n

(iv)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220CBD = 650<\/sup> (given)<\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

\u2220ABC + \u2220CBD = 1800<\/sup> [ Linear pair of angles]<\/p>\n\n\n\n

\u2220ABC + 650<\/sup> = 1800<\/sup><\/p>\n\n\n\n

\u2220ABC =1800<\/sup>\u2212650<\/sup>=1150<\/sup><\/p>\n\n\n\n

Again, reflex \u2220AOC = 2\u2220ABC [Degree measure theorem]<\/p>\n\n\n\n

x=2(1150<\/sup>) = 2300<\/sup><\/p>\n\n\n\n

The value of x is 2300<\/sup><\/p>\n\n\n\n

(v)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220OAB = 350<\/sup> (Given)<\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

\u2220OBA = \u2220OAB = 350<\/sup> [Angles opposite to equal radii]<\/p>\n\n\n\n

In\u0394AOB:<\/p>\n\n\n\n

\u2220AOB + \u2220OAB + \u2220OBA = 1800<\/sup> [angle sum property]<\/p>\n\n\n\n

\u2220AOB + 350<\/sup> + 350<\/sup> = 1800<\/sup><\/p>\n\n\n\n

\u2220AOB = 1800<\/sup> \u2013 350<\/sup> \u2013 350<\/sup> = 1100<\/sup><\/p>\n\n\n\n

Now, \u2220AOB + reflex\u2220AOB = 3600<\/sup> [Complex angle]<\/p>\n\n\n\n

1100<\/sup> + reflex\u2220AOB = 3600<\/sup><\/p>\n\n\n\n

reflex\u2220AOB = 3600<\/sup> \u2013 1100<\/sup> = 2500<\/sup><\/p>\n\n\n\n

By degree measure theorem: reflex \u2220AOB = 2\u2220ACB<\/p>\n\n\n\n

2500<\/sup> = 2x<\/p>\n\n\n\n

x = 2500<\/sup>\/2=1250<\/sup><\/p>\n\n\n\n

(vi)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220AOB = 60o<\/sup> (given)<\/p>\n\n\n\n

By degree measure theorem: reflex\u2220AOB = 2\u2220OAC<\/p>\n\n\n\n

60 o<\/sup> = 2\u2220 OAC<\/p>\n\n\n\n

\u2220OAC = 60 o<\/sup> \/ 2 = 30 o<\/sup> [Angles opposite to equal radii]<\/p>\n\n\n\n

Or x = 300<\/sup><\/p>\n\n\n\n

(vii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220BAC = 500<\/sup> and \u2220DBC = 700<\/sup> (given)<\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

\u2220BDC = \u2220BAC = 500 <\/sup>[Angle on same segment]<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

In \u0394BDC:<\/p>\n\n\n\n

Using angle sum property, we have<\/p>\n\n\n\n

\u2220BDC+\u2220BCD+\u2220DBC=1800<\/sup><\/p>\n\n\n\n

Substituting given values, we get<\/p>\n\n\n\n

500<\/sup> + x0<\/sup> + 700<\/sup> = 1800<\/sup><\/p>\n\n\n\n

x0 <\/sup>= 1800<\/sup>\u2212500<\/sup>\u2212700<\/sup>=600<\/sup><\/p>\n\n\n\n

or x = 60o<\/sup><\/p>\n\n\n\n

(viii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220DBO = 400<\/sup> (Given)<\/p>\n\n\n\n

Form figure:<\/p>\n\n\n\n

\u2220DBC = 900<\/sup> [Angle in a semicircle]<\/p>\n\n\n\n

\u2220DBO + \u2220OBC = 900<\/sup><\/p>\n\n\n\n

400<\/sup>+\u2220OBC=900<\/sup><\/p>\n\n\n\n

or \u2220OBC=900<\/sup>\u2212400<\/sup>=500<\/sup><\/p>\n\n\n\n

Again, By degree measure theorem: \u2220AOC = 2\u2220OBC<\/p>\n\n\n\n

or x = 2\u00d7500<\/sup>=1000<\/sup><\/p>\n\n\n\n

(ix)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220CAD = 28, \u2220ADB = 32 and \u2220ABC = 50 (Given)<\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

In \u0394DAB:<\/p>\n\n\n\n

Angle sum property: \u2220ADB + \u2220DAB + \u2220ABD = 1800<\/sup><\/p>\n\n\n\n

By substituting the given values, we get<\/p>\n\n\n\n

320<\/sup> + \u2220DAB + 500<\/sup> = 1800<\/sup><\/p>\n\n\n\n

\u2220DAB=1800<\/sup>\u2212320<\/sup>\u2212500<\/sup><\/p>\n\n\n\n

\u2220DAB = 980<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

\u2220DAB+\u2220DCB=1800 <\/sup>[Opposite angles of cyclic quadrilateral, their sum = 180 degrees]<\/p>\n\n\n\n

980<\/sup>+x=1800<\/sup><\/p>\n\n\n\n

or x = 1800<\/sup>\u2212980<\/sup>=820<\/sup><\/p>\n\n\n\n

The value of x is 82 degrees.<\/p>\n\n\n\n

(x)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220BAC = 350<\/sup> and \u2220DBC = 650<\/sup><\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

\u2220BDC = \u2220BAC = 350<\/sup> [Angle in same segment]<\/p>\n\n\n\n

In \u0394BCD:<\/p>\n\n\n\n

Angle sum property, we have<\/p>\n\n\n\n

\u2220BDC + \u2220BCD + \u2220DBC = 1800<\/sup><\/p>\n\n\n\n

350<\/sup> + x + 650<\/sup> = 1800<\/sup><\/p>\n\n\n\n

or x = 1800<\/sup> \u2013 350<\/sup> \u2013 650<\/sup> = 800<\/sup><\/p>\n\n\n\n

(xi)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220ABD = 400<\/sup>, \u2220CPD = 1100<\/sup> (Given)<\/p>\n\n\n\n

Form figure:<\/p>\n\n\n\n

\u2220ACD = \u2220ABD = 400<\/sup> [Angle in same segment]<\/p>\n\n\n\n

In \u0394PCD,<\/p>\n\n\n\n

Angle sum property: \u2220PCD+\u2220CPO+\u2220PDC=1800<\/sup><\/p>\n\n\n\n

400 + 1100<\/sup> + x = 1800<\/sup><\/p>\n\n\n\n

x=1800<\/sup>\u22121500 <\/sup>=300<\/sup><\/p>\n\n\n\n

The value of x is 30 degrees.<\/p>\n\n\n\n

(xii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220BAC = 520<\/sup> (Given)<\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

\u2220BDC = \u2220BAC = 520<\/sup> [Angle in same segment]<\/p>\n\n\n\n

Since OD = OC (radii), then \u2220ODC = \u2220OCD [Opposite angle to equal radii]<\/p>\n\n\n\n

So, x = 520<\/sup><\/p>\n\n\n\n

Question 5: O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that \u2220BOD = \u2220A.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

In \u0394OBD and \u0394OCD:<\/p>\n\n\n\n

OB = OC [Radius]<\/p>\n\n\n\n

\u2220ODB = \u2220ODC [Each 900<\/sup>]<\/p>\n\n\n\n

OD = OD [Common]<\/p>\n\n\n\n

Therefore, By RHS Condition<\/p>\n\n\n\n

\u0394OBD \u2245 \u0394OCD<\/p>\n\n\n\n

So, \u2220BOD = \u2220COD\u2026..(i)[By CPCT]<\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

By degree measure theorem: \u2220BOC = 2\u2220BAC<\/p>\n\n\n\n

2\u2220BOD = 2\u2220BAC [Using(i)]<\/p>\n\n\n\n

\u2220BOD = \u2220BAC<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

Question 6: In figure, O is the centre of the circle, BO is the bisector of \u2220ABC. Show that AB = AC.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Since, BO is the bisector of \u2220ABC, then,<\/p>\n\n\n\n

\u2220ABO = \u2220CBO \u2026..(i)<\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

Radius of circle = OB = OA = OB = OC<\/p>\n\n\n\n

\u2220OAB = \u2220OCB \u2026..(ii) [opposite angles to equal sides]<\/p>\n\n\n\n

\u2220ABO = \u2220DAB \u2026..(iii) [opposite angles to equal sides]<\/p>\n\n\n\n

From equations (i), (ii) and (iii), we get<\/p>\n\n\n\n

\u2220OAB = \u2220OCB \u2026..(iv)<\/p>\n\n\n\n

In \u0394OAB and \u0394OCB:<\/p>\n\n\n\n

\u2220OAB = \u2220OCB [From (iv)]<\/p>\n\n\n\n

OB = OB [Common]<\/p>\n\n\n\n

\u2220OBA = \u2220OBC [Given]<\/p>\n\n\n\n

Then, By AAS condition : \u0394OAB \u2245 \u0394OCB<\/p>\n\n\n\n

So, AB = BC [By CPCT]<\/p>\n\n\n\n

Question 7: In figure, O is the centre of the circle, then prove that \u2220x = \u2220y + \u2220z.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From the figure:<\/p>\n\n\n\n

\u22203 = \u22204 \u2026.(i) [Angles in same segment]<\/p>\n\n\n\n

\u2220x = 2\u22203 [By degree measure theorem]<\/p>\n\n\n\n

\u2220x = \u22203 + \u22203<\/p>\n\n\n\n

\u2220x = \u22203 + \u22204 (Using (i) ) \u2026..(ii)<\/p>\n\n\n\n

Again, \u2220y = \u22203 + \u22201 [By exterior angle property]<\/p>\n\n\n\n

or \u22203 = \u2220y \u2212 \u22201 \u2026..(iii)<\/p>\n\n\n\n

\u22204 = \u2220z + \u22201 \u2026. (iv) [By exterior angle property]<\/p>\n\n\n\n

Now, from equations (ii) , (iii) and (iv), we get<\/p>\n\n\n\n

\u2220x = \u2220y \u2212 \u22201 + \u2220z + \u22201<\/p>\n\n\n\n

or \u2220x = \u2220y + \u2220z + \u22201 \u2212 \u22201<\/p>\n\n\n\n

or x = \u2220y + \u2220z<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

\n\n\n\n

Exercise 16.5 Page No: 16.83<\/h4>\n\n\n\n

Question 1: In figure, \u0394ABC is an equilateral triangle. Find m\u2220BEC.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u0394ABC is an equilateral triangle. (Given)<\/p>\n\n\n\n

Each angle of an equilateral triangle is 60 degrees.<\/p>\n\n\n\n

In quadrilateral ABEC:<\/p>\n\n\n\n

\u2220BAC + \u2220BEC = 180o<\/sup> (Opposite angles of quadrilateral)<\/p>\n\n\n\n

60o<\/sup> + \u2220BEC = 180 o<\/sup><\/p>\n\n\n\n

\u2220BEC = 180 o<\/sup> \u2013 60 o<\/sup><\/p>\n\n\n\n

\u2220BEC = 120 o<\/sup><\/p>\n\n\n\n

Question 2: In figure, \u0394 PQR is an isosceles triangle with PQ = PR and m\u2220PQR=35\u00b0. Find m\u2220QSR and m\u2220QTR.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: \u0394PQR is an isosceles triangle with PQ = PR and m\u2220PQR = 35\u00b0<\/p>\n\n\n\n

In \u0394PQR:<\/p>\n\n\n\n

\u2220PQR = \u2220PRQ = 35o<\/sup> (Angle opposite to equal sides)<\/p>\n\n\n\n

Again, by angle sum property<\/p>\n\n\n\n

\u2220P + \u2220Q + \u2220R = 180 o<\/sup><\/p>\n\n\n\n

\u2220P + 35 o<\/sup> + 35 o<\/sup> = 180 o<\/sup><\/p>\n\n\n\n

\u2220P + 70 o<\/sup> = 180 o<\/sup><\/p>\n\n\n\n

\u2220P = 180 o<\/sup> \u2013 70 o<\/sup><\/p>\n\n\n\n

\u2220P = 110 o<\/sup><\/p>\n\n\n\n

Now, in quadrilateral SQTR,<\/p>\n\n\n\n

\u2220QSR + \u2220QTR = 180 o<\/sup> (Opposite angles of quadrilateral)<\/p>\n\n\n\n

110 o<\/sup> + \u2220QTR = 180 o<\/sup><\/p>\n\n\n\n

\u2220QTR = 70 o<\/sup><\/p>\n\n\n\n

Question 3: In figure, O is the centre of the circle. If \u2220BOD = 160o<\/sup>, find the values of x and y.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From figure: \u2220BOD = 160 o<\/sup><\/strong><\/p>\n\n\n\n

By degree measure theorem: \u2220BOD = 2 \u2220BCD<\/p>\n\n\n\n

160 o<\/sup><\/strong> = 2x<\/p>\n\n\n\n

or x = 80 o<\/sup><\/strong><\/p>\n\n\n\n

Now, in quadrilateral ABCD,<\/p>\n\n\n\n

\u2220BAD + \u2220BCD = 180 o<\/sup><\/strong> (Opposite angles of Cyclic quadrilateral)<\/p>\n\n\n\n

y + x = 180 o<\/sup><\/strong><\/p>\n\n\n\n

Putting value of x,<\/p>\n\n\n\n

y + 80 o<\/sup><\/strong> = 180 o<\/sup><\/strong><\/p>\n\n\n\n

y = 100 o<\/sup><\/strong><\/p>\n\n\n\n

Answer: x = 80 o<\/sup> <\/strong>and y = 100 o<\/sup><\/strong>.<\/p>\n\n\n\n

Question 4: In figure, ABCD is a cyclic quadrilateral. If \u2220BCD = 100o<\/sup> and \u2220ABD = 70o<\/sup>, find \u2220ADB.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

From figure:<\/p>\n\n\n\n

In quadrilateral ABCD,<\/p>\n\n\n\n

\u2220DCB + \u2220BAD = 180o<\/sup><\/strong> (Opposite angles of Cyclic quadrilateral)<\/p>\n\n\n\n

100 o<\/sup><\/strong> + \u2220BAD = 180o<\/sup><\/strong><\/p>\n\n\n\n

\u2220BAD = 800<\/sup><\/p>\n\n\n\n

In \u0394 BAD:<\/p>\n\n\n\n

By angle sum property: \u2220ADB + \u2220DAB + \u2220ABD = 180 o<\/sup><\/p>\n\n\n\n

\u2220ADB + 80o<\/sup> + 70 o<\/sup> = 180 o<\/sup><\/p>\n\n\n\n

\u2220ADB = 30o<\/sup><\/p>\n\n\n\n

Question 5: If ABCD is a cyclic quadrilateral in which AD||BC (figure). Prove that \u2220B = \u2220C.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: ABCD is a cyclic quadrilateral with AD \u2016 BC<\/p>\n\n\n\n

=> \u2220A + \u2220C = 180o<\/sup> \u2026\u2026\u2026(1)[Opposite angles of cyclic quadrilateral]<\/p>\n\n\n\n

and \u2220A + \u2220B = 180o<\/sup> \u2026\u2026\u2026(2)[Co-interior angles]<\/p>\n\n\n\n

Form (1) and (2), we have<\/p>\n\n\n\n

\u2220B = \u2220C<\/p>\n\n\n\n

Hence proved.<\/p>\n\n\n\n

Question 6: In figure, O is the centre of the circle. Find \u2220CBD.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: \u2220BOC = 100o<\/sup><\/p>\n\n\n\n

By degree measure theorem: \u2220AOC = 2 \u2220APC<\/p>\n\n\n\n

100 o<\/sup> = 2 \u2220APC<\/p>\n\n\n\n

or \u2220APC = 50 o<\/sup><\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

\u2220APC + \u2220ABC = 180 o<\/sup> (Opposite angles of a cyclic quadrilateral)<\/p>\n\n\n\n

50o<\/sup> + \u2220ABC = 180 o<\/sup><\/p>\n\n\n\n

or \u2220ABC = 130 o<\/sup><\/p>\n\n\n\n

Now, \u2220ABC + \u2220CBD = 180 o<\/sup> (Linear pair)<\/p>\n\n\n\n

130o<\/sup> + \u2220CBD = 180 o<\/sup><\/p>\n\n\n\n

or \u2220CBD = 50 o<\/sup><\/p>\n\n\n\n

Question 7: In figure, AB and CD are diameters of a circle with centre O. If \u2220OBD = 500<\/sup>, find \u2220AOC.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: \u2220OBD = 500<\/sup><\/strong><\/p>\n\n\n\n

Here, AB and CD are the diameters of the circles with centre O.<\/p>\n\n\n\n

\u2220DBC = 900<\/sup><\/strong> \u2026.(i)[Angle in the semi-circle]<\/p>\n\n\n\n

Also, \u2220DBC = 500<\/sup><\/strong> + \u2220OBC<\/p>\n\n\n\n

900<\/sup><\/strong> = 500<\/sup><\/strong> + \u2220OBC<\/p>\n\n\n\n

or \u2220OBC = 400<\/sup><\/strong><\/p>\n\n\n\n

Again, By degree measure theorem: \u2220AOC = 2 \u2220ABC<\/p>\n\n\n\n

\u2220AOC = 2\u2220OBC = 2 x 400<\/sup><\/strong> = 800<\/sup><\/strong><\/p>\n\n\n\n

Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(\u2220CAB) = 300<\/sup>. Find m(\u2220ACB) and m(\u2220ABC).<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: m(\u2220CAB)= 300<\/sup><\/strong><\/p>\n\n\n\n

To Find: m(\u2220ACB) and m(\u2220ABC).<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

\u2220ACB = 900<\/sup><\/strong> (Angle in semi-circle)<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

In \u25b3ABC, by angle sum property: \u2220CAB + \u2220ACB + \u2220ABC = 1800<\/sup><\/strong><\/p>\n\n\n\n

300<\/sup><\/strong> + 900<\/sup><\/strong> + \u2220ABC = 1800<\/sup><\/strong><\/p>\n\n\n\n

\u2220ABC = 600<\/sup><\/strong><\/p>\n\n\n\n

Answer: \u2220ACB = 900<\/sup><\/strong> and \u2220ABC = 600<\/sup><\/strong><\/p>\n\n\n\n

Question 9: In a cyclic quadrilateral ABCD if AB||CD and \u2220B = 70o<\/sup> , find the remaining angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

A cyclic quadrilateral ABCD with AB||CD and \u2220B = 70o<\/sup>.<\/p>\n\n\n\n

\u2220B + \u2220C = 180o<\/sup> (Co-interior angle)<\/p>\n\n\n\n

700<\/sup> + \u2220C = 1800<\/sup><\/p>\n\n\n\n

\u2220C = 1100<\/sup><\/p>\n\n\n\n

And,<\/p>\n\n\n\n

=> \u2220B + \u2220D = 1800<\/sup> (Opposite angles of Cyclic quadrilateral)<\/p>\n\n\n\n

700<\/sup> + \u2220D = 1800<\/sup><\/p>\n\n\n\n

\u2220D = 1100<\/sup><\/p>\n\n\n\n

Again, \u2220A + \u2220C = 1800<\/sup> (Opposite angles of cyclic quadrilateral)<\/p>\n\n\n\n

\u2220A + 1100<\/sup> = 1800<\/sup><\/p>\n\n\n\n

\u2220A = 700<\/sup><\/p>\n\n\n\n

Answer: \u2220A = 700<\/sup> , \u2220C = 1100 <\/sup>and \u2220D = 1100<\/sup><\/p>\n\n\n\n

Question 10: In a cyclic quadrilateral ABCD, if m \u2220A = 3(m\u2220C). Find m \u2220A.
Solution:<\/strong><\/p>\n\n\n\n

\u2220A + \u2220C = 180o<\/sup> \u2026..(1)[Opposite angles of cyclic quadrilateral]<\/p>\n\n\n\n

Since m \u2220A = 3(m\u2220C) (given)<\/p>\n\n\n\n

=> \u2220A = 3\u2220C \u2026(2)<\/p>\n\n\n\n

Equation (1) => 3\u2220C + \u2220C = 180 o<\/sup><\/p>\n\n\n\n

or 4\u2220C = 180o<\/sup><\/p>\n\n\n\n

or \u2220C = 45o<\/sup><\/p>\n\n\n\n

From equation (2)<\/p>\n\n\n\n

\u2220A = 3 x 45o<\/sup> = 135o<\/sup><\/p>\n\n\n\n

Question 11: In figure, O is the centre of the circle \u2220DAB = 50\u00b0. Calculate the values of x and y.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given : \u2220DAB = 50o<\/sup><\/p>\n\n\n\n

By degree measure theorem: \u2220BOD = 2 \u2220BAD<\/p>\n\n\n\n

so, x = 2( 500<\/sup>) = 1000<\/sup><\/p>\n\n\n\n

Since, ABCD is a cyclic quadrilateral, we have<\/p>\n\n\n\n

\u2220A + \u2220C = 1800<\/sup><\/p>\n\n\n\n

500<\/sup> + y = 1800<\/sup><\/p>\n\n\n\n

y = 1300<\/sup><\/p>\n\n\n\n

\n\n\n\n

Exercise VSAQs Page No: 16.89<\/h4>\n\n\n\n

Question 1: In figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If \u2220APB = 70\u00b0, find \u2220ACB.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

By degree measure theorem: \u2220AOB = 2 \u2220APB<\/p>\n\n\n\n

so, \u2220AOB = 2 \u00d7 70\u00b0 = 140\u00b0<\/p>\n\n\n\n

Since AOBC is a cyclic quadrilateral, we have<\/p>\n\n\n\n

\u2220ACB + \u2220AOB = 180\u00b0<\/p>\n\n\n\n

\u2220ACB + 140\u00b0 = 180\u00b0<\/p>\n\n\n\n

\u2220ACB = 40\u00b0<\/p>\n\n\n\n

Question 2: In figure, two congruent circles with centres O and O\u2019 intersect at A and B. If \u2220AO\u2019B = 50\u00b0, then find \u2220APB.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

As we are given that, both the triangle are congruent which means their corresponding angles are equal.<\/p>\n\n\n\n

Therefore, \u2220AOB = AO\u2019B = 50\u00b0<\/p>\n\n\n\n

Now, by degree measure theorem, we have<\/p>\n\n\n\n

\u2220APB = \u2220AOB\/2 = 250<\/sup><\/p>\n\n\n\n

Question 3: In figure, ABCD is a cyclic quadrilateral in which \u2220BAD=75\u00b0, \u2220ABD=58\u00b0 and \u2220ADC=77\u00b0, AC and BD intersect at P. Then, find \u2220DPC.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

\u2220DBA = \u2220DCA = 580<\/sup> \u2026(1)[Angles in same segment]<\/p>\n\n\n\n

ABCD is a cyclic quadrilateral :<\/p>\n\n\n\n

Sum of opposite angles = 180 degrees<\/p>\n\n\n\n

\u2220A +\u2220C = 1800<\/sup><\/p>\n\n\n\n

750<\/sup> + \u2220C = 1800<\/sup><\/p>\n\n\n\n

\u2220C = 1050<\/sup><\/p>\n\n\n\n

Again, \u2220ACB + \u2220ACD = 1050<\/sup><\/p>\n\n\n\n

\u2220ACB + 580<\/sup> = 1050<\/sup><\/p>\n\n\n\n

or \u2220ACB = 470<\/sup> \u2026(2)<\/p>\n\n\n\n

Now, \u2220ACB = \u2220ADB = 470<\/sup>[Angles in same segment]<\/p>\n\n\n\n

Also, \u2220D = 770<\/sup> (Given)<\/p>\n\n\n\n

Again From figure, \u2220BDC + \u2220ADB = 770<\/sup><\/p>\n\n\n\n

\u2220BDC + 470<\/sup> = 770<\/sup><\/p>\n\n\n\n

\u2220BDC = 300<\/sup><\/p>\n\n\n\n

In triangle DPC<\/p>\n\n\n\n

\u2220PDC + \u2220DCP + \u2220DPC = 1800<\/sup><\/p>\n\n\n\n

300<\/sup> + 580<\/sup> + \u2220DPC = 1800<\/sup><\/p>\n\n\n\n

or \u2220DPC = 920 <\/sup><\/p>\n\n\n\n

Question 4: In figure, if \u2220AOB = 80\u00b0 and \u2220ABC=30\u00b0, then find \u2220CAO.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: \u2220AOB = 800<\/sup> and \u2220ABC = 300<\/sup><\/p>\n\n\n\n

To find: \u2220CAO<\/p>\n\n\n\n

Join OC.<\/p>\n\n\n\n

Central angle subtended by arc AC = \u2220COA<\/p>\n\n\n\n

then \u2220COA = 2 x \u2220ABC = 2 x 300<\/sup> = 600<\/sup> \u2026(1)<\/p>\n\n\n\n

In triangle OCA,<\/p>\n\n\n\n

OC = OA[same radii]<\/p>\n\n\n\n

\u2220OCA = \u2220CAO \u2026(2)[Angle opposite to equal sides]<\/p>\n\n\n\n

In triangle COA,<\/p>\n\n\n\n

\u2220OCA + \u2220CAO + \u2220COA = 1800<\/sup><\/p>\n\n\n\n

From (1) and (2), we get<\/p>\n\n\n\n

2\u2220CAO + 600 <\/sup>= 1800<\/sup><\/p>\n\n\n\n

\u2220CAO = 600<\/sup><\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 16: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 16\u2013Circles<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 16\u2013Circles PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n