CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points? Ans. From the figure, we observe that when different pairs of circles are drawn, each pair have two common points (say A and B). Maximum number of common points are two in each pair of circles. Suppose two circles C (O, ) and C (O’, ) intersect each other in three points, say A, B and C. Then A, B and C are non-collinear points. We know that: There is one and only one circle passing through three non-collinear points. Therefore, a unique circle passes through A, B and C. O’ coincides with O and A contradiction to the fact that C (O’, ) C (O, ) Our supposition is wrong. Hence two different circles cannot intersect each other at more than two points. 2. Suppose you are given a circle. Give a construction to find its centre. Ans. Steps of construction: (a) Take any three points A, B and C on the circle. (b) Join AB and BC. (c) Draw perpendicular bisector say LM of AB. (d) Draw perpendicular bisector PQ of BC. (e) Let LM and PQ intersect at the point O. Then O is the centre of the circle. Verification: O lies on the perpendicular bisector of AB. OA = OB ……….(i) O lies on the perpendicular bisector of BC. OB = OC ……….(ii) From eq. (i) and (ii), we observe that OA = OB = OC = (say) Three non-collinear points A, B and C are at equal distance from the point O inside the circle. Hence O is the centre of the circle. 3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. Ans. Given: Let C (O, r) and C (O’, r') be two circles intersecting at A and B. AB is the common chord. To prove: OO’ is the perpendicular bisector of the chord AB, i.e., AM = MB and OMA = OMB = Construction: Join OA, OB, O’A, O’B. Proof: In triangles OAO’ and OBO’, OA = OB [Each radius] O’A = O’B [Each radius] OO’ = OO’ [Common] OAO’ OBO’ [By SSS congruency] AOO’ = BOO’ [By CPCT] AOM = BOM (i) Now in AOB, OA = OB And AOB = OBA [Proved earlier] Also AOM = BOM [From eq.(i)] Remaining AMO = BMO Since AMO + BMO = [Linear pair] 2AMO = AMO = BMO = OM AB OO’ AB Since OM AB M is the mid-point of AB, i.e., AM = BM Hence OO’ is the perpendicular bisector of AB.

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If two equal chords of a circle intersect within the circle prove that thesegments of one chord are equal to corresponding segments of the other chord

Posted by Ashish Kumar 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Alphonsa Jimmy 5 years ago

Drop a perpendicular from O to both chords AB and CD In △OMP and △ONP As chords are equal, perpendicular from centre would also be equal. OM=ON OP is common. ∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence) PM=PN ......................(1) AM=BM (Perpendicular from centre bisects the chord) Similarly ,CN=DN As AB=CD AB−AM=CD−DN BM=CN .........................(2) From (1) and (2) BM−PM=CN−PN PB=PC AM=DN (Half the length of equal chords are equal) AM+PM=DN+PN AP=PD Therefore , PB=PC and AP=PD is proved.

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An angle is 20 degree than three times the given angle. if the two angles are supplementary,then the angle are

Posted by Chetan Sangwan 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Aryan Raj 5 years ago

Let the given angle be x ..then that required angle 3x+20 Now x+3x+20 =180 4x=180-20=160 x=160/4 = 40 So angles are 40 degree and 140 degree respectively.

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A circular path of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distances on its boundary each having a toy telephone in his hand to talk to each other. Find the length of the string of each phone.

Posted by Abhishek Singh 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Alphonsa Jimmy 5 years ago

Let Ankur be represented as A, Syed as S and David as D. The boys are sitting at an equal distance. Hence, △ASD is an equilateral triangle. Let the radius of the circular park be r meters. ∴OS=r=20m. Let the length of each side of △ASD be x meters. Draw AB⊥SD ∴SB=BD= 2 1 SD= 2 x m In △ABS,∠B=90 o By Pythagoras theorem, AS 2 =AB 2 +BS 2 ∴AB 2 =AS 2 −BS 2 =x 2 −( 2 x ) 2 = 4 3x 2 ∴AB= 2 3 x m Now, AB=AO+OB OB=AB−AO OB=( 2 3 x −20) m In △OBS, OS 2 =OB 2 +SB 2 20 2 =( 2 3 x −20) 2 +( 2 x ) 2 400= 4 3 x 2 +400−2(20)( 2 3 x )+ 4 x 2 0=x 2 −20 3 x ∴x=20 3 m The length of the string of each phone is 20 3 m.

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Calculate the value of x

Posted by Abhay Agarwal 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Khushi Kaur 5 years ago

Find the value of X? Steps involved in finding the value of x calculator is as follows: Method 1: The required input value must be entered in the divisor and the product field. Method 2: Click the ‘SOLVE’ option to obtain the output. Method 3: The output field will present the x value or the dividend.

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What is formula of square root

Posted by Emilal Nayak Emilal Nayak 5 years ago

CBSE > Class 09 > Mathematics

2 answers

Khushi Kaur 5 years ago

The square root of the number y whose square is x. The square root is denoted by √ We find the square root of a number by the following methods: i) By Prime Factorisation ii) By Long Division iii) By Repeated subtraction method

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Abhishek Singh 5 years ago

There is no formulae. There are various methods to find square root. One of the best methods is "long division method". It is used to find perfect square as well as imperfect square

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Simplify (2√3/√3-1)-(7/√12-√5)+(2+√5/4+3√5)

Posted by World Wide Handsome💜 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Patel Het 5 years ago

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Means of ar in

Posted by Sandeep Mahanta 5 years ago

CBSE > Class 09 > Mathematics

2 answers

Utkarsh Goyal 5 years ago

Xxx

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Abhishek Singh 5 years ago

Sexual contact

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ABCD is a quadrilateral in which angleABC= 73° angleC= 97° and angleD=110°. If AE||DC,BE||AD and AE intersect BD at F, find angle BEF

Posted by Bhavay Goel 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Khushi Kaur 5 years ago

Given AE is parallel to DC and BE is parallel to AD, ∠ ABC = 73°, C = 97° To find : ∠ EBF Since AE is parallel to DC ∴ 110° + ∠ DAE = 180° (Co-interior ∠'s) Also, ∠ BEA = ∠ DCF = 97° (Corresponding B ∠'s) ⇒ ∠ BEA = 97° Now in Δ ABF, ∠ FAB + ∠ ABF + ∠ AFB = 180° (Angle Sum Property) ⇒ ∠ FAB + 73° + 97° = 180° ⇒ ∠ FAB = 180°-170° ⇒ ∠ FAB = 10° Now, AD is parallel to BE (Given) ∴ ∠ DAB + ∠ ABE = 180° ( Co-interior angles) ⇒ ∠ DAF + ∠ FAB + ∠ ABF + ∠ EBF = 180° ⇒ 70° + 10° + 73° + ∠ EBF = 180° ∠ EBF = 180°- 153° ∠ EBF = 27°

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What is the fourmula of a2-b2

Posted by Tanmay Saini 5 years ago

CBSE > Class 09 > Mathematics

4 answers

Abhishek Singh 5 years ago

(a+b)(a-b)

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K S 5 years ago

(a+b)(a-b)

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Naman Kumar 5 years ago

identity a2 – b2 = (a + b)(a – b)

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Bhavya Kumari 5 years ago

(a+b)(a-b)

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Let's have ad dance

Posted by Pranam Kala 5 years ago

CBSE > Class 09 > Mathematics

0 answers

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80/hour convert metre per second

Posted by Priya . 5 years ago

CBSE > Class 09 > Mathematics

3 answers

Abhishek Singh 5 years ago

22.22m/s

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Aditi Patil 5 years ago

22.22 m/s

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Pranam Kala 5 years ago

Multiple by 1000/3600

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In ∆ ABC, D and E are the midpoints of the sides AB and AC respectively. If DE is 5 cm, find the measure of the side BC.

Posted by Nihal Shankar 5 years ago

CBSE > Class 09 > Mathematics

3 answers

Bhavya Kumari 5 years ago

BC=10cm because of mid point theorem in which DC is half and parallel to BC

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Kripa Daga 5 years ago

10 cm

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Himanshu Gupta 5 years ago

According to the midpoint theorem DE =1\2 of BC 5 = 1\2 of BC BC = 5 * 2 Hence, BC = 10

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What is incentre

Posted by Ankush Ranjan 5 years ago

CBSE > Class 09 > Mathematics

2 answers

Bhavya Kumari 5 years ago

The centre of circle inscribed in triangle

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Gurtej Singh Bedi 2959 5 years ago

the centre of the incircle of a triangle or other figure.

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What is the percentage increase in dispensaries from 1951 to 2015 Dispensaries |29274|29709|29957

Posted by Sherap Yt 5 years ago

CBSE > Class 09 > Mathematics

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a-b-a^2+b^2

Posted by Neha Yadav 5 years ago

CBSE > Class 09 > Mathematics

2 answers

Rimjhim Goswami 5 years ago

(a-b)(a+b)(a-b)

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Priyanka Nautiyal 5 years ago

A-b

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Write the zero of polynomial p(x)=3x^2-1

Posted by Anshu Kumari 5 years ago

CBSE > Class 09 > Mathematics

5 answers

Adarsh Patil 5 years ago

Root -1/3

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T.Nikhil Chowdary Nikhil Chowdary 5 years ago

NIKHIL CHOWDARY is crrt

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T.Nikhil Chowdary Nikhil Chowdary 5 years ago

-1

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Account Deleted 5 years ago

Ghj

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Sasjasee Sasjasee 5 years ago

7777

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Puncuate prospero eas a duck of millan in the kingdom of nepal

Posted by Sujal Mane 5 years ago

CBSE > Class 09 > Mathematics

0 answers

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Evaluate 16 3/4

Posted by Mayur Sonkar 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Kripa Daga 5 years ago

16*3/4 •4*3 •12

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The coordinates of two points are A(4,5) and B(-3,6), then find (abscissa of A) - (abscissa of B)

Posted by Zainab Ahmed 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Ankush Ranjan 5 years ago

The abscissa of A is 4 and abscissa of B is -3 Thus, 4-(-3) = 7

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Express 0.00323232------ in the form of p/q

Posted by Anshul Kaushik 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Adarsh Patil 5 years ago

8/2475

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In the figure O is the centre of the circle and angle BCO=30.Find x and y

Posted by Aditya Raj 5 years ago

CBSE > Class 09 > Mathematics

1 answers

Naman Kumar 5 years ago

Given−BA, BD & BC are the chords of the circle with centre O.AE⊥BC & OD⊥AE.∠OCE=30o.To find out−∠BAE=x=? And ∠DBC=y=? Solution−We join AD & DC.∠ABD=21∠AOD=21×90o=45o.(∵ The angle at the centre of a circle subtended by a chordis double of that at the circumference.).∴ In ΔOEC we have ∠COE=180o−(∠OEC+∠OCE)=180o−(90o+30o)=60o. (angle sum property of triangles)∴ ∠DOC=∠DOE−∠COE=90o−60o=30o.Now ∠DBC=y=21∠DOC=21×30o=15o.(∵ The angle at the centre of a circle subtended by a chordis double of that at the circumference.).Again ∠ABE=y+45o=45o+15o=60o.∴ In ΔABE we have ∠BAE=x=180o−(∠ABE+∠AEB)=180o−(60o+90o)=30o. (angle sum property of triangles)So x=30o & y=15o

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In triangle abc angle a-b=33 and angle b-c=18 find the angles of the triangles

Posted by Lav Kumar Lav Kumar 5 years ago

CBSE > Class 09 > Mathematics