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Abhishek Singh 5 years ago

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  • 0 answers
  • 1 answers

Niharika Sharma 5 years ago

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.
  • 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.
  • 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.
  • 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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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.
  • 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

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
  • 2 answers

Utkarsh Goyal 5 years ago

Xxx

Abhishek Singh 5 years ago

Sexual contact
  • 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°
  • 4 answers

Abhishek Singh 5 years ago

(a+b)(a-b)

K S 5 years ago

(a+b)(a-b)

Naman Kumar 5 years ago

identity a2 – b2 = (a + b)(a – b)

Bhavya Kumari 5 years ago

(a+b)(a-b)
  • 0 answers
  • 3 answers

Abhishek Singh 5 years ago

22.22m/s

Aditi Patil 5 years ago

22.22 m/s

Pranam Kala 5 years ago

Multiple by 1000/3600
  • 3 answers

Bhavya Kumari 5 years ago

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

Kripa Daga 5 years ago

10 cm

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
  • 2 answers

Bhavya Kumari 5 years ago

The centre of circle inscribed in triangle
the centre of the incircle of a triangle or other figure.
  • 2 answers

Rimjhim Goswami 5 years ago

(a-b)(a+b)(a-b)
A-b
  • 1 answers

Kripa Daga 5 years ago

16*3/4 •4*3 •12
  • 1 answers

Ankush Ranjan 5 years ago

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

Adarsh Patil 5 years ago

We must take equation as 51+c+18+c+c=180 (By seeing the question) Then we get answer as C=37 B=55 C=88
124 is the third angle

Kripa Daga 5 years ago

a=88, b=55, c=37

Priyanshu Pishu 5 years ago

51 is a answer
  • 3 answers

Sumit Bharti 5 years ago

2x+3yka gunnakhand
Root 2+ Root15

Kripa Daga 5 years ago

(√2+√5)/√3 (√2+√5)√3/√3×√3 √6+√15/3
  • 2 answers

Adarsh Patil 5 years ago

50

Kripa Daga 5 years ago

2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,90,92,94,96,98,100

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