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Ask QuestionPosted by Vansh Kakkar 3 years, 11 months ago
- 2 answers
Yogita Ingle 3 years, 11 months ago
Say an angle is 8x then the complement angle is x then we know two complementary angles sum up to 90∘.
So,
8x+x=90∘
9x=90∘
x=10∘
Hence, the angle is 80∘.
Posted by Aiswarya Raanee .K 3 years, 11 months ago
- 3 answers
Posted by Radheshyam Thakur 3 years, 11 months ago
- 1 answers
Yogita Ingle 3 years, 11 months 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=90o
△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.
Posted by Gyan Prakash 3 years, 11 months ago
- 1 answers
Ayush Kumar 3 years, 11 months ago
Posted by Reshma . B 3 years, 11 months ago
- 1 answers
Ayush Kumar 3 years, 11 months ago
Posted by Biswajeet Lenka 3 years, 11 months ago
- 0 answers
Posted by Dhinchaik Pooja 3 years, 11 months ago
- 0 answers
Posted by Arpita Keshari 3 years, 11 months ago
- 0 answers
Posted by Devansh Singh 3 years, 11 months ago
- 5 answers
Posted by Harshavardhana M 3 years, 11 months ago
- 2 answers
Sujita N B 3 years, 11 months ago
Posted by Gyan Prakash 3 years, 11 months ago
- 1 answers
Posted by Anushka Ghatera 3 years, 11 months ago
- 2 answers
Yogita Ingle 3 years, 11 months ago
4/9 = ×\27
9x = (4 × 27)
x = (4 × 27)/9
x = (4 × 3)
x = 12
Posted by Sukhman Ghuman 3 years, 11 months ago
- 5 answers
Posted by C K Manoj 3 years, 11 months ago
- 1 answers
Yogita Ingle 3 years, 11 months ago
Point intersects x-axis, so y = 0
3x +2y=12
3x + 2(0) = 12
3x = 12
x = 12/3 = 4
Posted by V. H 3 years, 11 months ago
- 0 answers
Posted by Nainshi Pandey 3 years, 11 months ago
- 0 answers
Posted by Janani Surya 3 years, 11 months ago
- 1 answers
Gaurav Seth 3 years, 10 months ago
In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ∠DBC = 60° and ∠BAC = 40°. Find
(i) ∠BCD,
(ii) ∠CAD.
(i) We know that the angles in the same segment are equal
So we get
∠BDC = ∠BAC = 40o
Consider △ BCD
Using the angle sum property
∠BCD + ∠BDC + ∠DBC = 180o
By substituting the values
∠BCD + 40o + 60o = 180o
On further calculation
∠BCD = 180o – 40o – 60o
By subtraction
∠BCD = 180o – 100o
So we get
∠BCD = 80o
(ii) We know that the angles in the same segment are equal
So we get
∠CAD = ∠CBD
So ∠CAD = 60o
Posted by Neela Chandra Prasad Gunde 3 years, 11 months ago
- 1 answers
Yogita Ingle 3 years, 11 months ago
y=2x−4
⇒y=2x+4
When x=0,y=4
When y=0,x=−2
When x=1,y=6
| x | 0 | −2 | 1 |
| y=2x+4 | 4 | 0 | 6 |
Plot the points A(0,4),B(−2,0) and C(1,6) on the graph paper.
Join these points by a straight line BC
Posted by Trushti Parmar 3 years, 11 months ago
- 2 answers
Yogita Ingle 3 years, 11 months ago
0.75=75/100
1.2=12/10
First we have to make the denominators equal.
LCM of 10 and 100 is 100.
The two rational numbers are 120/100 and 75/100. The rational numbers between them can be 76/100, 77/100, 78/100, 79/100, etc. until 120/100.
Therefore, there are many rational numbers between them.
Posted by Ruchil Shah 3 years, 11 months ago
- 0 answers
Posted by Aleena Siju 3 years, 11 months ago
- 1 answers
Posted by Poonam Singh 3 years, 11 months ago
- 1 answers
Gaurav Seth 3 years, 10 months ago
STEPS: 1: On a numberline mark AB = 9.3 unit & BC= 1 unit.
2: Mark O the mid point of AC
3: Draw a semicircle with O as centre & OA as radius
4: At B draw a perpendicular BD.
5: BD = √9.3 unit
6: Now, B as centre, BD as radius, draw an arc, meeting the numberline at E.
Now, with BD or BE = √9.3 as radius , with 0 ( origin) of the number line as centre, draw an arc on the the number line, intersecting at point ‘k’. And this point ‘k' lies between integers 3 & 4, & represents √9.3
JUSTIFICATION:
BD = √ {(10.3/2)² - (8.3/2)²}
=> BD = √{10.3²- 8.3²)/4 }
=> BD = √{(10.3+8.3)(10.3–8.3)/4}
=> BD = √{18.6*2/4}
=> BF = √{37.2/4}
=> BD = √9.3 = BE
Posted by Abhay Chaubay 3 years, 11 months ago
- 1 answers
Darsh Jain 3 years, 11 months ago
Posted by Dhruvil Bardoliya 3 years, 11 months ago
- 1 answers
Posted by Pallavi Rajpurohit 3 years, 11 months ago
- 1 answers
Yogita Ingle 3 years, 11 months ago
Please refer to video for the figure of different pair of circles. There can be three possibilities.
(i) When two circles are intersecting each other, then there are two common points. So, the maximum number of common points in this case, is 2.
(ii) When two circles are just touching each other, then there is a single common point. So, the maximum number of common points in this case, is 1.
(iii) When two circles are not touching each other, then there is no common point. So, the maximum number of common points in this case, is 0.
Posted by Yashas Kumar 3 years, 11 months ago
- 1 answers
Gaurav Seth 3 years, 10 months ago
a² + b² + c² = ab + bc + ca
On multiplying both sides by “2”, it becomes
2 ( a² + b² + c² ) = 2 ( ab + bc + ca)
2a² + 2b² + 2c² = 2ab + 2bc + 2ca
a² + a² + b² + b² + c² + c² – 2ab – 2bc – 2ca = 0
a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0
(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca) = 0
(a – b)² + (b – c)² + (c – a)² = 0
=> Since the sum of square is zero then each term should be zero
⇒ (a –b)² = 0, (b – c)² = 0, (c – a)² = 0
⇒ (a –b) = 0, (b – c) = 0, (c – a) = 0
⇒ a = b, b = c, c = a
∴ a = b = c.
</article>Posted by Adeel Raza 3 years, 11 months ago
- 1 answers
Gaurav Seth 3 years, 10 months ago
1. In which quadrant or on which axis do each of the points (– 2, 4), (3, – 1), (– 1, 0), (1, 2) and (– 3, – 5) lie? Verify your answer by locating them on the Cartesian plane.
Solution:
- (– 2, 4): Second Quadrant (II-Quadrant)
- (3, – 1): Fourth Quadrant (IV-Quadrant)
- (– 1, 0): Negative x-axis
- (1, 2): First Quadrant (I-Quadrant)
- (– 3, – 5): Third Quadrant (III-Quadrant)
Posted by Rashik Tech 12 3 years, 11 months ago
- 2 answers
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Ashish Kumar 3 years, 11 months ago
1Thank You