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Posted by Hansika Rajput 1Â week ago

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Mahi Sharma 4Â days ago

On simple form
It is a data which is used to represent coordinates or one or more than one number to determine the position of the points

Mahi Sharma 4Â days ago

In geometry, a coordinate system is a system that uses one or more numbers, or coordinates, to uniquely determine the position of the points or other geometric elements on a manifold such as Euclidean space.

Posted by Lakshita Bora 1Â week, 1Â day ago

- 2 answers

Rahul Gurjar 1Â week ago

25+12+17Ã·2=27
=âˆš27(27-25)(27-17)(27-12)
=âˆš27(2)(10)(15)
=âˆš9Ã—3Ã—2Ã—5Ã—2Ã—5Ã—3
=3Ã—3Ã—2Ã—5 =90
Altitude to the longest side
90=Â½Ã—12
90Ã—2Ã·12=h
180Ã·12=h
15=h

Posted by Diya Sahu 1Â week ago

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Posted by R K 1Â week, 1Â day ago

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Maneesh R.K 1Â week ago

2.111
2.112
2.113
2.114
2.115
2.116
2.117
2.118
2.119
2.120
2.121
2.122
2.123
2.124
2.125
2.126
2.150

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Himanshi Goyal 2Â days, 22Â hours ago

Chapter 1
Chapter 3
Chapter 6
Chapter 7
Chapter 12
Chapter 14

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Posted by Prathibha. D Prathibha. D 1Â week, 3Â days ago

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Posted by Vrinda Sharma 1Â week, 3Â days ago

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Sujeet Roy 1Â week, 3Â days ago

1 , 3 , 4 , 6 , 7 , 12 , 14 syllabus he aur 1 , 6 , 12 , 3 padhke jha

Posted by Sohani Khatun 1Â week, 3Â days ago

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Vanshika Agarwal 1Â week, 3Â days ago

Either the question will be
ABCD is a rhombus and p, q ,r and s are the mid-points of the sides AB,BC, CD and DA respectively . Show that the quadrilateral PQRS is a rectangle.
Either it will be
ABCD is a rectangle and p, q ,r and s are the mid-points of the sides AB,BC, CD and DA respectively . Show that the quadrilateral PQRS is a rhombus.

Posted by Shivang Shakya 1Â week, 3Â days ago

- 4 answers

Mahi Sharma 4Â days ago

In figure 6.31 if AB is parallel to CD,Angle APQ = 50Â° and angle PRD=127Â°
Find x and y

Shreyansh Singh 1Â week ago

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Posted by Mukes Akash 1Â week, 4Â days ago

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Posted by Mariyam Shahid 1Â week, 4Â days ago

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Vandana Yashu 1Â week, 2Â days ago

Given:AC=BD----1
AB bisects angle A
Therefore angle BAC =angle BAD----2
In âˆ†ABC and âˆ†ABD
AC=AB(by equation 1)
Angle BAC=angle BAD (by equation 2)
AB=AB (common)
Therefore by SAS axiom
âˆ†ABC~âˆ†ABD
By CPCT
BC=BD

Vandana Yashu 1Â week, 2Â days ago

AC =BD and AB bisects angle A .Show that âˆ† ACB ~âˆ†ABD ,what can you say about BC and BD?

Posted by Neil Ahuja 1Â week, 4Â days ago

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Posted by Sahil Kumar 1Â week, 5Â days ago

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Utkarsh Agarwal 1Â week, 5Â days ago

xÂ² = 2 then x Â³ =2âˆš2
EXPLANATION : Let the no.'x' be âˆš2
Now when you put the value of X in the first eq. you will get âˆš2 *2 =2
Now put this value of X in the second eq. you will get âˆš2*âˆš2*âˆš2 = 2âˆš2

Posted by Likith Sai 1Â week, 6Â days ago

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Lucky Singh 1Â week, 6Â days ago

- Don't post personal information, mobile numbers and other details.
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Posted by Anil Jaithalia 1Â week, 6Â days ago

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Posted by Khushi Dhillon 1Â week, 6Â days ago

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Sia ðŸ¤– 1Â week, 6Â days ago

Given: A square ABCD.

To Prove : (i) AC = BD and (ii) Diagonals bisect each other at right angles.

Proof :

- In {tex}\triangle{/tex}ADB and {tex}\triangle{/tex}BCA, we have

AD = BC ...[As sides of a square are equal]

{tex}\angle{/tex}BAD = {tex}\angle{/tex}ABC ...[All interior angles are of 90^{o}]

AB = BA ...[Common]

{tex}\triangle{/tex}ADB {tex}\cong{/tex} {tex}\triangle{/tex}BCA ...[By SAS rule]

AC = BD ...[c.p.c.t.] - Now in {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}COD, we have

AB = CD ...[Sides of a square]

{tex}\angle{/tex}AOB = {tex}\angle{/tex}COD ...[Vertically opp. angles]

{tex}\angle{/tex}OBA = {tex}\angle{/tex}ODC ...[Alternate interior angles are equal]

{tex}\triangle{/tex}AOB {tex}\cong{/tex} {tex}\triangle{/tex}COD ...[By ASA rule]

OA = OC and OB = OD ...[c.p.c.t.] ...(1)

Now consider {tex}\triangle{/tex}s AOD and COD.

AD = CD ...[Sides of square]

OA = OC ...[As proved above]

OD = OD ...[Common]

{tex}\triangle{/tex}AOD {tex}\cong{/tex} {tex}\triangle{/tex}COD ...[By SSS rule]

{tex}\angle{/tex}AOD = {tex}\angle{/tex}COD ...[c.p.c.t.]

But {tex}\angle{/tex}AOD + {tex}\angle{/tex}COD = 180° ...[linear pair]

or {tex}\angle{/tex}AOD + {tex}\angle{/tex}AOD = 180° ...[As {tex}\angle{/tex}AOD = {tex}\angle{/tex}COD]

or 2{tex}\angle{/tex}AOD = 180° {tex}\therefore{/tex} {tex}\angle{/tex}AOD = 90° ...(2)

From equation (1) and (2) it is clear that diagonals of a square bisect each other at right angles.

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Maneesh R.K 1Â week ago

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