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  • 2 answers

Maneesh R.K 3 years ago

Ex 5/7

Maneesh R.K 3 years ago

Give the sum
  • 2 answers

Mahi Sharma 2 years, 11 months 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 2 years, 11 months 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.
  • 2 answers

Rahul Gurjar 3 years ago

25+12+17÷2=27 =√27(27-25)(27-17)(27-12) =√27(2)(10)(15) =√9×3×2×5×2×5×3 =3×3×2×5 =90 Altitude to the longest side 90=½×12 90×2÷12=h 180÷12=h 15=h

Md Abdal 3 years ago

Longest altitude always forms from the shortest side
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Maneesh R.K 3 years 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

Sachin Negi 3 years ago

2.1/1× 17/17 and 2.2/1×17/17
  • 2 answers

Harjot Singh 3 years ago

The data we collect ourself

Sachin Negi 3 years ago

Primary data also called internal data
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Himanshi Goyal 2 years, 11 months ago

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

Kanchan Gadwal 3 years ago

Chapter 1 , 3, 4,6,7,12,14
  • 3 answers

Himanshi Goyal 2 years, 11 months ago

12.5
12.5

Anak Yadav 3 years ago

Answer is 12.5
  • 3 answers

Ayush Sahu 3 years ago

sare padh l9

Huda Khan 3 years ago

7 bhi bht important h

Sujeet Roy 3 years ago

1 , 3 , 4 , 6 , 7 , 12 , 14 syllabus he aur 1 , 6 , 12 , 3 padhke jha
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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.
Ur question is wrong
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Mahi Sharma 2 years, 11 months ago

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

Sia ? 3 years ago

Please write full question. 

  • 2 answers
Plz complete your question

Sia ? 3 years ago

Complete your question.

  • 4 answers

Vandana Yashu 3 years ago

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

Vandana Yashu 3 years ago

AC =BD and AB bisects angle A .Show that ∆ ACB ~∆ABD ,what can you say about BC and BD?
Reason- How a triangle can be ABCD?
Ur question is wrong
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Shubham Kumar 3 years ago

2√2
Answer-3

Utkarsh Agarwal 3 years ago

x² = 2 then x ³ =2√2 EXPLANATION : Let the no.'x' be √2 Now when you put the value of X in the first eq. you will get √2 *2 =2 Now put this value of X in the second eq. you will get √2*√2*√2 = 2√2

Bhagya Singla 3 years ago

2x=2 x=2/2 x=1 Then, 3x Put (x=1) 3(1)= 3
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Sia ? 3 years ago

Given: A square ABCD.

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

  1. 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 90o]
    AB = BA ...[Common]
    {tex}\triangle{/tex}ADB {tex}\cong{/tex} {tex}\triangle{/tex}BCA ...[By SAS rule]
    AC = BD ...[c.p.c.t.]
  2. 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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