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Pooja Arya 2 years, 11 months ago

Class 9 8.2 is about mid point theorem It says .. In a triangle, the line segment that joins the mid point of any two sides is parallel to the third side also half of it
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Preeti Dabral 2 years, 11 months ago

Given: Let ABCD is a quadrilateral.
Let its diagonal AC and BD bisect each other at right angle at point O.

{tex}\therefore {/tex}OA = OC, OB = OD
And {tex}\angle {/tex}AOB = {tex}\angle {/tex}BOC = {tex}\angle {/tex}COD = {tex}\angle {/tex}AOD = {tex}{90^ \circ }{/tex}
To prove: ABCD is a rhombus.
Proof: In {tex}\triangle{/tex}AOD and {tex}\triangle{/tex} BOC,
OA = OC [Given]
{tex}\angle {/tex}AOD = {tex}\angle {/tex}BOC [Given]
OB = OD [Given]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOD {tex} \cong{/tex} {tex}\triangle{/tex}COB [By SAS congruency]
{tex} \Rightarrow {/tex} AD = CB [By C.P.C.T.]……….(i)
Again, In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}COD,
OA = OC [Given]
{tex}\angle {/tex} AOB = {tex}\angle {/tex}COD [Given]
OB = OD [Given]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOB  {tex} \cong {/tex} {tex}\triangle{/tex}COD [By SAS congruency]
{tex} \Rightarrow {/tex} AB = CD [By C.P.C.T.]……….(ii)
Now In {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}BOC,
OA = OC [Given]
{tex}\angle {/tex} AOB = {tex}\angle {/tex}BOC [Given]
OB = OB [Common]
{tex}\therefore {/tex} {tex}\triangle{/tex}AOB {tex} \cong{/tex} {tex}\triangle{/tex}COB [By SAS congruency]
{tex} \Rightarrow {/tex} AB = BC [By C.P.C.T.]……….(iii)
From eq. (i), (ii) and (iii),
AD = BC = CD = AB
And the diagonals of quadrilateral ABCD bisect each other at right angle.
Therefore, ABCD is a rhombus.

  • 1 answers

Sehajpreet Kaur 3 years ago

In Δ ABC, P and Q are the mid-points of sides AB and BC respectively. ∴ PQ || AC and PQ = 12 AC ( Using mid-point theorem) …..(1) In ΔADC, R and S are the mid - points of CD and AD respectively. ∴ RS || AC and RS = 12 AC ( Using mid-point theorem) …..(2) From equation (1) and (2) , we obtain PQ || RS and PQ = RS Since in quadrilateral PQRs, one pair of opposite sides is equal and parallel to each other , it is a parallelogram. Let the diagonals of rhombus ABCD intersect each other at point O. In Quadrilateral OMQN, MQ || ON (∵ PQ || AC) QN || OM (∵ QR || BD) Therefore, OMQN is a parallelogram. ∴∠MQN=∠NOM and ∠PQR=∠NOM However, ∠NOM=90∘ (Diagonals of a rhombus are perpendicular to each other) ∴∠PQR=90∘ Clearly , PQRS is a parallelogram having one of its interior angles as 90∘ Hence, PQRS is a rectangle.
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Debayan Kar 3 years ago

In history ch5

Ronak Verma 3 years ago

Scientific forest
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Tanveer Kaur 3 years ago

Bedouins .berbers. massai .Somali. turkana.

Debayan Kar 3 years ago

Maasai
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Debayan Kar 3 years ago

50-√32×√50 =50-4√2×5√2 =50-20×√2×√2 =50-20×2 =50-40 =10
10
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Aditya Maurya 2 years, 11 months ago

Ese bakwas karke
Padhai karke

Debayan Kar 3 years ago

Practice any CBSE sample paper
padhke
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Sehajpreet Kaur 3 years ago

In Δ ABC, P and Q are the mid-points of sides AB and BC respectively. ∴ PQ || AC and PQ = 1 2 AC ( Using mid-point theorem) …..(1) In Δ ADC, R and S are the mid - points of CD and AD respectively. ∴ RS || AC and RS = 1 2 AC ( Using mid-point theorem) …..(2) From equation (1) and (2) , we obtain PQ || RS and PQ = RS Since in quadrilateral PQRs, one pair of opposite sides is equal and parallel to each other , it is a parallelogram. Let the diagonals of rhombus ABCD intersect each other at point O. In Quadrilateral OMQN, MQ || ON ( ∵ PQ || AC) QN || OM ( ∵ QR || BD) Therefore, OMQN is a parallelogram. ∴ ∠ M Q N = ∠ N O M and ∠ P Q R = ∠ N O M However, ∠ N O M = 90 ∘ (Diagonals of a rhombus are perpendicular to each other) ∴ ∠ P Q R = 90 ∘ Clearly , PQRS is a parallelogram having one of its interior angles as 90 ∘ Hence, PQRS is a rectangle.
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Akshadeep Meshram 2 years, 11 months ago

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Mohak Yadav 3 years ago

Question is wrong
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Shivang Sharma 2 years, 11 months ago

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Lavesh Sahu 3 years ago

17/35
  • 3 answers

Debayan Kar 3 years ago

Stop akshadeep

Debayan Kar 3 years ago

In rural areas Farm laborers who could not buy land and necessary items which are daily needs for livelihood due to less income as more indebtedness

Akshadeep Meshram 2 years, 11 months ago

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  • 2 answers

Debayan Kar 3 years ago

C.deputy minsters

Sumeet Singh 2 years, 11 months ago

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  • 3 answers

Smriti Gupta 2 years, 11 months ago

35.5 u

Srishti Singh 3 years ago

For calculating this firstly u have to see it's mass number and percentage Formula- (Percentage×mass number/100) Calculation:- (75% × 35/100) + (25% × 37/100) =} (2625÷100)+(925÷100) =} (3550÷100) = 35.5u Answer :-. 35.5u *THANK YOU* 😊
Exam hone wala hai kya tera?
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Deepika Saini 3 years ago

Title bar just locate at the top of writer window. It contain the name of currently document which is opened

Priyanshu Kumar 3 years ago

what do you mean by keyboarding skills
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Kaif Alam 2 years, 5 months ago

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Vinay Yadav 3 years ago

Social science का sample paper class 9 का Answer
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Just go to this link
https://drive.google.com/file/d/1tQ0igeXX1Qscpz6uhyHXxDkJ6ZdThN7O/view?usp=drivesdk
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Deepika Saini 3 years ago

Impress working area known as slide
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B. Wamsi 3 years ago

Answer Kya ha
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Debayan Kar 3 years ago

1. Material required are ruler , notebook, pencil with compass and eraser

Abhishek Ranjan 3 years ago

Plz give m answer
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Shivam Sharma 3 years ago

We get food from plants and animals

Gauri Gauri 3 years ago

We get food from plants and animals
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Shivam Sankpal 2 years, 11 months ago

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Kashvi Jain 3 years ago

Line chart Bar chart Pie chart Area chart
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Palash Sharma 3 years ago

Lan - Local area network Wan - wide area network Lan it is used for small area or local area Wan is used for wide area network

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