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Bhanu Pratap Patel 1 day, 19 hours ago

4
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Nihar Ranjan Bastia 1 day, 2 hours ago

T.s.a of cuboid without base area

Sambit Kumar Das 1 day, 17 hours ago

2(lb+bh+hl) it is total surface area of cuboid (T.S.A). OK!?
  • 2 answers

Disha Kashyap 😊 1 day, 5 hours ago

Hiu

Sambit Kumar Das 1 day, 17 hours ago

What diameter
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Preeti Dabral 6 days, 1 hour ago

Given: ABCD is a square, where {tex}\angle{/tex}PQR = 90and PB = QC = DR
To prove: (i) QB = RC (ii) PQ = QR
(iii) {tex}\angle{/tex}QPR = 45o
Proof:

  1. Here,
    BC = CD … (Sides of square)
    CQ = DR … (Given)
    BC = BQ + CQ
    {tex}\therefore{/tex} CQ = BC − BQ
    {tex}\therefore{/tex} DR = BC – BQ .......(1)
    Also,
    CD = RC + DR
    {tex}\therefore{/tex} DR = CD - RC = BC - RC ........(2)
    From (1) and (2), we have,
    BC - BQ = BC - RC
    {tex}\therefore{/tex} BQ = RC
  2. Now in {tex}\triangle{/tex}RCQ and {tex}\triangle{/tex}QBP
    we have, PB = QC … (Given)
    BQ = RC … [from (i)]
    {tex}\angle{/tex}RCQ = {tex}\angle{/tex}QBP … 90o each
    Hence by SAS(Side-Angle-Side) congruence rule,
    {tex}\triangle{/tex}RCQ {tex}\cong{/tex} {tex}\triangle{/tex}QBP
    {tex}\therefore{/tex} QR = PQ … (by cpct)
  3. {tex}\triangle{/tex}RCQ {tex}\cong{/tex} {tex}\triangle{/tex}QBP and QR = PQ … [from (2)]
    {tex}\therefore{/tex} In {tex}\triangle{/tex}RPQ,
    {tex}\angle{/tex}QPR = {tex}\angle{/tex}QRP ={tex}\frac{1}{2}{/tex}(180o - 90o) = {tex}\frac{90}{2}{/tex}= 45o
    {tex}\therefore{/tex} {tex}\angle{/tex}QPR = 45

Nilanjna Singh 17 hours ago

Thanks
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Preeti Dabral 1 week ago


Given: In right triangle ABC, right angled at C. M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B.
To Prove:

  1. {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD
  2. {tex}\angle{/tex}DBC is a right angle
  3. {tex}\triangle{/tex}DBC {tex}\cong{/tex} {tex}\triangle{/tex}ACB
  4. CM = {tex}\frac {1} {2}{/tex}AB

Proof:

  1. In {tex}\triangle{/tex}AMC and {tex}\triangle{/tex}BMD
    AM = BM ...[As M is the mid-point]
    CM = DM ...[Given]
    {tex}\angle{/tex}AMC = {tex}\angle{/tex}BMD ...[Vertically opposite angles]
    {tex}\therefore{/tex} {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD proved ...[SAS property] ...(1)
  2. {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD ...[From (1)]
    {tex}\angle{/tex}ACM = {tex}\angle{/tex}BDM ...[c.p.c.t.]
    These are alternate interior angles and they are equal.
    {tex}\therefore{/tex} AC {tex}\|{/tex} BD
    As AC {tex}\|{/tex} BD and transversal BC intersects them
    {tex}\therefore{/tex} {tex}\angle{/tex}DBC + {tex}\angle{/tex}ACB = 180° ...[Sum of the consecutive interior angles of the transversal]
    {tex}\angle{/tex}DBC + 90° = 180°
    {tex}\angle{/tex}DBC = 180° - 90° = 90°
    {tex}\angle{/tex}DBC is a right angle proved.
  3. {tex}\triangle{/tex}AMC {tex}\cong{/tex} {tex}\triangle{/tex}BMD ...[From (1)]
    {tex}\therefore{/tex} AC = BD ...[c.p.c.t.] ...(2)
    In DDBC and DACB
    BC = CB ...[Common]
    {tex}\angle{/tex}DBC = {tex}\angle{/tex}ACB ...[each = 90° as proved above]
    BD = CA ...[From (2)]
    {tex}\therefore{/tex} {tex}\triangle{/tex}DBC {tex}\cong{/tex} {tex}\triangle{/tex}ACB ...[SAS property]
  4. DDBC {tex}\cong{/tex} DACB ...[As proved in (iii)]
    {tex}\therefore{/tex} DC = AB ...[c.p.c.t.]
    {tex}\therefore{/tex} 2CM = AB ...[DM = CM = {tex}\frac {1} {2}{/tex} DC]
    {tex}\therefore{/tex} CM ={tex}\frac {1} {2}{/tex} AB
  • 1 answers

Preeti Dabral 6 days, 23 hours ago


Let a square ABCD in which L, M, N & O are the midpoints.
In {tex}\triangle{/tex}AML and {tex}\triangle{/tex}CNO
AM = CO (AB = DC and M and O are the midpoints)
AL = CN (AD = BC and L and N are the midpoints)
{tex}\angle{/tex} MAL =  {tex}\angle{/tex}NCO (all angles of a square = 90°)
by AAS criteria
{tex}\triangle{/tex}AML {tex}\cong{/tex} {tex}\triangle{/tex}CNO
{tex}\therefore{/tex} ML = ON (CPCT)
similarly {tex}\triangle{/tex}MBN {tex}\cong{/tex}  {tex}\triangle{/tex}LDO
now,
in {tex}\triangle{/tex}AML,
{tex}\angle{/tex}AML = {tex}\angle{/tex}ALM (AM = AL)
= 45°
similarly in  {tex}\triangle{/tex}LDO
 {tex}\angle{/tex}DLO = 45°
{tex}\therefore{/tex} {tex}\angle{/tex} MLO = 90°
by the properties of SQUARE
All sides are equal and angles are 90°

  • 1 answers

Preeti Dabral 6 days, 22 hours ago

Factor Theorem is generally applied to factoring and finding the roots of polynomial equations. It is the reverse form of the remainder theorem. Problems are solved based on the application of synthetic division and then to check for a zero remainder.

  • 2 answers

Nandini.M ......... 2 days, 18 hours ago

Yes ur right tanishk Verma

Tanishk Verma 1 week, 3 days ago

31 amu
  • 3 answers

Tahalka .. 1 week, 3 days ago

3..

Tanishk Verma 1 week, 3 days ago

By using (a+b) (a-b) = a²-b² (√8)²-(√5)² = 8-5 = 3

Mohd Rihaan 1 week, 3 days ago

0
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  • 4 answers

Tahalka .. 1 week, 3 days ago

72°each

Aditya Yadav 1 week, 5 days ago

Sum of all angles in a quadrilateral is 360° 3x+144°=360 3x=360°-144° x=216°/3 x=72° Ans

Rani Sahu 2 weeks ago

3 equal side = 72°

Laksh Rathi 2 weeks ago

Hehe boi
  • 1 answers

Suprabha Barik 1 week, 6 days ago

If each pair of opposite side of a quadrilateral is equal then it is a parallelogram.
O 5
  • 1 answers
I
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Aditya Yadav 1 week, 5 days ago

720° rotation in one second Then 720°×60second =43200° One turns means 360° then 43200/360=120 turns
  • 1 answers

Aditya Yadav 1 week, 5 days ago

Wait bro......

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