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  • 1 answers

Sia ? 6 years, 4 months ago


In equilateral {tex}\Delta ABC{/tex}, AD{tex}\perp{/tex}BC
and AB=BC=CA 
Also BD={tex}\frac{BC}{2}{/tex}.....(1)
So, {tex}\Delta ABD{/tex} is right-angled triangle
By the Pythagoras theorem,
AB= AD+ BD2
{tex}\Rightarrow{/tex} AB= AD+ ({tex}\frac{BC}{2}{/tex})2
{tex}\Rightarrow{/tex} AB= AD{tex}\frac{BC^2}{4}{/tex}
{tex}\Rightarrow{/tex} 4AB= 4AD+ BC2
{tex}\Rightarrow{/tex} 4AB= 4AD+ AB2 (AB = BC)
{tex}\Rightarrow{/tex} 4AB= 4AD+ AB2 
{tex}\Rightarrow{/tex} 4AB- AB= 4AD2
{tex}\Rightarrow{/tex} 3AB= 4AD2
Hence Proved.

  • 3 answers

Ayush Dubey 6 years, 11 months ago

111 find number of terms then use formula of middle term it is 16th Find 16th term

Anshika Pal???? 6 years, 11 months ago

111 is the answer

. . 6 years, 11 months ago

121
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  • 1 answers

Poonam Jyoti? 6 years, 11 months ago

Ram please improve your hand writing Because Board ka Liya difficult ho jyaga
  • 1 answers

Rahul Raj 6 years, 11 months ago

It's too simple always remember just opposite more and less means lower limit and cf
  • 1 answers

Gaurav Seth 6 years, 11 months ago

Angle BAC = Angle BDC (Angles in the same segment)

Angle BDC= 50 degree

In triangle BDC,

Angle BDC + Angle BCD + Angle DBC = 180 degree (Angle sum property)

50 degree + Angle BCD + 60 degree = 180 degree

110 degree + Angle BCD = 180 degree

Angle BCD = 70 degree

  • 5 answers

Aayushi Tyagi 6 years, 11 months ago

Use formula an =a+(n-1)×d 129=-1+(n-1)×5 130=5n-5 135÷5=n 27=n

Aayushi Tyagi 6 years, 11 months ago

Its 27

. . 6 years, 11 months ago

Ram here l is not 149 its 129

Himanshi Sharma 6 years, 11 months ago

N=31

Ram Kushwah 6 years, 11 months ago

Here a=-1, d=4-(-1)=4+1=5 ,   l=149

so 149=-1+(n-1)x5

149=-1+5n-5

5n=149+6=155

so {tex}\begin{array}{l}n=\frac{155}5=31\\\end{array}{/tex}

  • 2 answers

Himanshi Sharma 6 years, 11 months ago

Ya priyanka is ryt..

Priyanka S 6 years, 11 months ago

Use BPT theorem u will get the answer
  • 4 answers

. . 6 years, 11 months ago

Is it right or not .plz verify it

. . 6 years, 11 months ago

Vol.=48510 cm cube TSA=8080.47 and CSA=5462.47

Shriya ?? 6 years, 11 months ago

but is ka answer kya hai

. . 6 years, 11 months ago

Put on formulas and then calculate
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Poonam Jyoti? 6 years, 11 months ago

Assumed 3rout 3 is rational But we know rout 3 is irrational So, We assumption is wrong 3rout3 is irrational
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Sia ? 6 years, 4 months ago

We have, {tex}(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0{/tex}

Here, {tex}A=(c^2-ab), B=-2(a^2-bc),C=b^2-ac{/tex} 

Now, {tex}B^2-4AC =0{/tex}

{tex}4(a^2-bc)^2-4(c^2-ab)(b^2-ac)=0{/tex}

{tex}4(a^2-bc)^2=4(c^2-ab)(b^2-ac){/tex}

{tex}a^4-2a^2bc+b^2c^2=b^2c^2-ac^3-ab^3+a^2bc{/tex}

{tex}a^4-2a^2bc+ac^3+ab^3-a^2bc=0{/tex}

{tex}a^4-3a^2bc+ac^3+ab^3=0{/tex}

{tex}a(a^3-3abc+c^3+b^3)=0{/tex}

{tex}a=0 \ or \ a^3-3abc+c^3+b^3=0{/tex}

{tex}a=0\ or\ a^3+b^3+c^3=3abc{/tex}

Hence proved

  • 1 answers

Puja Sahoo 6 years, 11 months ago

Yu can check it in google.......?
  • 3 answers

Manan Agarwal 6 years, 11 months ago

Refer to maths ncert at back in summary the correct formula is given of mode

Rahul Raj 6 years, 11 months ago

l+(f1-f0)/(2f1-f0-f2) ×h

Rohit Attri 6 years, 11 months ago

L+(n÷2+cf)÷f× h
  • 2 answers

@ Aashu 6 years, 11 months ago

Its for equilateral triangle ..given in rs agarwall

Anshika Pal???? 6 years, 11 months ago

Its ok i got equilateral
  • 1 answers

Rahul Raj 6 years, 11 months ago

It is simple bro... first make cf table and then choose median class (class whose cf is nearly greater than n/2) and find f i.e. frequency of class preciding median class and put it on formula
  • 2 answers

Puja Sahoo 6 years, 11 months ago

He's right......

Rahul Raj 6 years, 11 months ago

F1 is the frequency of modal class. F0 is the frequency of class preciding modal class.F2 is the frequency of class after modal class.
  • 2 answers

Gaurav Seth 6 years, 11 months ago

R= 8cm
R=13cm
angle ODB= 90 degrees ( BD is tangent)
angle APB =90 degrees ( angle in semicircle)
angle DB0 =ANGLE PBA (COMMON)

SO, BY AA criteria tri APB similar to tri DBO 
BY CPST ...... BD/BP=BO/AB= OD/ AP
OD/AP= BO/AB
8/AP=13/26
AP=16 cm

Himanshu Choubey 5 years, 8 months ago

Answer is 19
  • 4 answers

Gaurav Seth 6 years, 11 months ago

Volume of roof= l x b x h 
= 22X20 X h

height of roof= height of rainfall

volume of cylinder= 22/ 7 x 1x1 x35/10

and
volume of water on roof = volume of 
Water in cylindrical tank 

22 X 20 X h = 22/7 1 x 1x35/10
22 X 20 X h = 11
h = 11/440
h = 1/40 m
h = 0.025 m
= 2.5 cm 

so the rainfall = 2.5 cm

Rahul Raj 6 years, 11 months ago

Sorry π(100)^2×350 cm^3

Rahul Raj 6 years, 11 months ago

Is it correct?

Rahul Raj 6 years, 11 months ago

Volume of rainwater (in cm)=π(50)^2×350 cm^3
  • 1 answers

Sia ? 6 years, 6 months ago

Consider the following figure:

We know that the regular hexagon is a combination of 6 equilateral triangles.
{tex}\therefore{/tex} Area of a hexagon = 6 {tex}\times{/tex} Area of {tex}\Delta O A B{/tex}
{tex}\Rightarrow \text { Area of } \Delta O A B = \frac { 24 \sqrt { 3 } } { 6 } = 4 \sqrt { 3 } \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow \frac { \sqrt { 3 } } { 4 } ( \text { side } ) ^ { 2 } = 4 \sqrt { 3 }{/tex}
{tex}\Rightarrow ( \text { side } ) ^ { 2 } = 4 \sqrt { 3 } \times \frac { 4 } { \sqrt { 3 } } = 16{/tex}
{tex}\Rightarrow{/tex} side = 4 cm
{tex}\Rightarrow{/tex} Radius of a circle = 4 cm
{tex}\therefore{/tex} Area of a circle {tex}= \pi r ^ { 2 } = 3.14 \times 4 \times 4 = 50.24 \mathrm { cm } ^ { 2 }{/tex}

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