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Hlo guys......
  • 4 answers

Sanjana Dhakad 31 minutes ago

Would you like to give my answer

Sanjana Dhakad 32 minutes ago

Hiii any problem

Yash Raj 33 minutes ago

hlw. 👍

Priya Kumari 34 minutes ago

Hii...!!!
What is value of under root 3
  • 1 answers

Priya Kumari 33 minutes ago

1.732 approx
What is perfect squire no.
  • 1 answers

Samridhi Yadav 49 minutes ago

Square of any square root
Prove that n cube - n is divisible by 6
  • 1 answers

Kaku Goyal an hour ago

Let n = 1 So 1^3 -1 = 0, which is divisible by 6. Let n = 2 So 2^3 - 2 = 6, which is divisible by 6. Let n = 3 So 3^3 - 3 = 24, which is divisible by 6. Hence proved
find The sum all the 11th term of an AP whose midel term is 30
  • 1 answers

Gaurav Seth 3 hours ago

given  n =11

middle term = (a1 + a11)/2 = 30

a1 + a11 = 60  .......... i

Sn = n/2(a1 + a11)  ......... ii

     = 11/2 (60)    ... from  i

     = 11 x 30

     = 330

Check whether the given number is composite or not? 2×3×5×7×11+2.
  • 5 answers

Chetna Pandey 23 minutes ago

Yes it is composite😶

Ankit Tarar an hour ago

Firstly take 11 common from this equation than you get == 11(2×3×5×7×1+2). So it is a composite no. Because it has 11 as common and it has three common factor.

Deepak R 3 hours ago

2×3×5×7×11+2 2(1×3×5×7×11+1) 2(15×78) =2340 Take LCM of 2340 LCM=2×2×5×117

Deepak R 3 hours ago

Yes

Nitin Kumar 3 hours ago

Help me my unknown friends.😂👼💁
Is anybody online? 🤗🤔🙌🙋🙋🙌👏👋👐🗯
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Mahima Pandey 3 hours ago

✋✋✋✋

Mahima Pandey 3 hours ago

Good morning

Nitin Kumar 4 hours ago

How sre u my friends🙋👬👭👫

Nitin Kumar 4 hours ago

Hiiiu

Mahima Pandey 4 hours ago

Yes
Sec0 + tan0 +1 / sec0 + tan0 _1
  • 3 answers

Nitin Kumar 3 hours ago

Thnx👍
Rationalise denomenator karo answer aa jayegi .....😏😏😏

Nitin Kumar 5 hours ago

Is ka karna kya hh
If 7 times the 7th term of an AP is equal to 11times the 11th term,then find the 18th term?
  • 1 answers

Gaurav Seth 10 hours ago

Let 1st term of AP be 'a',

Let the common difference be 'd',

So,as per question, 7 x 7th term = 11 x 11th term

or, 7 [a+6d]  = 11 [a+10d]

or, 7a+42d = 11a +110d

or, 4a + 68d = 0

or, a + 17d = 0 .......(1)

Now, 18th term = a+17d

or, t18 = 0  (from 1)

I have dought about frequency in more than and less than table
  • 5 answers

Simar Simar 2 hours ago

Very good harsh katiyar👍

Harsh Katiyar 4 hours ago

Ye. Less. Then. Ka. Aa. Jaye. Fir. More. Then. Ka. Puch. Lena. Ok

Harsh Katiyar 4 hours ago

Is. Type. Ke. Questions. Me. Interval. Se. Upper. Limit. Likh. Lo. Fir. Sabhi. Upper. Limit. Ki. Frequency. Likho. Jaise. Class. Me. Test. Hota. Hai. To. 5. Se. Kam. Kitne. Log. Aaye 8se. Numer. Kitne. Ke. Aaye. Aise. Hi. Given. Condition. Ki. Frequency. Nikalo. Fir. Cf. Nikalo. Aur. Lower. Limit. Aur. Cf ki. Coordinate. Banakar. Grap. Pe. Show. Karo

Drashti Vaish 11 hours ago

What doubt ????

Himanshi Sharma 11 hours ago

Question ko elaborate karke btaiye kya puchna h
solve this equation by step Wise k. 2.2-14*2+8=0
  • 2 answers

Himanshi Sharma 11 hours ago

Solve kerke es equation ko badi ajeeb si value aa rhi h 8.95 something😄😄😄😂😂😁😁😀😀

Himanshi Sharma 11 hours ago

K ki value btaani h kya
PQ is a chord of length 8 cm of a circle of radius 5 cm the tangent at P and Q intersect at a point P find the length TP
  • 1 answers

Gaurav Seth 11 hours ago

 

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Pratik Sabre asked in Math

 

PQ is a chord of length 8cm to a circle of radius 5cm. The tangents at P and Q intersect at a point T. Find the length TP. 

 

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Rayya Halim answered this
<big id="4134851_vote_span">1732 helpful votes in Math, Class XII-Science</big>

Given, PQ is the chord of the circle and PT and QT are the tangents drawn at the end points of the chord PQ. PQ = 8cm and OP = 5cm.

OT ⊥ PQ,

∴ PR = RQ =  (Perpendicular from the centre of the circle to a chord bisect the chord)

In right ΔOPR,

OP2 = PR2 + OR2

⇒ OR2 = OP2 – PR2

⇒ OR2 = (5cm)2 – (4cm2) = (25 – 16)cm2

⇒ OR2 = 9cm2

⇒ OR = 3cm

In right ΔPTR,

PT2 = TR2 + PR2 ...(1)

We know that, the tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠OPT = 90º

In right ΔOPT,

OT2 = PT2 + OP2 ...(2)

From (1) and (2), we get

OT2 = (TR2 + PR2) + OP2

⇒ (TR + OR)2 = (TR2 + PR2) + OP2

⇒ TR2 + OR2 + 2 × TR × OR = TR2 + PR2 + OP2

⇒ 9 + 6 TR = 16 + 25

⇒ 6TR = 25 + 16 – 9 = 32

⇒ TR = 

∴ PT2 = TR2 + PR2 (Using (1))

⇒ PT2 =  + (4cm)2

⇒ PT2 = 

⇒ PT 

Thus, the length of tangent PT is 

 

two dice
  • 1 answers

Drashti Vaish 11 hours ago

What two dice
Cbse sample paper ka question 25 mein doubt hai
  • 2 answers

Nitin Kumar 5 hours ago

Kon se subject aur vase doubt kya h

Drashti Vaish 11 hours ago

Kya doubt hai
If the points A(x,y) , B(a,0) and C(0,b) are collinear ,prove that x/a + y/b = 1
  • 3 answers

Ram Kushwah 12 hours ago

The area of triangle ABC=0

1/2(x(0-b)+a(b-y) +0(y-0))=0

0r -bx+ ab-ay +0=0

bx +by=ab

both side divide by ab

bx/ab + by/ab = ab/ab

x/a +y/b=1

 

 

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Subscribe Royal fun unlimited on YouTube
Miner segment
  • 1 answers
Subscribe Royal fun unlimited on YouTube
Yrr plzz help me 😐😐 Consider the no 4 ki power n where n is natural no . Check whether there is any value of n for which 4 ki power n ends with the digit zero
  • 2 answers

Aayushi Tyagi 13 hours ago

Thxs aditya bhardwaj

Aditya Bhardwaj 14 hours ago

No because if we take any natural no. Randomly as the power of 4 then it should always be end with 6 or 4
Kya kisi ke pass maths ki 2019 ke sple papers hein ??
  • 6 answers

Ansh Giri 4 minutes ago

Thamks friends..

Nitin Kumar 5 hours ago

You can see in the cbse.nic .in . Ki site pr maths ka marking scheme ok Ansh

Drashti Vaish 10 hours ago

Aap mycbsegide aap par dekh sakte hai

Chetna Pandey 13 hours ago

Check out this app for sample paper woth solution in mathematics section.

Ansh Giri 14 hours ago

Kya tum uske sample paper 1 ke question 6 section A ka solution bata sakte ho plz.

Nitin Kumar 14 hours ago

Yes
If a+b+c=1 and ab+bc+ca=1/3 find a:b:c
  • 1 answers

Kaku Goyal 13 hours ago

(a+b+c)^2=1^2; = a^2+ b^2+ c^2+ 2ab+ 2bc+ 2ca= 1; = ab + bc + ca = 1- a^2- b^2 -c^2 /2; = 1/3 a^2 + b^2 + c^2 = -2/3 + 1 = 1/3 ab + bc + ca = a^2 + b^2 + c^2 On comparison, a^2 = ab, b^2 = bc, c^2 = ac So a = b, b = c, c = a a : b : c = 1 : 1 : 1
If triangle ABC ~ PQR , perimeter of ABC is 32 cm , perimeter of PQR is 48 cm and PR is 6cm find AC ? Koi Answer jaldi bhezdo plzzz..
  • 5 answers

Ansh Giri 14 hours ago

Kya kisi ko iss question ka solution pata hai ? Plz bhez do

Ansh Giri 14 hours ago

Par sides similar nahi hoti vo to proportion mein hoti hai ??

Ansh Giri 14 hours ago

Code language mein answer bheja hai kya ? Plz mujhe is question ka solution chaiye.

Lokeshwari P 14 hours ago

May be 6cm becoz ABC~PQR there fore AC~PR

Harsh Joon 14 hours ago

Hjgtkk

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