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Show that T, T, T, .... T.... form an AP where Tis defined as: T= 9 - 5n. Also find the sum of the first 15 terms.

Posted by satvik Kumar (Feb 22, 2017 12:04 p.m.) (Question ID: 2440)

• We have to prove something also

Posted by satvik Kumar (Feb 22, 2017 12:47 p.m.)
• Please read the question carefully.. Question also says that Show that T1 , T2 , T3 , .... Tn .... form an AP where Tn is defined as: Tn = 9 - 5n

Posted by satvik Kumar (Feb 22, 2017 12:46 p.m.)
• Just calculate the 1st 3 or 4 terms & show the constant common difference.

T1 = 9 - 5 = 4

T2  = 9 - 10 = -1

T3 = 9 - 15 = -6

T = 9 - 20 = -11

Now, if one checks the difference in any of the consecutive terms, it is -5 which is constant number.

Also, Tn = 9 - 5n (given) => Tn-1 = 9 - 5(n - 1)

Further, Tn  - Tn-1  = (9 - 5n) - {9 - 5(n - 1)}

= 9 - 5n - (9 - 5n + 5) = 9 - 5n - 9 + 5n - 5 = - 5 (Common Difference between any 2 consecutive terms for any value of n)

Hence, one can say that it is forming an AP.

Answered by Manish Gandhi (Feb 22, 2017 1:38 p.m.)
Thanks (0)
• Given:     Tn = 9 - 5n

T1 = a = 9 - 5 x 1 = 9 - 5 = 4

T2 = 9 - 5 x 2 = 9 - 10 = -1

Therefore, d = -1 - 4 = -5

S15 = n/2[2a + (n - 1)d]

= 15/2[2 x 4 + (15 - 1) x (-5)]

= 15/2[8 - 70]

= 15/2 x [-62]

= -465

Therefore, the sum of 15 terms of given AP is -465.

Answered by Rashmi Bajpayee (Feb 22, 2017 12:30 p.m.)
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If the ratio of sum of the first n terms of  an a.p is  m: n2, show that the ratio of its mth  and nth term is   (2m-1 ) :  (2n-1)....

Posted by Ranveer Raj (Feb 21, 2017 11:37 p.m.) (Question ID: 2424)

• Ans. Let First Term of AP = a

Common Difference = d

Then

$$S_m = {m\over 2} [{2a+(m-1)d}]$$

And

$$S_n = {n\over 2} [{2a+(n-1)d}]$$

=> $${S_m\over S_n} = { m^2\over n^2}$$

=> $${2m [2a+(m-1)d] \over 2n[2a+(n-1)d]} = {m^2\over n^2}$$

=> $${ [2a+(m-1)d] \over [2a+(n-1)d]} = {m\over n}$$

=> $$2an+mnd - nd = 2am+ mnd -md$$

=> $$2an-2am = nd -md$$

=> $$2a(n-m) = (n -m)d$$

=> 2a = d       ......(1)

=> $${a_m\over a_n} ={ a+(m-1)d \over a+(n-1)d}$$

=> $${a_m\over a_n} ={ a+(m-1)2a\over a+(n-1)2a}$$

=> $${a_m\over a_n} ={ 2m-1\over 2n-1}$$

Hence Proved

Answered by Naveen Sharma (Feb 22, 2017 8:41 a.m.)
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For an AP Sm=20 and Sn=10 and n-m=1,then prove that n=10÷a where a=d.

Posted by Sagar Maheshwari (Feb 21, 2017 11:49 a.m.) (Question ID: 2392)

• $$I \space think \space in \space question S_m = 10 \space and \space S_n = 20$$

Ans. Given : $$S_m = 10$$

$$S_n = 20 \space \space \space \space ..... (1)$$

=> n-m = 1

=> m = n-1

The sum of m terms i.e (n-1) terms

$$S_{n-1} =S_m = 10 \space \space \space \space \space \space ...... (2)$$

=> $$n^{th} term = S_n - S_{n-1}$$

=>$$n^{th} term = 20 - 10 = 10$$

=> $$a + (n-1)d = 10$$

=> $$a + (n-1)a = 10 \space \space \space \space \space [as \space a =d ]$$

=> $$a+ na -a = 10$$

=> $$na = 10$$

=> $$n = {10\over a}$$

Hence proved

Answered by Naveen Sharma (Feb 21, 2017 12:42 p.m.)
Thanks (1)

How to improve my minor mistake in maths.

Posted by Rockstar Aadi (Feb 21, 2017 6:33 a.m.) (Question ID: 2379)

• Work without tension. Enjoy doing maths. Never think it as meri attma hamesha satayegee. Best of luck.

Answered by Amrit Joseph (Feb 21, 2017 10:05 p.m.)
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• Practice maths for 3 hours daily with full concentration

Answered by pranjal boora (Feb 21, 2017 9:34 a.m.)
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The radii of circular ends of a bucket are 5.5  cm and 15.5 cm and its height is 24 cm. Find the surface area of the bucket.

Posted by Shruti Sharma (Feb 19, 2017 5:37 p.m.) (Question ID: 2336)

• Slant height (l) = {h2 + (R - r)2)}1/2

= {242 + (15.5 - 5.5)2 }1/2

= {576 + 100}1/2

= 26 cm

CSA of bucket = πl(R + r) + πr2

= 22/7{26(15.5 + 5.5) + (5.5)2}

= 22/7{30.25 + 546}

= 22/7 × 576.35

= 1811.07 sq. cm

Answered by Rashmi Bajpayee (Feb 19, 2017 6:12 p.m.)
Thanks (2)

For external boards will our sa1 result affect our sa2 result for the final cgpa?  Sa1 marks will be added to our sa2 marks to get the final cgpa??

Posted by Jagriti Khanna (Feb 19, 2017 12:02 p.m.) (Question ID: 2327)

• So then if I got a cgpa of 8.6 in the first term can I get a 10cgpa in the final result?

Posted by Jagriti Khanna (Feb 21, 2017 10:46 p.m.)
• Yes , sa 1 is important .

If you want to get good cgpa then you have tobringo

Answered by Uditkumar Swami (Feb 19, 2017 5:57 p.m.)
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The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm.find the area of a circle.

Posted by John Dungdung (Feb 19, 2017 7:32 a.m.) (Question ID: 2317)

Posted by Rashmi Bajpayee (Feb 19, 2017 3:57 p.m.)

The sum of the number and its reciprocal is 2 ×1/30.find the number

Posted by John Dungdung (Feb 19, 2017 7:28 a.m.) (Question ID: 2316)

what is 8/15 of an hour ( in minutes )

Posted by Jithin Joseph (Feb 18, 2017 11:34 p.m.) (Question ID: 2310)

• 8/15 of an hour

= 8/15 of 60 min

= 8/15 × 16 min

= 8 × 4 min

= 32 minutes

Answered by Rashmi Bajpayee (Feb 19, 2017 4:03 p.m.)
Thanks (0)
• 32minuts

Answered by Sakshi Gupta (Feb 18, 2017 11:39 p.m.)
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If 3+5+7+........ to n terms/ 5+8+11+.........to 10 terms = 7. Find 'n' ?

Posted by Harmehar Malhotra (Feb 18, 2017 4:48 p.m.) (Question ID: 2291)

• Both the numerator and denominator are the sums of two series up to 'n' terms:-

Since the sum = (n/2)(2a+(n-1)d)

For series in the numerator :-

a=3, d=2, n=n

Therefore numerator = (n/2)*(6+(n-1)2) = (n/2)*(2n+4)

And for the denominator series:-

a=5, d=3, n=10

Therefore denominator = (10/2)*(10+(10-1)3) = (n/2)*(3n+7) = 185

Therefore Acc. to question:-

(n/2)*(2n+4) / (185)   = 7

= n2 +2n- 1295 = 0

= n2 + 37n - 35n - 1295 = 0

=n=35 ( Since ncan not be equal to 37)

Answered by Devanshu Agarwal (Feb 18, 2017 6:26 p.m.)
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(x÷x+1)2.  -.  5(x÷x+1). +. 6 = 0

Posted by pritish pandey (Feb 17, 2017 2:48 p.m.) (Question ID: 2239)

• Ans. $$({x \over (x+1)})^2 - 5 ({x \over (x+1)}) + 6 = 0$$

$$Put \space ({x \over (x+1)}) = y, We \space get$$

=> $$y^2 -5y +6 =0$$

=> $$y^2 - 3y -2y +6 = 0$$

=> $$y(y-3) -2(y-3) = 0$$

=> $$(y-3)(y-2) = 0$$

=> y = 3 , 2

Now, $${x\over (x+1) } = 3 \space \space or \space \space {x\over (x+1) } = 2$$

=> x = 3x +3       or     x = 2x + 2

=> x = -3/2         or    x = -2

Values of x = -3/2, -2

Answered by Naveen Sharma (Feb 17, 2017 3:26 p.m.)
Thanks (1)

9x2   -.   9(a+b)x. +. (2a2 .  +. 5ab. +. 2b2)

Posted by pritish pandey (Feb 17, 2017 2:45 p.m.) (Question ID: 2238)

• Ans. $$9x^2 - 9(a+b)x + (2a^2 + 5ab+2b^2) = 0$$

=> $$9x^2 - 3(3a+3b)x + (2a^2 + 4ab+ab +2b^2 ) = 0$$

=> $$9x^2 - 3[(2a+b)+(a+2b)]x + (2a^2 + 4ab+ab +2b^2 ) = 0$$

=> $$9x^2 - 3[(2a+b)+(a+2b)]x + [2a(a + 2b)+b(a +2b)] = 0$$

=> $$9x^2 - 3[(2a+b)+(a+2b)]x + [(2a+b)(a +2b)] = 0$$

=> $$9x^2 - 3(2a+b)x-3(a+2b)x + (2a+b)(a +2b) = 0$$

=> $$[3x-(a+2b)][3x - (2a+b)] = 0$$

=> $$[3x-(a+2b)] = 0 \space \space or \space \space [3x - (2a+b)] = 0$$

=> $$3x=(a+2b) \space \space or \space \space 3x = (2a+b)$$

=> $$x= {(a+2b) \over 3} \space \space or \space \space x = {(2a+b) \over 3}$$

Answered by Naveen Sharma (Feb 17, 2017 3:38 p.m.)
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A trader buys a number of goods for rupees 900 and 5 goods are found to be faulty. he sells the remaining goods at 80 rupees per piece in  the whole transaction. find the number of good he bought

Posted by Harmehar Malhotra (Feb 16, 2017 2:43 p.m.) (Question ID: 2199)

• I think your question is "A trader buys a number of goods for rupees 900 and 5 goods are found to be faulty. he sells the remaining goods at Rs 2 per piece more than what he paid for it . He got a profit of 80 rupees in the whole transaction. find the number of good he bought"

Ans. Let Number of Article Bought = x

Total Amount Paid =  900         .................. (1)

Amount Paid for one article = $$900 \over x$$

Faulty Article = 5
Remaining Article that He Sold = (x - 5)

Price for One article = $${900 \over x } + 2 = {900 +2x \over x}$$

Total Amount he Gain For all Article = $${900 +2x \over x} \times (x-5) = {(900x +2x^2 -4500 -10x)\over x}$$       ....... (2)

Profit = Total amount gain - total amount paid

So, from (2) and (1), We get

=> Profit = $${(900x +2x^2 -4500 -10x)\over x} - {900} = {(900x +2x^2 -4500 -10x-900x )\over x}$$

=> $${2x^2 -4500 -10x \over x}$$       ...... (3)

Given Profit  =  80      ..... (4)

So from (3) and (4)

=> $${2x^2 -4500 -10x \over x} = 80 =>{2x^2 -4500 -10x = 80x }$$

=> $${2x^2 -4500 -10x - 80x } => {2x^2 -90x -4500 }$$

Divide it by 2, we get

=> $${x^2 -45x -2250 }$$

=> x2 - 75x + 30x - 2250 = 0

=> x(x-75) +30(x-75) =0

=> (x-75) (x+30) = 0

=> x = 75, -30  => 

x cant not be -30,

So Number of Article Bought = 75

Answered by Naveen Sharma (Feb 16, 2017 3:06 p.m.)
Thanks (1)

The sum of first q terms is 162. The ratio of its 6th term to its 13th term is 1:2 . Find the first and 15th term of the AP.

Posted by Sudhanshu Yadav (Feb 15, 2017 6:59 p.m.) (Question ID: 2154)

• Check the question. Sum of how many terms

Posted by Rashmi Bajpayee (Feb 15, 2017 7:17 p.m.)

If a2,b2,c2 are in A.P to prove that                  1/b+c,1/c+a,1/a+b are in A.P

Posted by dev saini (Feb 15, 2017 5:21 p.m.) (Question ID: 2148)

• If a2, b2, c2 are in AP, then

2b2 = a2 + c2

b2 + b2 = a2 + c2

b2 - a2 = c2 - b2

(b - a)(b + a) = (c - b)(c + b)

(b - a)/(c + b) = (c - b)/(b + a)

Dividing both sides by (c + a),

(b - a)/{(c + b)(c + a)} = (c - b)/{(b + a)(c + a)}

{(b + c) - (c + a)}/{(b + c)(c + a) = {(c + a) - (a + b)}/{(a + b)(c + a)}

{1/(c + a)} - {1/(b + c)} = {1/(a + b)} - {1/(c + a)}

{2/(c + a)} = {1/(a + b)} - {1/(b + c)}

Therefore 1/(a + b), 1/(b + c), 1/(c + a) are in AP.

Answered by Rashmi Bajpayee (Feb 15, 2017 6:08 p.m.)
Thanks (1)
• Ans. Given : a2 b2 c2 are in A.P.

To Prove $${1 \over b +c } ,{1 \over c+a}, {1\over a+b}$$ Are in A.P.

Proof : 2b2 = a2 + c2   [as a2 b2 c2 are in A.P.]

=> b2 + b2 = a2 + c2

=> b2 - a2 = c2 - b2

=> (b-a) (b+a) = (c-b) (c+b)

=> $${(b-a) \over (c+b) } = {(c-b) \over (b+a)}$$

Divide both side by $$1\over (c+a)$$, We get

=> $${(b-a) \over (c+b) \times (c+a) } = {(c-b) \over (b+a)\times(c+a)}$$

=> $${(b-a+c-c) \over (c+b) \times (c+a) } = {(c-b+a-a) \over (b+a)\times(c+a)}$$

=> $${(b+c) - (c+a) \over (c+b) \times (c+a) } = {(c+a) -(a+b)) \over (b+a)\times(c+a)}$$

=> $${1 \over(c+a)} - {1\over (c+b) } = {1\over (a+b)} -{ 1\over(c+a)}$$

=> $${2 \over(c+a)} = {1\over (a+b)} + {1\over (c+b) }$$

Hence by this equation we, can say that $${1 \over b +c } ,{1 \over c+a}, {1\over a+b}$$are in A.P.

Answered by Naveen Sharma (Feb 15, 2017 5:57 p.m.)
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A line through the centre O of a circle of radius 7 cm cuts the tangent, at a point P on the circle at Q, such that PQ=24 cm. Find OQ.

Posted by satvik Kumar (Feb 15, 2017 2:31 p.m.) (Question ID: 2145)

Posted by satvik Kumar (Feb 15, 2017 3:36 p.m.)
• Ans. Given Radius of Circle (OP) = 7 cm

PQ = 24 cm

In Right Angle Triangle OPQ

=> (OQ)2 = (OP)2 + (PQ)2

=> (OQ)2 = 49 + 576

=> (OQ)2 = 625

=> OQ = 25 cm

Answered by Naveen Sharma (Feb 16, 2017 11:21 a.m.)
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PB is a tangent to the circle with centre O at B. AB is a chord of length 24 cm at a distance of 5cm from the centre. If the tangent is of length 20cm ,find the length of PO.

Posted by Har Shita (Feb 14, 2017 2:38 p.m.) (Question ID: 2108)

• Ans. In Right Angle Triangle  OMB,

OM = 5 cm   [Given]

MB = AB/2 = 24/2 = 12 cm       [perpendicular from the center to chord divided it into two equal part]

OB2 = OM2 + MB2     [By Pythagoras Theorem]

=> OB2 = 25 + 144 = 169

=> OB = 13 cm

Now in Right Angle Triangle OBP

OB = 13 cm          [Calculated Above]

PB = 20 cm          [Given]

PO = OB2 + PB2        [By Pythagoras Theorem]

=> PO2 = 169 + 400

=> PO = $${ \sqrt{569}}$$

Answered by Naveen Sharma (Feb 14, 2017 4:06 p.m.)
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The product of two successive integral multiples of 5 is 300. Determine the multiples.

Posted by Vedant Bothikar (Feb 14, 2017 12:02 p.m.) (Question ID: 2101)

• 15 and 20

Answered by Nisha Goel (Feb 14, 2017 12:35 p.m.)
Thanks (1)
• Ans. Let First Number = 5x              [because it is multiple of 5.]
then Next Number = 5x + 5

According to ques.

5x ( 5x + 5 ) = 300

=> 25x2 + 25x = 300

=> 25x2 + 25x - 300 = 0

divide equation by 25, we get

=> x2 + x - 12 = 0

=> x2 + 4x - 3x - 12 = 0

=> x( x+4) - 3(x+4) = 0

=> (x+4) (x-3) = 0

Either (x + 4 ) = 0  or (x-3) = 0

=> x = -4  or x = 3

So, Multiple of 5 are : 15, 20

Answered by Naveen Sharma (Feb 14, 2017 12:35 p.m.)
Thanks (1)

Three positive intiger are in A.P  a1, a2 ,a3.   a1+a2+a3=33 and product of a1*a2*a3=1155 find value of infringers a1, a2,  a3 ?

Posted by Pankaj Asery (Feb 14, 2017 9:57 a.m.) (Question ID: 2098)

• Given, a1 + a2 + a3 = 33 ..........(i)

and  a1 x a2 x a3 = 1155 ..........(ii)

Since a1, a2, a3 are in AP, therefore, a1 + a3 = 2a2 ..........(iii)

Substituting the value of a1 + a3 in eq. (i), we get 2a2 + a2= 33

=>  a2= 11

Putting the value of a2 in eq. (i), we get,

a1 + a3 = 22

=>   a1 = (22 -  a3) .........(iv)

Putting the value of a2 in eq. (ii), we get, a1 x a3 = 105   .........(v)

Substituting the value of a1 in eq. (v), we get, (22 - a3) x a3 = 105

=> a32 - 22a3 + 105 = 0

=>  a32 - 15a- 7a3 +105 = 0

=>  a3(a3 - 15) - 7(a3 - 15) = 0

=>  a3 - 15 = 0 and  a3 - 7 = 0

=>  a3 = 15 and a3 = 7

Putting these values in eq.(ii) we get, a1 = 7 and a1 = 15

Therefore, the required three positive integers are 7, 11 and 15.

Answered by Rashmi Bajpayee (Feb 14, 2017 12:28 p.m.)
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Solve the equation:

22x+3=65(2x-2)+122.

Posted by Komaldeep Kaur (Feb 13, 2017 11:10 p.m.) (Question ID: 2090)

• Ans. Equation is in the Form = 22x+3  = 65 (2x - 2 ) + 122

=> 22x . 23 = 65 ( 2x - 2 ) + 122

=> 8(2x)2 = 65 ( 2x - 2) + 122

Put 2x = y , We get

=> 8y2 = 65(y-2) + 122

=> 8y2 = 65y - 130 + 122

=> 8y2 - 65y + 8 = 0

=> 8y2 - 64y - y + 8 = 0

=> 8y ( y - 8) - 1(y-8) = 0

=> (8y-1) ( y - 8) = 0

Either 8y -1 = 0 or y - 8 = 0

So, y = 1/8 or 8

=> 2x = 1/8             or 2x = 8

=> 2x = (2)-3      or 2x = (2)3

On comparing, We get

x = -3, 3

Answered by Naveen Sharma (Feb 14, 2017 5:20 p.m.)
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If Sdenotes the sum of first n terms of A.P. Prove that S30 = 3(S20- S10).

Posted by satvik Kumar (Feb 13, 2017 8:04 p.m.) (Question ID: 2082)

• R.H.S.

3[20/2{2a + (20 - 1)d} - 10/2{2a + (10 - 1)d]

= 3[10{2a + 19d - 5{2a + 9d}]

= 15[4a + 38d - 2a - 9d]

= 30/2[2a + 29d]

= 30/2[2a + (30 - 1)d]

= S30

Answered by Rashmi Bajpayee (Feb 13, 2017 8:57 p.m.)
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sector of circle explain

Posted by LEEENA WATKAR (Feb 12, 2017 5:45 p.m.) (Question ID: 2055)

• The region enclosed by an arc of the circle is called sector.

Answered by Shweta Gulati (Feb 12, 2017 8:06 p.m.)
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A ladder rests against a vertical wall at an inclination a to the horizontal . Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle b to the ground . Show that Cos a minus cos b by sin a minus sin b equal to p by q.

Posted by aishwarya lal (Feb 12, 2017 5:28 p.m.) (Question ID: 2053)

• A  ladder rests against a wall at an angle α to the horizontal. Its foot is pulled away from the wall through a distance a, so that it slides down a distance b down the wall making an angle β with the horizontal. Show that

a/b = cosα - cos β / sin β - sinα

( Its the same question you have asked. I have only taken 'a' instead of 'p' and 'b' instead of 'q'.)

Answered by Shweta Gulati (Feb 12, 2017 8:10 p.m.)
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Find the coordinate of the point l m n whitch divide the line segment joining a( -2,2)  l m n b(2,8) into four equal parts

Posted by Shashant Ahuja (Feb 11, 2017 9:46 a.m.) (Question ID: 1999)

• Points L, M and N divide the given line segment in equal parts, then

For point L, the ratio will be 1 : 3.

Then, Coordinates of L = $\left(\frac{1×2+3×\left(-2\right)}{1+3},\frac{1×8+3×2}{1+3}\right)=\left(\frac{-1}{2},\frac{7}{2}\right)$

M is the mid-point of AB, then

Coordinates of M = $\left(\frac{-2+2}{2},\frac{2+8}{2}\right)=\left(0,5\right)$

For point N, the ratio will be 3 : 1, then

Coordinates of N = $\left(\frac{3×2+1×\left(-2\right)}{3+1},\frac{3×8+1×2}{3+1}\right)=\left(1,\frac{13}{2}\right)$

Answered by Rashmi Bajpayee (Feb 13, 2017 5:13 p.m.)
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• Ans. a(-2,2) b(2,8)

as l m n divides the segment in four equal parts. Then m ll be mid-point of ab

Co-ordinate of m = (-2+2/2), (2+8/2)= (0,5)

l ll be mid-point of ma.

co-ordinate of l = (-2+0/2), (2+5/2) = (-1,7/2)

n ll be mid-point of mb.

Coordinate of n = (0+2/2), (5+8/2) = (1,13/2)

All four parts al=lm=mn=nb are equal

Answered by Naveen Sharma (Feb 13, 2017 5:13 p.m.)
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A  ladder of length 6m makes an angle of 45 with the floor while leaning against one wall of a room if the foot of the ladder is fixed on the floor and it is made to lean against the opposite wall of the room it makes an angle 60 with the floor find the distance between these two walls of the room

Posted by Shashant Ahuja (Feb 13, 2017 5:14 p.m.) (Question ID: 1998)

Posted by Shashant Ahuja (Feb 11, 2017 10:30 a.m.)

Posted by Shashant Ahuja (Feb 11, 2017 10:26 a.m.)
• Ans. Let AB and CD are two wals of room and Ladder is fixed at O.

In Triangle ABO : cos 60 = BO/6

=> 1/2 = BO/6

cross multiply. We get

BO = 3m

In Triangle CDO : cos 45 = DO/6

=> 1/√2  = DO/6

=> DO = 3√2 m

Distance b/w walls = BO + DO = 3 + 3√2  = 3 + 3(1.41) = 3 + 4.23 = 7.23m

Answered by Naveen Sharma (Feb 13, 2017 5:14 p.m.)
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(a2+b2)x2-2b(a+c)x+(b2+c2)=0

For equal roots ...To find b2=????

Posted by aryan raj (Feb 11, 2017 9 a.m.) (Question ID: 1996)

• Ans. On comparing with standard form.of quadratic equation

Ax2 + Bx + C = 0, We get

A = (a2 + b2)

B = -2b(a+c)

C = (b2 + c2)

as equation has equal roots,So

D = 0

=> B2 - 4AC = 0

=> [-2b(a+c)]2 - 4(a2 + b2)(b2 +c2) =0

=> 4b2(a2 + c2+2ac) = 4(a2b2 + a2c2 + b4 + b2c2)

divide by 4 both sides,

=> a2b2 + b2c2 - 2acb2 - a2b2 + a2c2 + b4 - b2c2 = 0

=> (ac)+ (b2)2 - 2(ac)(b2) =0

=> (ac - b2)=0

squareroot both side

=> ac - b2 = 0

=> b2 = ac

Hence proved

Answered by Naveen Sharma (Feb 11, 2017 10:31 a.m.)
Thanks (0)

had shreya sccored 10 marks more in her english test out of 30 marks ,5 times if these marks would have been the square of her actual marks how many marks did she get in the test

Posted by Shashant Ahuja (Feb 10, 2017 8:21 p.m.) (Question ID: 1987)

• Ans. Let her actual marks = x

If she score 10 more then marks = x + 10

According to ques

5(x+10) = x2

=>5x +50 = x2

=> x2 - 5x - 50 = 0

=> x2 - 10x + 5x - 50 = 0

=> x(x-10) +5(x-10) = 0

=> (x-10)(x+5) = 0

either (x-10) = 0   or  (x+5) = 0

x = 10 or x = -5

x can not be -5.

So x = 10

Actual marks = 10

Answered by Naveen Sharma (Feb 10, 2017 9:30 p.m.)
Thanks (0)

Prove that the tangent drawn at the midpoint of an arc of a circle is parallel to the chord joining the end points of the arc.

Posted by satvik Kumar (Feb 10, 2017 6:42 p.m.) (Question ID: 1984)

• The line segment joining the centre of a circle to the mid-point of a chord bisects the chord.

So, <ODB= 90$°$

Posted by Shweta Gulati (Feb 11, 2017 2:22 p.m.)
• How is angle ODB=90°

Posted by satvik Kumar (Feb 11, 2017 1:07 p.m.)
• Let XY be the tangent at the mid point of the arc A

Let T be the point of contact

Let AB be the chord

Take D as mid point of AB.

Construction join OA ,OB and OT , where O is the centre of the circle .

Proof : ∠OTY=90 degrees

∠ODB=90 degrees

Since the alternate interior angles are equal.

AB is parallel to XY

Hence proved .

Answered by Krish Gulati (Feb 14, 2017 1:03 p.m.)
Thanks (0)
1. State the theorem which is used for deviding a line segment in given ratio.
2. In a circle chord CD is parallel to the tangent at a point A prove that angel ACD is isosceles.
Posted by vinod sharma (Feb 10, 2017 1:14 p.m.) (Question ID: 1978)

• Construction-From center O draw a straight line connecting point of contact A on tangent to a point M on chorday

Proof-since angle between radius  and tangent at point of contact is 90 thus angle formed between  radius and tangent at A is 90

Now chord//tangent

Thus angle OMD is also 90

This proves that OM   is perpendicular to chord

A perpendicular from centre to chord bisects the chord

Thus CM=DMITRY

CM=DMITRY

angleAMC=AND

AM=AM

AS TWO SIDES ARE EQUAL TRIANGLE ACD IS ISOSCELES

Answered by ashita singh (Feb 11, 2017 8:53 a.m.)
Thanks (0)
• Ans. 1) Section Formula : The Co-ordinates of the point P(x,y) which divides the line joining the point  Internally in the Ration m:n  is given by .

2. Given : Chord CD is parallel to Tangent FAE.

To Prove : Triangle ACD is Isosceles.

Proof :

Answered by Naveen Sharma (Feb 10, 2017 3:28 p.m.)
Thanks (0)

A horse is tied to the peg at one corner of a square shaped grass field of side 15 m by means of a long rope. Find

(1) the area of that part of the field in which the horse can graze.

(2) the increase in the grazing area if the rope were 10 m long instead of 5 m.

Posted by jota singh (Feb 10, 2017 12:49 p.m.) (Question ID: 1977)

• Ans. (1)The horse can graze a sector of 90° in a circle of 5 m radius.
Area that can be grazed by horse = Area of sector OACB

=>

(2) if Rope is 10m Long =

increase in area = 78.5 - 19.625 = 58.875m2

Answered by Naveen Sharma (Feb 10, 2017 1:06 p.m.)
Thanks (1)

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