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Show that T_{1 }, T_{2 }, T_{3 }, .... T_{n }.... form an AP where T_{n }is defined as: T_{n }= 9  5n. Also find the sum of the first 15 terms.

We have to prove something also

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
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Answers:

Just calculate the 1st 3 or 4 terms & show the constant common difference.
T_{1} = 9  5 = 4
T_{2 }= 9  10 = 1
T_{3} = 9  15 = 6
T_{4 } = 9  20 = 11
Now, if one checks the difference in any of the consecutive terms, it is 5 which is constant number.
Also, T_{n} = 9  5n (given) => T_{n1} = 9  5(n  1)
Further, T_{n }  T_{n1 } = (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.
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Given: T_{n} = 9  5n
T_{1} = a = 9  5 x 1 = 9  5 = 4
T_{2} = 9  5 x 2 = 9  10 = 1
Therefore, d = 1  4 = 5
S_{15} = 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.
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If the ratio of sum of the first n terms of an a.p is m^{2 }: n^{2}, show that the ratio of its m^{th }and nth term is (2m1 ) : (2n1)....

Answers:

Ans. Let First Term of AP = a
Common Difference = d
Then
\(S_m = {m\over 2} [{2a+(m1)d}]\)
And
\(S_n = {n\over 2} [{2a+(n1)d}]\)
=> \({S_m\over S_n} = { m^2\over n^2}\)
=> \({2m [2a+(m1)d] \over 2n[2a+(n1)d]} = {m^2\over n^2}\)
=> \({ [2a+(m1)d] \over [2a+(n1)d]} = {m\over n}\)
=> \( 2an+mnd  nd = 2am+ mnd md\)
=> \( 2an2am = nd md\)
=> \( 2a(nm) = (n m)d\)
=> 2a = d ......(1)
=> \({a_m\over a_n} ={ a+(m1)d \over a+(n1)d}\)
=> \({a_m\over a_n} ={ a+(m1)2a\over a+(n1)2a}\)
=> \({a_m\over a_n} ={ 2m1\over 2n1}\)
Hence Proved
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For an AP Sm=20 and Sn=10 and nm=1,then prove that n=10÷a where a=d.

Answers:

\(I \space think \space in \space question S_m = 10 \space and \space S_n = 20 \)
Ans. Given : \(S_m = 10 \)
\(S_n = 20 \space \space \space \space ..... (1) \)
=> nm = 1
=> m = n1
The sum of m terms i.e (n1) terms
\(S_{n1} =S_m = 10 \space \space \space \space \space \space ...... (2) \)
=> \(n^{th} term = S_n  S_{n1}\)
=>\(n^{th} term = 20  10 = 10\)
=> \(a + (n1)d = 10\)
=> \(a + (n1)a = 10 \space \space \space \space \space [as \space a =d ] \)
=> \(a+ na a = 10 \)
=> \(na = 10 \)
=> \(n = {10\over a}\)
Hence proved
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How to improve my minor mistake in maths.

Answers:

Work without tension. Enjoy doing maths. Never think it as meri attma hamesha satayegee. Best of luck.
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Practice maths for 3 hours daily with full concentration
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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.

Answers:

Slant height (l) = {h^{2} + (R  r)^{2})}^{1/2}
= {24^{2} + (15.5  5.5)^{2} }^{1/2}
= {576 + 100}^{1/2}
= 26 cm
CSA of bucket = πl(R + r) + πr^{2}
= 22/7{26(15.5 + 5.5) + (5.5)^{2}}
= 22/7{30.25 + 546}
= 22/7 × 576.35
= 1811.07 sq. cm
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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??

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

Yes , sa 1 is important .
If you want to get good cgpa then you have tobringo
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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.

Please check your question. If radius is already given then you can find the area easily.
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Answers:
The sum of the number and its reciprocal is 2 ×1/30.find the number

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

Answers:

8/15 of an hour
= 8/15 of 60 min
= 8/15 × 16 min
= 8 × 4 min
= 32 minutes
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32minuts
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If 3+5+7+........ to n terms/ 5+8+11+.........to 10 terms = 7. Find 'n' ?

Answers:

Both the numerator and denominator are the sums of two series up to 'n' terms:
Since the sum = (n/2)(2a+(n1)d)
For series in the numerator :
a=3, d=2, n=n
Therefore numerator = (n/2)*(6+(n1)2) = (n/2)*(2n+4)
And for the denominator series:
a=5, d=3, n=10
Therefore denominator = (10/2)*(10+(101)3) = (n/2)*(3n+7) = 185
Therefore Acc. to question:
(n/2)*(2n+4) / (185) = 7
= n^{2} +2n 1295 = 0
= n^{2} + 37n  35n  1295 = 0
=n=35 ( Since ncan not be equal to 37)
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(x÷x+1)^{2. }. 5(x÷x+1). +. 6 = 0

Answers:

Ans. \(({x \over (x+1)})^2  5 ({x \over (x+1)}) + 6 = 0\)
\(Put \space ({x \over (x+1)}) = y, We \space get \)
=> \(y^2 5y +6 =0\)
=> \(y^2  3y 2y +6 = 0\)
=> \(y(y3) 2(y3) = 0\)
=> \((y3)(y2) = 0\)
=> 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
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9x^{2 }. 9(a+b)x. +. (2a^{2 . }+. 5ab. +. 2b^{2})

Answers:

Ans. \(9x^2  9(a+b)x + (2a^2 + 5ab+2b^2) = 0\)
=> \(9x^2  3(3a+3b)x + (2a^2 + 4ab+ab +2b^2 ) = 0\)
=> \(9x^2  3[(2a+b)+(a+2b)]x + (2a^2 + 4ab+ab +2b^2 ) = 0\)
=> \(9x^2  3[(2a+b)+(a+2b)]x + [2a(a + 2b)+b(a +2b)] = 0\)
=> \(9x^2  3[(2a+b)+(a+2b)]x + [(2a+b)(a +2b)] = 0\)
=> \(9x^2  3(2a+b)x3(a+2b)x + (2a+b)(a +2b) = 0\)
=> \([3x(a+2b)][3x  (2a+b)] = 0\)
=> \([3x(a+2b)] = 0 \space \space or \space \space [3x  (2a+b)] = 0\)
=> \(3x=(a+2b) \space \space or \space \space 3x = (2a+b) \)
=> \(x= {(a+2b) \over 3} \space \space or \space \space x = {(2a+b) \over 3}\)
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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

Answers:

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 = \({900 +2x \over x} \times (x5) = {(900x +2x^2 4500 10x)\over x} \) ....... (2)
Profit = Total amount gain  total amount paid
So, from (2) and (1), We get
=> Profit = \({(900x +2x^2 4500 10x)\over x}  {900} = {(900x +2x^2 4500 10x900x )\over x} \)
=> \({2x^2 4500 10x \over x} \) ...... (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
=> \({x^2 45x 2250 } \)
=> x^{2}  75x + 30x  2250 = 0
=> x(x75) +30(x75) =0
=> (x75) (x+30) = 0
=> x = 75, 30 => \(\)
x cant not be 30,
So Number of Article Bought = 75
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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.

Check the question. Sum of how many terms
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Answers:
If a^{2},b^{2},c^{2} are in A.P to prove that 1/b+c,1/c+a,1/a+b are in A.P

Answers:

If a^{2}, b^{2}, c^{2} are in AP, then
2b^{2} = a^{2} + c^{2}
b^{2} + b^{2} = a^{2} + c^{2}
b^{2}  a^{2} = c^{2}  b^{2}
(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.
Thanks (1) 
Ans. Given : a^{2} b^{2} c^{2} are in A.P.
To Prove : \({1 \over b +c } ,{1 \over c+a}, {1\over a+b} \) Are in A.P.
Proof : 2b^{2} = a^{2} + c^{2} [as a^{2} b^{2} c^{2} are in A.P.]
=> b^{2} + b^{2} = a^{2} + c^{2}
=> b^{2}  a^{2} = c^{2}  b^{2}
=> (ba) (b+a) = (cb) (c+b)
=> \({(ba) \over (c+b) } = {(cb) \over (b+a)}\)
Divide both side by \(1\over (c+a)\), We get
=> \({(ba) \over (c+b) \times (c+a) } = {(cb) \over (b+a)\times(c+a)}\)
=> \({(ba+cc) \over (c+b) \times (c+a) } = {(cb+aa) \over (b+a)\times(c+a)}\)
=> \({(b+c)  (c+a) \over (c+b) \times (c+a) } = {(c+a) (a+b)) \over (b+a)\times(c+a)}\)
=> \({1 \over(c+a)}  {1\over (c+b) } = {1\over (a+b)} { 1\over(c+a)}\)
=> \({2 \over(c+a)} = {1\over (a+b)} + {1\over (c+b) }\)
Hence by this equation we, can say that \({1 \over b +c } ,{1 \over c+a}, {1\over a+b} \)are in A.P.
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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.

Figure please!!..
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Answers:

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
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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.

Answers:

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]
OB^{2} = OM^{2} + MB^{2 } [By Pythagoras Theorem]
=> OB^{2} = 25 + 144 = 169
=> OB = 13 cm
Now in Right Angle Triangle OBP
OB = 13 cm [Calculated Above]
PB = 20 cm [Given]
PO^{2 } = OB^{2} + PB^{2} [By Pythagoras Theorem]
=> PO^{2} = 169 + 400
=> PO = \( { \sqrt{569}}\)
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The product of two successive integral multiples of 5 is 300. Determine the multiples.

Answers:

15 and 20
Thanks (1) 
Ans. Let First Number = 5x [because it is multiple of 5.]
then Next Number = 5x + 5According to ques.
5x ( 5x + 5 ) = 300
=> 25x^{2} + 25x = 300
=> 25x^{2} + 25x  300 = 0
divide equation by 25, we get
=> x^{2} + x  12 = 0
=> x^{2} + 4x  3x  12 = 0
=> x( x+4)  3(x+4) = 0
=> (x+4) (x3) = 0
Either (x + 4 ) = 0 or (x3) = 0
=> x = 4 or x = 3
So, Multiple of 5 are : 15, 20
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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 ?

Answers:

Given, a_{1} + a_{2} + a_{3} = 33 ..........(i)
and a_{1 }x a_{2} x a_{3} = 1155 ..........(ii)
Since a_{1}, a_{2}, a_{3} are in AP, therefore, a_{1} + a_{3} = 2a_{2} ..........(iii)
Substituting the value of a_{1} + a_{3} in eq. (i), we get 2a_{2} + a_{2}= 33
=> a_{2}= 11
Putting the value of a_{2} in eq. (i), we get,
a_{1} + a_{3} = 22
=> a_{1} = (22  a_{3}) .........(iv)
Putting the value of a_{2} in eq. (ii), we get, a_{1} x a_{3} = 105 .........(v)
Substituting the value of a_{1} in eq. (v), we get, (22  a_{3}) x a_{3} = 105
=> a_{3}^{2}  22a_{3 }+ 105 = 0
=> a_{3}^{2}  15a_{3 } 7a_{3} +105 = 0
=> a_{3}(a_{3 } 15)  7(a_{3 } 15) = 0
=> a_{3}  15 = 0 and a_{3}  7 = 0
=> a_{3} = 15 and a_{3} = 7
Putting these values in eq.(ii) we get, a_{1} = 7 and a_{1} = 15
Therefore, the required three positive integers are 7, 11 and 15.
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Solve the equation:
2^{2x+3=65(2}x^{2)+122.}

Answers:

Ans. Equation is in the Form = 2^{2x+3 } = 65 (2^{x}  2 ) + 122
=> 2^{2x} . 2^{3} = 65 ( 2^{x}  2 ) + 122
=> 8(2^{x})^{2} = 65 ( 2^{x}  2) + 122
Put 2^{x} = y , We get
=> 8y^{2} = 65(y2) + 122
=> 8y^{2} = 65y  130 + 122
=> 8y^{2}  65y + 8 = 0
=> 8y^{2}  64y  y + 8 = 0
=> 8y ( y  8)  1(y8) = 0
=> (8y1) ( y  8) = 0
Either 8y 1 = 0 or y  8 = 0
So, y = 1/8 or 8
=> 2^{x} = 1/8 or 2^{x} = 8
=> 2^{x} = (2)^{3} or 2^{x} = (2)^{3}
On comparing, We get
x = 3, 3
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If S_{n }denotes the sum of first n terms of A.P. Prove that S_{30 }= 3(S_{20} S_{10}).

Answers:

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]
= S_{30}
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sector of circle explain

Answers:

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

Answers:

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'.)
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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

Answers:

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 = $\left(\frac{1\times 2+3\times \left(2\right)}{1+3},\frac{1\times 8+3\times 2}{1+3}\right)=\left(\frac{1}{2},\frac{7}{2}\right)$
M is the midpoint 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 = $\left(\frac{3\times 2+1\times \left(2\right)}{3+1},\frac{3\times 8+1\times 2}{3+1}\right)=\left(1,\frac{13}{2}\right)$
Thanks (0) 
Ans. a(2,2) b(2,8)
as l m n divides the segment in four equal parts. Then m ll be midpoint of ab
Coordinate of m = (2+2/2), (2+8/2)= (0,5)
l ll be midpoint of ma.
coordinate of l = (2+0/2), (2+5/2) = (1,7/2)
n ll be midpoint of mb.
Coordinate of n = (0+2/2), (5+8/2) = (1,13/2)
All four parts al=lm=mn=nb are equal
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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

Please

Please draw the figure
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Answers:

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
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(a^{2}+b^{2})x^{2}2b(a+c)x+(b^{2}+c^{2})=0
For equal roots ...To find b^{2}=????

Answers:

Ans. On comparing with standard form.of quadratic equation
Ax^{2} + Bx + C = 0, We get
A = (a^{2} + b^{2})
B = 2b(a+c)
C = (b^{2} + c^{2})
as equation has equal roots,So
D = 0
=> B^{2}  4AC = 0
=> [2b(a+c)]^{2}  4(a^{2} + b^{2})(b^{2} +c^{2}) =0
=> 4b^{2}(a^{2} + c^{2}+2ac) = 4(a^{2}b^{2} + a^{2}c^{2} + b^{4} + b^{2}c^{2})
divide by 4 both sides,
=> a^{2}b^{2} + b^{2}c^{2}  2acb^{2}  a^{2}b^{2} + a^{2}c^{2} + b^{4}  b^{2}c^{2} = 0
=> (ac)^{2 }+ (b^{2})^{2}  2(ac)(b^{2}) =0
=> (ac  b^{2})^{2 }=0
squareroot both side
=> ac  b^{2} = 0
=> b^{2} = ac
Hence proved
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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

Answers:

Ans. Let her actual marks = x
If she score 10 more then marks = x + 10
According to ques
5(x+10) = x^{2}
=>5x +50 = x^{2}
=> x^{2}  5x  50 = 0
=> x^{2}  10x + 5x  50 = 0
=> x(x10) +5(x10) = 0
=> (x10)(x+5) = 0
either (x10) = 0 or (x+5) = 0
x = 10 or x = 5
x can not be 5.
So x = 10
Actual marks = 10
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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.

The line segment joining the centre of a circle to the midpoint of a chord bisects the chord.
So, <ODB= 90$\xb0$

How is angle ODB=90°
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Answers:

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 .
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 State the theorem which is used for deviding a line segment in given ratio.
 In a circle chord CD is parallel to the tangent at a point A prove that angel ACD is isosceles.

Answers:

ConstructionFrom center O draw a straight line connecting point of contact A on tangent to a point M on chorday
Proofsince 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
IN TRIANGLE ACM AND ADM
CM=DMITRY
angleAMC=AND
AM=AM
THUS TRIANGLE ACM AND ADM ARE CONGURENT thus AC=AD
AS TWO SIDES ARE EQUAL TRIANGLE ACD IS ISOSCELES
Thanks (0) 
Ans. 1) Section Formula : The Coordinates of the point P(x,y) which divides the line joining the point $A({x}_{1},{y}_{1})andB({x}_{2},{y}_{2})$ Internally in the Ration m:n is given by $\left(\left(\frac{m{x}_{2}+n{x}_{1}}{m+n}\right),\left(\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)\right)$.
2. Given : Chord CD is parallel to Tangent FAE.
To Prove : Triangle ACD is Isosceles.
Proof : $\angle CAF=\angle ACD...\left(1\right)(Alternateangles)$
$\angle CAF=\angle ADC...\left(2\right)(AlternateSegmenttheorem)$
$From\left(1\right)and\left(2\right),weget$
$\angle ACD=\angle ADC\Rightarrow \u2206ACDisisoscelestriangle.$
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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.

Answers:

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=> $\frac{90}{360}{\mathrm{\pi r}}^{2}=\frac{1}{4}(3.14)\times (5{)}^{2}=19.625{m}^{2}\phantom{\rule{0ex}{0ex}}$
(2) if Rope is 10m Long = $\frac{90}{360}{\mathrm{\pi r}}^{2}=\frac{1}{4}(3.14)\times (10{)}^{2}=78.5{m}^{2}\phantom{\rule{0ex}{0ex}}$
increase in area = 78.5  19.625 = 58.875m^{2}
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