QP = 3.8
QP = PT
(Length of tangents from external points are equal)
\(\Rightarrow\)  PT = 3.8 cm
PR = PT = 3.8 cm
\(\Rightarrow\)  QR = 7.6 cm

x²-2ax-(4b²-a²)=0
Given equation can be written as
x²-2ax-+a²-4b²=0
or (x-a)²(2b)²=0
(x-a+2b)(x-a-2b)=0
x=a-2b,x=a+2b

ac=\(\left( \sqrt { 3 } \right) \left( 6\sqrt { 3 } \right) \) =18, take factors such that their sum should be 11 and product is 18.

1. Draw a circle of centre O and radius 4 cm.
2. Take a point P outside the circle and draw a secant PAB, intersect~ng the circle at A and B.
3. Produce AP to C such that AP = CP.
4. Draw a semi-circle with CB as diameter.
5. Draw \(PD\bot CB\)  intersecting the semi-circle at D.
6. With P as centre and PD as radius, draw arcs to intersect the given circle at T and T'.
7. Join PT and PT'. Thus, PT and PT' are the required tangents.

Let d be the common difference of an AP,
Given, \(a = \frac {1}{2}\) and \({a}_{8}=\frac{17}{6}\)
We know that,
\(a+(n-1)d={ a }_{ 8 }\)
\(\therefore\)  \({ a }_{ 8 }=a+\left( 8-1 \right) d=\frac { 17 }{ 6 } \)
\(\Rightarrow\)   \(a+7d=\frac { 17 }{ 6 } \)   
\(\Rightarrow\)  \(\frac { 1 }{ 2 } +7d=\frac { 17 }{ 6 } \)               ​​​\(\left[ \because a=\frac { 1 }{ 2 } \right] \)
\(\Rightarrow\)  \(7d=\frac { 17 }{ 6 } -\frac { 1 }{ 2 } \)
\(\Rightarrow\)   \(7d=\frac { 17 }{ 6 } -\frac { 1 }{ 2 } \)
\(\Rightarrow\)   \(7d=\frac { 14 }{ 6 } \)
\(\Rightarrow\)  \(d=\frac { 1 }{ 3 } \)
Hence, the common difference is 1/3.

Let PQ touch the circle at the point R.
We known that tangents drawn from an external point to a circle are equal in length.
\(\therefore\)    AB = AC = 5 cm
\(\Rightarrow\)  AP+BP = AQ + QC = 5 cm
\(\Rightarrow\)  AP + PR = AQ + QR = 5 cm               ..........(i)
[\(\because\) BP = PR and QC = QR]
Now, perimeter of APQ = AP+PQ+AQ
  =AP + RP + QR + AQ
= 5 + 5  [from Eq. (i)]
= 10 cm

12 m

100m

A line which touches a circle only in one point, is called a tangent line to a circle.

Here, let PQ be the tower of height h m . A and B be the positions of the sun's elevation, such that
AB=50 m, ㄥPAQ=30^o, ㄥPBQ=60^o
Now, consider rt. ㄥed ΔPBQ
\(\frac { PQ }{ BQ } =tan60^{ 0 }\Rightarrow \frac { PQ }{ BQ } =\sqrt { 3 } \)
⇒ h=\(\sqrt { 3 } \) BQ   ..........(i)
\(\frac { PQ }{ AQ } \)=tan 30⁰
⇒ \(\frac { PQ }{ AB+BQ } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ \(\frac { h }{ 50+BQ } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ \(\sqrt { 3 } \)h=50+BQ
⇒ \(\sqrt { 3 } \)h=50+\(\frac { h }{ \sqrt { 3 } } \)  [using (i)]
⇒ 3h=50\(\sqrt { 3 } \)+h
⇒ 2h=50\(\sqrt { 3 } \)
⇒ h=25\(\sqrt { 3 } \)
Hence, the height of the tower is 25\(\sqrt { 3 } \) m.

\(\sqrt{2}\)

Let S_n = 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.
Therefore,   \(\mathrm{S}_{n}=\frac{n(1+n)}{2} \text { or } \mathrm{S}_{n}=\frac{n(n+1)}{2}\)
So, the sum of first n positive integers is given by
\(\mathrm{S}_{n}=\frac{n(n+1)}{2}\) 

\(45^{\circ}\)

a = -1

\(AB=2\sqrt{2r_1r_2-r^{2}_{2}}\)

0.6, 1.7, 2.8, ........, to 100 terms
Here, a = 0.6, d = 1.7 - 0.6 = 1.1, n = 100, S₁₀=?
S_n = \(\frac{n}{2}\)[2a + (n - 1)d]
 S₁₀ = \(\frac{100}{2}\)[2 x 0.6 + (100 - 1)1.1]
= 50[1.2 + 108.9]
= 50 x 110.1 = 5505

Let AB be a diameter of a given circle and LM and PQ be the tangent lines drawn to the circle at points A and B, respectively.

To prove LM || PQ
Proof We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OA \(\perp\) PQ and OB \(\perp\) LM
\(\Rightarrow\) AB \(\perp\) PQ
and AB \(\perp\) LM
\(\Rightarrow\) \(\angle\)PAB = 90°
and  \(\angle\)ABM = 90°
\(\Rightarrow\) \(\angle\)PAB = \(\angle\)ABM
[each = 90°]
But these are alternate angles.
\(\therefore\) PQ || LM
Hence, the tangents drawn at the ends of a diameter of a circle are parallel.
Hence proved.

320

Given equation is \(x^{ 2 }-55x+750=0\)
\(\Rightarrow x^{ 2 }+-30x-25x+750=0\)
\(\left[ \because 30X-25=750\\ and-30-25=-55 \right] \)
\(\Rightarrow x(x-30)-25(x-30)=0\)
\(\Rightarrow x(x-30)(x-25)=0\)
\(\Rightarrow x-30=0\Rightarrow x=30\)
and x-25=0  \(\Rightarrow x=25\)
Hence, the roots of the equation  \(x^{ 2 }-55x+750=0\) are 25 and 30.

Join OC and OD.
ஃ  OC ⊥ BC
[tangent to any circle is perpendicular to its radius at point of contact]
ஃ \(\angle \)OCB = 90°
In \(\triangle\)OCB and \(\triangle\)ODB, we have
CB = DB
[tangent from an external point]
OB = OB   [common side]
OC = OD   [radii of a circle]
ஃ By SSS congruency, we have
\(\triangle\)OCB ≅\(\triangle\)ODB
⇒  \(\angle \)OBC = \(\angle \)OBD
= \(1\over2\) \(\angle \)CBD = \(1\over2\) x 120° = 60°
ஃ \(BC\over BO\)=cos 60°
⇒ \(BC\over BO\) = \(1\over2\)
⇒ BO = 2BC
Also,  BO = BC + BC
⇒ BO = BC + BD    [∵ BC = BD]

180m

Let aeroplane is at A and B, C are two consecutive kilometre stones such that
ㄥXAB=α and ㄥYAC=β
∴ ㄥABD=α and ㄥACD=β
Let BD=x km
AD is height of aeroplane
In ΔADB, \(\frac { AD }{ BD } \)=tanα
⇒ \(\frac { AD }{ x } =tan\alpha \quad \Rightarrow \quad x=\frac { AD }{ tan\alpha } \)       ........(i)
In ∆ADC, \(\frac { AD }{ DC } \)=tanβ
⇒ \(\frac { AD }{ 1-x } =tan\beta \)
⇒ \(\frac { AD }{ 1-\frac { AD }{ tan\alpha } } =tan\beta \)      [From (i)]
⇒ \(\frac { AD\quad tan\alpha }{ tan\alpha -AD } =tan\beta \)
⇒ AD tanα=tanα.tanβ-AD tanβ
⇒ AD tanα+AD tanβ=tanα.tanβ
⇒ AD=\(\frac { tan\alpha .tan\beta }{ tan\alpha +tan\beta } \) Hence proved.

Let C is cloud and R is its reflection.

ㄥDAC=30^o, ㄥDAR=60^o, let CD=x m
∴ Height of the cloud above the lake
=(x+20)m
∴ ER=(20+x)m
Now In right ΔADC,
\(\frac { CD }{ AD } =tan30^{ 0 } \Rightarrow \frac { x }{ AD } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ AD=\(\sqrt { 3 } \)x
In right, ΔADR,
\(\frac { DR }{ AD } \)=tan 60^o
⇒ \(\frac { DE+ER }{ AD } =\sqrt { 3 } \)
⇒ \(\frac { 20+20+x }{ \sqrt { 3 } x } =\sqrt { 3 } \)  (using (i))
40+x=3x
⇒ x=20 m
Now In right ΔADC, \(\frac { AC }{ DC } \)=cosec 30^o
\(\frac { AC }{ 20 } \)=2 ⇒ AC=40 m
∴ Distance of the cloud from A=40 m

AF = AD
BE = BD,
(tangents from external points)
CE = CF
AB = AC
AD + BD = AF + FC
\(\Rightarrow\)  BD = FC    (\(\because\)  AD = AF)
BE = EC (\(\because\)  BD = BE, CE = CF)
\(\therefore\)  E bisects BC.

Given, total amount Sandeep had=RS.12000
In the beginning of every month, he spend=RS.500
So, in the beginning of 1st month, he had amount, t₁=RS.12000
In the beginning of 2nd month, he had amount, t₂=12000-500=RS.11500
In the beginning of 3rd  month, he had amount, t₃=1500-500=RS.11000
In the beginning of 4th month, he had amount, t₄=11000-500=RS.10500 and so on.
Now, the list of amount is 12000, 11500, 11000, 10500, ...Here, t₂ - t₁ = t₃ - t₂ = t₄ - t₃ =-500
i.e. t_k+1-t_k is the same everytime.
So, the above list of numbers forms an AP.

\(\frac { 9 }{ 8 } \)

The given equation is \(\sqrt { 3 } { x }^{ 2 }+11x+6\sqrt { 3 } =0\)
On comparing with \({ ax }^{ 2 }+bx+c=0\)
\(a=\sqrt { 3 } ,b=11\quad c=6\sqrt { 3 } \)
On substituting the values of a,b and c in the quadratic formula.
\(x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ x=\frac { -11\pm \sqrt { { \left( 11 \right) }^{ 2 }-4\left( \sqrt { 3 } \right) \left( 6\sqrt { 3 } \right) } }{ 2(\sqrt { 3 } ) } \\ x=\frac { -11\pm \sqrt { 121-72 } }{ 2\sqrt { 3 } } \\ =\frac { -11\pm \sqrt { 49 } }{ 2\sqrt { 3 } } =\frac { -11\pm 7 }{ 2\sqrt { 3 } } \\ x=\frac { -11+7 }{ 2\sqrt { 3 } } =\frac { -4 }{ 2\sqrt { 3 } } =\frac { -2 }{ \sqrt { 3 } } \\ x=\frac { -11-7 }{ 2\sqrt { 3 } } =\frac { -18 }{ 2\sqrt { 3 } } =\frac { -9 }{ \sqrt { 3 } } \)
Hence, \(\frac { -2 }{ \sqrt { 3 } } \) and \(\frac { -9 }{ \sqrt { 3 } } \) are the required solutions of the given equation.

1039.2 m

Yes, the new triangle is also equilateral.

Here, a = 2, d = 8 and \({S}_{n}=90\)
\(\therefore\)  \({S}_{n}=90\)
\(​​​​​​\Rightarrow \frac { n }{ 2 } \left[ 2\times 2\left( n-1 \right) 9 \right] =90\)
\(\left[ \because { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} \right] \)
\(\Rightarrow \quad \frac { n }{ 2 } \left( 4+8n-8 \right) =90\)
\(\Rightarrow \quad \frac { n }{ 2 } \left( 8n-4 \right) =90\)
\(\Rightarrow \quad \quad n\left( 4n-2 \right) =90\)
\(\Rightarrow \quad \quad 4{ n }^{ 2 }-2n-90=0\)
\(\therefore \quad n=\frac { -\left( -2 \right) \pm \sqrt { { \left( -2 \right) }^{ 2 } } -4\times 4\left( -90 \right) }{ 2\times 4 } \)
   [\(\because\) By quadratic formula, ]
\(=\frac { 2\pm \sqrt { 4+1440 } }{ 8 } =\frac { 2\pm \sqrt { 1444 } }{ 8 } \)
\(=\frac { 2\pm 38 }{ 8 } =\frac { 40 }{ 8 } ,\frac { -36 }{ 8 } =5,\frac { -9 }{ 2 } \)
Since, n cannot be negative.
\(\therefore\)  n = 5
Now, \({ a }_{ 5 }=2+\left( 5-1 \right) 8\)  
\(\left[ \because { a }_{ n }=a+\left( n-1 \right) d \right] \)   
\(=2+32=34\)   
Hence, n =5 and \({ a }_{ 5 }=34.\)

346.4 m/s

Given: A circle C (O, r) with centre O. AB is a tangent to the circle at the point p. 
To Prove: OP ⊥ AB 
Const: Take any point Q on AB other than P and join OQ. 

OR > OP    .........(ii)
Thus, we find that among all such line segments, the line segment OP is the shortest, which is possible only when OP is perpendicular to AB. 
Hence, OP ⊥ AB. 

Let present age of his son = x yr
One year ago, his son's age = (x - 1) yr
One year ago, man's age = 8(x - 1) yr = (8x - 8) yr
Present age of man = (8x - 8 + 1) yr = (8x - 7) yr
According to the question,
8x - 7 = x² ⇒ x² - 8x + 7 = 0
which is the required quadratic equation.
Now, x² - 7x - x + 7 = 0 [by factorisation]
⇒ x(x - 7) - l(x - 7) = 0
⇒ (x - 7)(x - 1) = 0
⇒ x - 7 = 0 or x- 1 = 0
⇒ x = 7 or x = 1
But x = 1is not possible because if x = 1, then present age of the son and father are same. So, x = 7.
Hence, present age of his son = 7 yr and present age of man = 8 x 7 - 7 = 49 yr.

yes