Given, AP = 8 and OA = 6 cm
\(\angle OAP\) = 90° (as radius \(\bot\) tangent)
\(\Rightarrow\)  OP² = AP² + OA²
= 8² + 6²
= 64 + 36
OP² = 100
\(\Rightarrow\) OP = 10 cm
Now in triangle OPB,
OP = 10 cm, OB = 4 cm
\(\Rightarrow\) OP²  = BP² + OB²
\(\Rightarrow\) 100 = BP² + 16
BP² = 100 - 16 = 84
or BP = \(\sqrt{84}\)
\(=2\sqrt{21}\) cm.

Let the first term of an AP. be a, common difference,
d =-6
a₁₆ = a + (16- 1)(- 6)
= a-90
a₁₂ = a + (12-1)(-6)
= a-66
a₁₆ - a₁₂ = (a - 90) - (a - 66)
=a-90-a+66
=-24

1. Draw a line segment AC = 4.2 cm.
2. With A as a centre and radius 2.5 cm, draw two arcs, one above AC and one below AC.
3. With C as a centre and radius 3.5 cm, two arcs are drawn intersecting previous arcs at Band D.
4.Join AB, BC, AD and CD. Thus, ABCD is the required quadrilateral.
5. taking A as a centre and radius 6.3 cm, draw an arc' which intersects AC produced at C'.
6. Through C', draw C'B' and C'D' parallel to CB and CD, respectively.
Hence, AB'C'D' is the required quadrilateral similar to quadrilateral ABCD.

Given, radius of circle = 5 cm
1.Draw a circle with 0 as centre and radius 5 cm.
2. Draw any chord PQ through the given point P on the circle.
3.Take a point R on the circle and join P and Q to a pointR.
4. Construct \(\angle QPY=\angle PQX\)  on the opposite side of the chord PQ.
5. Produce yP to X to get YPX, as the required tangent.

50 m

\(P\ge -\frac { 4 }{ 3 } \)

Since, the tangents drawn from an external point to a circle are equal. 
ஃ EK = EM = 9 cm 
⇒ ED + DK = EF + FM = 9 cm 
⇒ ED + DH  = EF + FH = 9 cm
Now, perimeter of \(\triangle\)EDF 
= ED + DF + EF 
= (ED + DH) + (EF + FH)
= 9 + 9 = 18 cm

Hypotenuse is the largest side in a right triangle.

10.38m

12m

52m

\(x^{2} - 25x + 150 = 0 , x = 15 , 10\)

Given: \(\angle \)EBD = 146°
To find: \(\angle \)x.
Proof: \(\angle \)DBC + \(\angle \)EBD = 180°   [linear pair]
⇒  \(\angle \)DBC + 146° = 180°
⇒ \(\angle \)DBC = 180° - 146° = 34°  ........(i)
⇒ \(\angle \)x = \(\angle \)DBC = 34°  [From (i)]
[Angles in the alternate segment]

S_p = ap² + bp
S₁ = a x 1² + b x 1 = a + b = a₁
S₂ = a x 2² + b x 2 = 4a + 2b
a₁ + a₂ = 4a + 2b
a + b + a₂ = 4a + 2b
 a₂ = 3a + b
Now, d = a₂ - a₁ = (3a + b) - (a + b) =2a.

OR bisects \(\angle \)PRQ \(\angle \)PRO = \(\angle \)QRO = 60^o
In right \(\triangle\)OPR (∵ OP⊥ PR)
\(PR\over OR\) = cos 60^o ⇒ OR = 2PR   ....(i)
Similarly \(QR \over OR\) = \(1\over2\)⇒ OR = 2QR   .....(ii)
(i) + (ii), we get
2OR = 2PR  + 2QR
⇒ OR = PR + RQ

The odd numbers between 0 and 50 are 1,3, 5,., 49 which form an AP.
Here, first term (a) = 1, last term (l) = 49
and common difference (d) = 3 - 1 = 2.
Let there be n numbers in the AP.
Then, nth term (a_n) = a + (n - 1) d = l
\(\Rightarrow\) 1 + (n - 1) (2) = 49
\(\Rightarrow\) n - 1 = 24 \(\Rightarrow\) n = 25
Now, sum of n terms, \(S_n=\frac{n}{2}(a+l)\)
\(\therefore\) Sum of 25 terms, \(\begin{aligned} S_{25} & =\frac{25}{2}(1+49)=\frac{25}{2} \times 50 \\ \end{aligned}\)
\(\begin{aligned} & =25 \times 25=625 \end{aligned}\)

Px(x-3)+9=.0 \(\Rightarrow \) px²-3px+p=0 Here a=p,b=-3p,c=9
For equal roots D=0 \(\Rightarrow D=b^{ 2 }-4ac\)
\(\Rightarrow (-3p)^{ 2 }-4\times p\times 9=0\quad \Rightarrow 9p^{ 2 }-36p=0\)
\(\Rightarrow 9p(p-4)=0\quad \Rightarrow 9p=0\)  or p=4
but p \(\neq 0\)  \([\therefore \) In quadratic equation a \(\neq \) 0 ]

Let Ist term of AP = a and common difference = d.
Now,    a₆ = -10 \(\Rightarrow\)  a +  5d  = -10                 ..(i)
Also,     a₁₀ = -26  \(\Rightarrow\)    a + 9d = -26           ...(ii)
Subtract (i) from (ii),    
a + 9d = - 26
a + 5d = -10
-  -          +
 4d = -16   \(\Rightarrow\) d = - 4
Substituting in (i), we get
a + 5 x ( -4 ) = - 10  \(\Rightarrow\) a = 10
Now,  a₁₅ = a + 14d = 10 + 14 X  - 4 = - 46

This is of the form ax²+bx+c=0, where a=4, b=4\(\sqrt { 3 } \) and c=3.
Discriminant(D)=b²-4ac=\({ \left( 4\sqrt { 3 } \right) }^{ 2 }\)-4 x 4 x 3=48-48=0
Roots are \(\alpha =\frac { -b+\sqrt { D } }{ 2a } =\frac { -4\sqrt { 3 } +0 }{ 8 } =\frac { -4\sqrt { 3 } }{ 8 } =\frac { -\sqrt { 3 } }{ 8 } \)
and \(\beta =\frac { -b-\sqrt { D } }{ 2a } =\frac { -4\sqrt { 3 } -0 }{ 8 } =\frac { -4\sqrt { 3 } }{ 8 } =\frac { -\sqrt { 3 } }{ 2 } \)
Hence, the roots are \(\frac { -\sqrt { 3 } }{ 2 } ,\frac { -\sqrt { 3 } }{ 2 } \)

Here, we have
a₂ - a₁ = -3.2 - (-1.2) = -3.2 + 1.2 = -2
a₃ - a₂ = -5.2 - (-3.2) = -5.2 + 3.2 = -2
a₄ - a₃ = -7.2 - (-5.2) = -7.2 + 5.2 = -2
and so on.
Since, the difference of any two consecutive terms same. Therefore, the given list of numbers forms a AP and its common difference (d) is - 2.
Now, next three terms of this AP are,
a₅ = a₄ + d = -7.2 + (-2) = -9.2
a₆ = a₅ + d = -9.2 + (-2) = -11.2
and a₇ = a₆ + d = -11.2 + (-2) = -13.2

Steps of Construction:
1. Draw a line segment AB of 7 cm
2. Taking A and B as centre draw two circle of 3 cm and 2 cm radius

3. Bisect the line AB. Let mid-point of AB is C.
4. Taking C as centre draw a circle of radius AC which will intersect the circle at point P, Q, R and S.
5. Join BP, BQ, AS and AR. These are the required tangents.

Here a = 101, d = 7, a_n = 997
\(\Rightarrow\) a_n = a + (n - l)d.
997 = 101 + (n - 1) X 7
\(\therefore\)  n = 129

Given, two concentric circles of radii 2.8 cm and 1.8 cm with common centre say, O
  
 Steps of Construction
1. Draw two circles with common centre 0 and radii 2.8 cm and 1.8 cm.
2. Take a point P on the outer circle and join OP.
3. Now, bisect OP. Let mid-point of OP be M.
4. Taking M as centre and PM as radius, draw a dotted circle, which intersects the inner circle at two points say A and B.
5. Join APand BP. Then, AP and BP are required tangents. On measuring the lengths, we get
 PA = PB
= 2.14 cm
Calculation
Join OA. Then, OA = 1.8 cm   [radius of inner circle C1]
and      OP = 2.8 cm    [radius of outer circle C2]
Hence,    \(\angle PAO=90°\)
  [angle in semi-circle of constructed circle]
So, in \(\Delta PAO\) , by Pythagoras theorem,
 OP²=OA² + AP²          => (2.8)² = (1.8)² + AP²
=> 7.84 = 3.24 +AP²      => AP² =7.84-3.24 = 4.6
=> AP= 2.14 cm            => PA = PB = 2.14 cm
Hence, the length of tangents is 2.14 cm

d = 6; Next two terms are 24 and 30

Given: Let PQ and GH are the common tangents to both circles and common tangent at C meets PQ at E and GH at F. 
To Prove: EF bisects PQ and GH. 
Proof: Tangents'drawn from an external point to a circle are equal. 
EP = EC and EQ = EC
⇒ EP = EQ
⇒ PQ is bisected by EF at E. 
Similarly, GH is bisected by EF at F. 
ஃ The common tangent EF drawn at 'C' bisects the other two common tangents. 

Let CD be a tower of height x m and BC is a pole and ㄥBAD=600 and ㄥPCA=450 ㄥPCA=ㄥCAD=45⁰
In right ΔCDA, \(\frac { CD }{ DA } \)=tan 45^o
⇒ \(\frac { x }{ DA } \)=1 ⇒ DA= x m
In right ΔBDA, \(\frac { BD }{ DA } \)=tan 60^o
⇒ \(\frac { 5+x }{ x } =\sqrt { 3 } \)
⇒ 5+x=\(\sqrt { 3 } \)x ⇒ 5=(\(\sqrt { 3 } \)-1)x
⇒ x=\(\frac { 5 }{ \sqrt { 3 } -1 } m=\frac { 5(\sqrt { 3 } +1) }{ 2 } m=\frac { 5(1.73+1) }{ 2 } m=\frac { 13.65 }{ 2 } \)m=6.82 m

Given: A circle with centre O. AB is the diameter of this circle. I is tangent to the circle. AH and BK are perpendicular to I from A and B at H  and K respectively. 
To prove: AH + BK = AB 
Proof: AH and HP are tangents from the external point H 
ஃ  AH = HP        .....(i)
and BK, KP are tangent from the external point K
ஃ   BK = KP      ......(ii)
Adding (i) and (ii) we get
AH + BK = HP + PK = HK ......(iii)
AB ⊥  AH
AB ⊥ BK
[Tangent makes 90° angle with radius at the point of contact]
⇒ \(\angle \)1 = \(\angle \)2 = 90°
Given that AH ⊥ i ⇒ \(\angle \)3 = 90°
and BK ⊥ i ⇒ \(\angle \)4 = 90°
∵ \(\angle \)1 = \(\angle \)2 = \(\angle \)3 = \(\angle \)4 = 90°​​​​​​​
⇒ AHKB is a rectangle
⇒  AB = HK        ......(iv)
[Opposite sides of a rectangle are equal]
From (iii) and (iv) ⇒ PH + PK = AB
AH + BK = AB from (i) and (ii) Hence proved.

Let Varun's present age be x years and Swati's present age be y years
Case I : 7 years ago
Varun's age was (x-7) years and Swati's age was (y-7) years
ATQ   (x-7) = 5(y-7)²   \(\Rightarrow x=5(y-7)^{ 2 }+7\)  ...(i)
Case II : 3 years hence 
Varun's age will be (x+3) years and Swati's age will be (y+3) years 
A.T.Q \(y+3=\frac { 2 }{ 5 } (x+3)\)
\(\Rightarrow y+3=\frac { 2 }{ 5 } \left[ 5(y-7)^{ 2 }+7+3 \right] \) [Using eq.(i)]
\(\Rightarrow y+3=\frac { 2 }{ 5 } \times 5(y-7)^{ 2 }+\frac { 2 }{ 5 } \times 10\Rightarrow y+3=22(y^{ 2 }-14y+49)+4\)
\(\Rightarrow y+3=2y^{ 2 }-28y+98+4\Rightarrow 2y^{ 2 }-29y+99=0\)
\(\Rightarrow 2y^{ 2 }-18y-11y+99=0\Rightarrow 2y(y-9)-11(y-9)=0\)
\(\Rightarrow (y-9)(2y-11)=0\Rightarrow y=9,y=\frac { 11 }{ 2 } \) (rejecting)
\(\therefore y=9\)  \(\therefore x=5(9-7)^{ 2 }+7\) [From (i)]
x = 27
Present age of Swati = 9 years and Varun =27 years.

Let numerator be
denominator is x + 2.
\(\Rightarrow fraction=\frac { x }{ x+2 } \)
\(\Rightarrow \frac { x }{ x+2 } +\frac { x+2 }{ x } =\frac { 34 }{ 15 } \)
\(\Rightarrow 15(x^{ 2 }+x^{ 2 }+4x+4)=34(x^{ 2 }+2x)\)
\(\Rightarrow 30x^{ 2 }+60x+60=34x^{ 2 }+68x\)
\(\Rightarrow 4x^{ 2 }+8x-60=0\)
\(x^{ 2 }+2x-15=0\)
\(\Rightarrow x^{ 2 }+5x-3x-15=0\)
\(\Rightarrow x(x+5)-3(x+5)=0\)
\(\Rightarrow (x+5)(x-3)=0\)
x=3
Fraction = \(\frac { 3 }{ 5 } \)

Given, OB = 5 cm and radius OA = 3 cm
Use Pythagoras theorem in right \(\Delta\)OAB to find AB
4 cm

(i) Station P should send the team, its distance covered 14.66km.

(ii) Presence of mind and ability to take prompt decisions.

Given equation is 
\((x+1)^{ 2 }=2(x-3)\)
\(\Rightarrow x^{ 2 }-2x=(-2)(3-x)\)
\([\because (a+b)^{ 2 }=a^{ 2 }+2ab+b^{ 2 }]\)
\(\Rightarrow x^{ 2 }+1+6=0\)
\(\Rightarrow x^{ 2 }+7=0\)
Which is of the form \(ax^{ 2 }+bx+c=0\) ,where \(a\neq 0andb=0\)
Hence,it is a quadratic equation

Since OABC is a rhombus. 
∴ OA = AB = BC = CO 
Also, OB = OA = OC [radii Of a circle] 
∴  We have OA = OB = AB 
and 0B = OC = BC 
⇒ \(\triangle\)OAB and \(\triangle\)OBC are equilateral triangles and OB is the diagonal of rhombus OABC. 
ஃ Area (rhombus OABC) = 2 x area (\(\triangle\)OAB) 
[.∴ diagonal divide rhombus in two triangles of equal areas] 
= \(​​\left( 2\times \frac { \sqrt { 3 } }{ 4 } \times { 10 }^{ 2 } \right) \)cm²
[ ∵ area of equilateral triangle = \(\frac { \sqrt { 3 } }{ 4 } \) (side)²]
= \(\left( 2\times \frac { \sqrt { 3 } }{ 4 } \times { 10 }0 \right) \)cm²
= 50\(\sqrt3\) cm²

Let AB be the tower of height h m and CD be the cliff of height 60\(\sqrt m\) m, such that 
\(\angle \)XDA = \(\angle \)DAE = 45^o and \(\angle \)XDB = \(\angle \)DBC = 60^o

In rt.  \(\triangle\)DCB,
\(DC\over BC\)=tan 60^o
\(\frac { 60\sqrt { 3 } }{ BC } =\sqrt { 3 } \)
BC = 60m
Now BC = AE = 60m
and DE = DC - CE (or AB)
= \(60\sqrt { 3 } -h\)
Again in rt. \(\triangle\)DEA
\(\angle \)E = 90^o
ஃ \(DE\over AE\)=tan 45^o
\(\frac { 60\sqrt { 3 } -h }{ 60 } =1\)
⇒ 60\(\sqrt3\) -  h = 60
⇒ h = 60\(\sqrt3\) - 60 
⇒ h = 60(\(\sqrt3\)-1)
Hence the required height of the tower is 60(\(\sqrt3\)-1)m

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

Number of houses preceding no.X
= X - 1
Sum of numbers of houses preceding  X
= 1 + 2 + 3 + ... + X - 1
Number of houses following X = 49 - X
Sum of numbers on the houses following
X = ( X + 1 ) + ( X + 2 ) + ... + 49
\(={49-X\over2}(X+1+49)\)
\(=\left( 49-X\over2 \right)[X+50]\)
\(={{2450-X-{X}^{2}}\over{2}}\)
Now, \({{{X}^{2}-X}\over{2}}={{2450-{X}-{X}^{2}}\over{2}}\)
\(\Rightarrow\)   2X² = 2450
\(\Rightarrow\)  X² = 1225
X = \(\sqrt{1225}\) = 35
\(\therefore\)     X = 35