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.

12

The justification will be given by using the basic proportionality theorem and similar triangles.

QM = 3cm, RN = 5cm and PL = 7cm

\({40\over\sqrt{3}}m\)

28 units, 7 units

Here, a_n = n²+1
Put n = 1, 2, 3, ..., we obtain
 a₁ = 1²+1 = 2
a₂ = 2²+1 = 5
 a₃ = 3²+1 = 10
Now, list of numbers becomes 2, 5, 10, ......
\(\therefore \) a₂ - a₁ = 5-2 = 3
a₃-a₂ = 10-5 = 5
Since 3 ​​​​\(\neq \) 5 i.e., a₂-a₁\(\neq \) a₃-a₂
Thus, it does not form an A.P.

Here, the height of the pole be h  m and length of its shadow is  \(\sqrt { 3 } \)  h .
Let ፀ be the elevation of the sun. Consider right-angle ΔPQR
tanፀ=\(\frac { h }{ \sqrt { 3 } h } =\frac { 1 }{ \sqrt { 3 } } \)
= tan 30\(\unicode{xb0} \)
ፀ = 30\(\unicode{xb0} \)
Hence, the angle of elevation of the sun is 30\(\unicode{xb0} \).

Let a and d be the first term and common difference of the required A.P.
Here,   a₅ = 22
\(\Rightarrow\)   a + 4d = 22                   ...(i)
And    a₉ = 6a₂
\(\Rightarrow\)   a + 8d = 6 ( a + d )
\(\Rightarrow\)   a + 8d = 6a + 6d
\(\Rightarrow\)  5a = 2d                  ...(ii)
From (i) and (ii), we have
a + 2(5a) = 22
\(\Rightarrow\)  11a = 22
\(\Rightarrow\)   a = 2
From (ii), we have 5(2) = 2d    \(\Rightarrow\)   d = 5
Hence, the required A.P. is 2, 7, 12, 17, ...

16, 14 - Creativity involves in the use of skills and imagination to produce something new and innovative.
Vision - ability to think about future.

13

52 m

9

Let AB is the tower, AB = 15 m, BC = 15 m 
In right \(\Delta\)ABC,  \(\tan { \theta =\frac { AB }{ BC } } \Rightarrow \tan { \theta } =\frac { 15 }{ 15 } \)
\(\Rightarrow\) \(\tan { \theta } =1 \Rightarrow \theta ={ 45 }^{ o }\)

Given: PQ is a tangent to the circle with centre O from a point P. QCR is a diameter of the circle
and \(\angle \)POR = 110°. 
To find: \(\angle \)OPQ 

Sol. POR = 110°
QR is the diameter of the circle.
⇒ \(\angle \)1 + \(\angle \)2 = 180°  [Linear pair axiom]
⇒ \(\angle \)1 + 110° = 180° ⇒ \(\angle \)1 = 70°
\(\angle \)OQP = 90°
(Tangent makes 90° angle with the radius at the point of contact).
In \(\triangle\)OPQ
\(\angle \)1 + \(\angle \)OQP + \(\angle \)QPO = 180°      [Angle sum properety]
⇒ 70° + 90° + \(\angle \)QOP = 180°
⇒  \(\angle \)OPQ = 180° - 160°
⇒ \(\angle \)OPQ = 20°

Let AD = AF = x    [Tangent form external point are equal]
BD = BE = y and CE = CF = z
According to the question,
AB = x + y = 24cm    ....(i)
BC = y + z = 16cm    .....(ii)
Ac = x + z = 20cm    ....(iii)
subtacting (iii) for (i), we get
y - z = 4   ......(iv)
Adding (ii) and (iv), we get
2y = 20 ⇒ y = 10 cm
Substituting the value of y in (ii) and (i)
we get z = 6 cm; x = 14 cm
ஃ AD = 14 cm, BE = 10 cm
and CF = 6 cm

PA = 12 cm, OA = 5 cm, OB = 3 cm
OP² = OA² + AP² = OB² + BP²
⇒ 25+14 = 9+BP2
⇒ 169-9 = BP²
⇒ BP = \(\sqrt{160}\) cm = 12.65 cm.(Approx.)

LHS=2x²+5x-12
When x=-4
LHS =2(-4)²+5(-4)-12
=32-20-12=0=RHS
x=-4 is a solution of the given equation.

\(kx(x-2\sqrt { 5 } )+10=0\)
\(\Rightarrow kx^{ 2 }-2\sqrt { 5 } kx+10=0\)
\(a=k,b=-2\sqrt { 5 } k,c=10\)
Given, roots are equal, D = b²- - 4ac = 0
\(\Rightarrow (-2\sqrt { 5k^{ 2 } } -4\times k\times 10=0\)
\(\Rightarrow 20k^{ 2 }-40k=0\)
\(\Rightarrow 20k(k-2)=0\)
k(k-2)=0
k=0
k=2

Here, a = 12, l = 264, d = 4
\(n=\frac { l-a }{ d } +1=\frac { 264-12 }{ 4 } +1\)
\(=\frac { 252 }{ d } +1=63+1=64\)
There are 64 multiples of 4 that lie between 11 and 266.

Given, sequence 3,6,9,12,...,111.
Here, 6 - 3 = 9 - 6 = 12 - 9 = 3
So, it is an AP with first term, a=3 and common difference, d = 3. Let there be n terms in the given sequence.
Then, nth term =1 11
⇒ a + (n-1) d = 111
⇒ 3 + (n-1)\(\times \) 3 = 111
⇒ 3 (1 + n-1) = 111
⇒ \(n=\frac { 111 }{ 3 } \Rightarrow n=37\)

Let the present age of Anjali be x yr.
Anjali's age 3yr ago=(x-3) yr
and Anjali's age 5 yr from now=(x+5)yr
According to the question,
\(\frac { 1 }{ x-3 } +\frac { 1 }{ x+5 } =\frac { 1 }{ 3 } \)
\( \Rightarrow \frac { x+5+x-3 }{ (x-3)(x+5) } =\frac { 1 }{ 3 }\)
\(\Rightarrow \frac { 2x+2 }{ { x }^{ 2 }-3x+5x-15 } =\frac { 1 }{ 3 } \)
\(\Rightarrow 3(2x+2)={ x }^{ 2 }+2x-15\)
\(\Rightarrow 6x+6={ x }^{ 2 }+2x-15\)
\( \Rightarrow { x }^{ 2 }-4x-21=0\)
which is the required quadratic equation.
Now, by factorization method, we get
x²-7x+3x-21=0
x(x-7)+3(x-7)=0
(x-7)(x+3)=0
x=7 or x=-3
But x=-3 is not possible because age cannot be negative.
x=7
Hence, Anjali's present age is 7 yr.

346.4 m/s

\(\frac { 20 }{ \sqrt { 3 } } \)

Tangent

r₁ = 5 cm, r₂ = ?,
AB = 24 cm
∵ AB is tangent to circle
C(C, r₁) at C
ஃ OC ⊥ AB
In circle C(O, r₂), AB is a chord and OC ⊥ AB
ஃ AC = BC
[∵ Perpendicular from the centre bisects the chord]
In right \(\triangle\)OCA
OC² + AC² = AC²
⇒ 5² + (12)² = (r₂)²
⇒ 25 + 144 = (r₂)² ⇒ (r₂)² = 169
r₂ = 13 cm

Let the two consecutive integers be x and x+1
ATQ x²+(x+1)2=365
\(\Rightarrow\) x²+x2+2x+1=365 \(\Rightarrow\) 2x²+2x-364=0
\(\Rightarrow\) x²+x-182=0 \(\Rightarrow\) x²+14x-13x-182=0
\(\Rightarrow\) x(x+14)-13(x+14)=0 \(\Rightarrow\) (x-13)(x+14)=0
\(\Rightarrow\) x=13, -14 (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and 13+1=14

8 cm and 8 cm

Given, in ΔBC, L.B = 90°
BC  = 6cm
and AB = 8 cm
By Pythagoras theorem,
AC² = AB² + BC²
= AC =\(\sqrt{AB^2+BC^2}\)
= \(\sqrt{8^2+6^2}\)
=\(\sqrt{64+36}\)
= \(\sqrt{100}\)
Let AC, AB and BC touch the circle at F, E and D, respectively.
Join OD and OE.
Now, in quadrilateral ODBE,
ㄥD = ㄥB = ㄥE = 90°
ㄥDOE = 90°
BD = BE = r cm
=> ODBE is a square of side r cm
Also, AF = AE
CF=CD
Now, AE = AF = AB - BE = 8 - r
and CF = CD = BC - BD = 6 - r
From Eq. (i), we get
AC = 10 cm
=> AF + CF = 10
=> 8- r + 6 -r = 10
=> 14 - 2r = 10
=> - 2r = 10 - 14 = - 4
=> r = 2
Hence, radius of circle is 2 cm.

\(\frac {n-1}{\sqrt{a_1}+\sqrt{a_n}}\)

29.28 m/s

Let AB=h m be height of tower and BC=30 m be length of its shadow when sun's elevation is 300.Let BD=x m be length of shadow when sun's elevation is 60^o
As ㄥACB=30^o and ㄥADB=60^o
Consider rt . ΔABC, we have
\(\frac { AB }{ BC } \)=tan300
⇒ \(\frac { h }{ 30 } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow h=\frac { 30 }{ \sqrt { 3 } } \)
⇒ AB=\(\frac { 30 }{ \sqrt { 3 } } \)m
Consider rt . ΔABD, we have
\(\frac { AB }{ BD } \)=tan600 ⇒ \(\frac { h }{ BD } =\sqrt { 3 } \)
⇒ \(\frac { \frac { 30 }{ \sqrt { 3 } } }{ BD } =\frac { \sqrt { 3 } }{ 1 } \)    [∵ h=\(\frac { 30 }{ \sqrt { 3 } } \)]
∴ Length of shadow=10 m

12 m, 7 m

Steps of Constructions:
1. Draw a line segment BC = 6 cm
2. Draw a perpendicular bisector of BCwhich cuts the line BCat Q.
3. Cut the line OA = 4 cm.

4. Join A to B and C
5. Triangle ABC is the given triangle.
6. Draw a ray BXmaking an acute angle.
7. Mark the four points B1, B2,B3and B4on the ray BX. Join B4C Draw a line parallel through B3 to B4 intersecting extended line segment AB at A'.
Hence  A'BC' is a required triangle.