\(\sqrt { (3-9)^{ 2 }+(k-2)^{ 2 } } =10\)
\(\left[ Dist\quad =\sqrt { (x_{ 1 }-x_{ 2 }) } ^{ 2 }+(y_{ 1 }-y_{ 2 } \right] \)
\(\Rightarrow (-6)^{ 2 }+k^{ 2 }-4k+4=100\)
\(\Rightarrow k^{ 2 }-4k+40=100\)
\(\Rightarrow k^{ 2 }-4k-60=0\)
\(\Rightarrow k^{ 2 }-10k+6k-60=0\)
\(\Rightarrow k(k-10)+6(k-10)=0\)
\(\Rightarrow (k-10)(k+6)=0\)
k=10,-6

Here a₁₃ = 4a₃
Let the firs term be a and the common difference be d.
a + 12d = 4(a + 2d)
\(\Rightarrow\)  3a = 4d          .....(i)
a₅ = 16
\(\Rightarrow\) a + 4d = 16     .....(ii)
Solving (i) and (ii), we get
a = 4 and d = 3
\(\\ { S }_{ 10 }=\frac { 1 }{ 2 } [2\times 4+(10-1)3]\)
= 5 [8 + 27] = 5 \(\times\)35
= 175

Given quadratic equation is px²+\(\left( \sqrt { 3 } -\sqrt { 2 } \right) x-1\)=0
and x=\(\frac { 1 }{ \sqrt { 3 } } \) is a root of this equation.
On putting x=\(\frac { 1 }{ \sqrt { 3 } } \)  in a given equation we get
\(p{ \left( \frac { 1 }{ \sqrt { 3 } } \right) }^{ 2 }+\left( \sqrt { 3 } -\sqrt { 2 } \right) \frac { 1 }{ \sqrt { 3 } } -1=0\)
\(\frac { p }{ 3 } +\frac { \sqrt { 3 } -\sqrt { 2 } -\sqrt { 3 } }{ \sqrt { 3 } } =0\)
\(\frac { p }{ 3 } -\frac { \sqrt { 2 } }{ \sqrt { 3 } } =0\)
\(p=\frac { \sqrt { 2 } }{ \sqrt { 3 } } \times 3=\frac { 3\sqrt { 2 } }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } }\)
\(=\frac { 3\sqrt { 2 } \times \sqrt { 3 } }{ 3 } =\sqrt { 2 } \times \sqrt { 3 } =\sqrt { 6 } \)
Hence, the value of p is \(\sqrt { 6 } \)

16.67 m

Given that, OP 10 cm and OQ = 8 cm 
and  RP 24 cm. Join OR. 

In  OPR, we have OP PR
OR = \(\sqrt { { PR }^{ 2 }+{ OP }^{ 2 } } \)
= \(\sqrt { { 24 }^{ 2 }+{ 10 }^{ 2 } } =\sqrt { 576+100 } \)
= \(\sqrt { 676 } =26\quad cm\)
In \(\triangle\)OQR, we have OQ ⊥ QR
OR² = RQ² + OQ²
⇒ RQ² = OR² - OQ²
= 26² - 8² = 676 - 64
= 612
⇒  RQ = \(\sqrt { 612 } =6\sqrt { 17 } cm\)

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.

The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :
23, 21, 19, . . ., 5
It forms an AP . Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = – 2, a_n = 5
As, a_n = a + (n – 1) d
We have,  5 = 23 + (n – 1) (– 2)
i.e.,   – 18 = (n – 1) (– 2)
i.e.,   n = 10
So, there are 10 rows in the flower bed.

Let A be the window and CE be the opposite house
CD=AB=60 m [Opposite side of a rectangle] ........(i)
In rt. ΔABC, tan45^o=\(\frac { 60 }{ BC } \)
⇒ 1=\(\frac { 60 }{ BC } \)
⇒ BC=60 m        .........(ii)
AD=BC [Opposite sides of a rectangle]
∴ AD=60 m [From (ii)]        ......(iii)
In rt. ΔADE, tan600=\(\frac { DE }{ AD } \)
⇒ \(\sqrt { 3 } =\frac { DE }{ 60 } \)    [From (iii)]
⇒ DE=60\(\sqrt { 3 } \) m
∴ Height of the opposite house
CE=CD+DE=60+60\(\sqrt { 3 } \)
=60(1+\(\sqrt { 3 } \))m.

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

\(\angle ROQ={ 180 }^{ ° }-{ 60 }^{ ° }={ 120 }^{ ° }\)
\(\angle QSR=\frac { 1 }{ 2 } \angle ROQ=\frac { 1 }{ 2 } \times { 120 }^{ ° }\)
\(={ 60 }^{ ° }\)

Given, PQ = 6.4 cm and
 \(\frac { QR }{ PQ } =\frac { 5 }{ 8 } or\quad \frac { QR }{ PR } =\frac { 5 }{ 3 } \)
Steps of Construction
1. Draw a line segment of length PQ = 6.4 cm.
2. Draw any ray QX making an acute \(\angle PQX\) with PQ.
3. Draw a ray parallel to Q)( by making \(\angle QPY=\angle PQX\)
4. Mark five points A₁ , A₂ , A _{3 ,} A₄ and A₅ on QX and three points B₁, B₂ , and, B₃ on PY such that
QA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = PB₁ = B₁B₂ = B₂B₃        
5.Join A₅ to B₃.
Then it intersect PQ at a point R.Thus, R is the point of dividing PQ such that \(\frac { QR }{ PQ } =\frac { 5 }{ 8 } or\quad \frac { QR }{ PR } =\frac { 5 }{ 3 } \) 

Let PR = 10 m be the length of rope and PQ = h m be the height of pole.
Given, angle of elevation \(\angle PQR=45°\)
In right angled \(\Delta PQR,\)
\(sin45°=\frac { PQ }{ PR } \Rightarrow \frac { 1 }{ \sqrt { 2 } } =\frac { h }{ 10 } \left[ \because sin45°=\frac { 1 }{ \sqrt { 2 } } \right]\)
\(\therefore h=\frac { 10 }{ \sqrt { 2 } } =\frac { 10 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =5\sqrt { 2 } m\)
Hence, the height of pole is \(=5\sqrt { 2 } m\).

As the tangents drawn from an external point to the circle are equal 
ஃ AR = AP = 2 cm 
BP = BQ = 3 cm 
CQ = CR = 4 cm
ஃ AB = AP + BP
= 2 + 3 = 5 cm 
BC = BQ + CQ 
= 3 + 4 = 7
CA = CR + AR 
= 4 + 2 = 6 cm
Perimeter of \(\triangle\)ABC 
= AB + BC + CA = 18 cm

False

4 cm

52m

Let us assume that
AB = h m be the height of the pole
As AC = 20 m is the length of the rope, and ㄥACB = 30^o (given)
Consider a rt, angle ΔABC, we have
sin 30^o =\(\frac { AB }{ AC } \)
⇒ \(\frac { 1 }{ 2 } =\frac { h }{ 20 } \)    (cross-multiply)
⇒ 20=2 x h
⇒ h=\(\frac { 20 }{ 2 } \)=10
⇒ h=10 m
Hence, the height of the pole is 10 m

77.94 m

23

6(a² + b² + 3ab)

Roots are real and distinct 

0

\(83\over149\)

Given quadratic equation  9x²-6x-2=0
Here a=9, b=-6, c=-2
D=b²-4ac
D=(-6)²-4X9X(-2)=36+72=108 > 0
Given quadratic has two unequal real roots.

7, 13, 19, …, 205
For this A.P.,
a = 7
d = a₂ − a₁ = 13 − 7 = 6
Let there are n terms in this A.P.
a_n = 205
We know that
a_n = a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, this given series has 34 terms in it.

\(\sqrt{85}\)

\(6\frac { 2 }{ 3 } \)cm

19.5 cm and 22.5 cm

Let AB= hm be height of the hill. BC = x km and BD = (x+1) km As ∠ACB = 60⁰ and ∠ADB = 30⁰ Consider rt. △ABC, we have
\(\frac{AB}{BC}\) =tan 60^o
\(\frac{h}{x}=\sqrt{3}\)
⇒ h=\(\sqrt{3}x\)  --- (i)

Consider rt. △ABD, we have
\(\frac{AB}{BD} \)=tan 30^o
⇒  \(\sqrt{3}\)h=x+1
⇒  \(\sqrt{3}\)ｘ\(\sqrt{3}\)x=x+1  [using(i)]
⇒ 3x=x+1
2x=1 ⇒ \(x=\frac{1}{2}\)
Putting the value of x in eq.(i), we have
h=\(\sqrt{3}\)ｘ\(\frac{1}{2}\)=\(\frac{1.732}{2}\)
=0.866 km
=866 metre
∴ Height of the hill = 866 metre.

Rs.800, Rs.750, Rs.700,........., Rs.450

Given: A boy is standing at a distance of 100 m from the bird flying at an elevation of 30^o. A girl is standing on the roof of 20 m high building finds the angle of elevation of the bird to be 45^o Boy and girl are on opposite side of the bird.
To find: Distance between the bird and the girl i.e., BE.
Solution: In ΔACB,
⇒ \(\frac { h }{ 100 } \) = sin 300 ⇒ h = \(\frac { 1 }{ 2 } \) x 100 = 50 m
⇒ BF = h - 20 = (50 - 20) m = 30 m
In ΔBFE,\(\frac { 30 }{ BE } =sin{ 45 }^{ 0 }\Rightarrow \frac { 30 }{ BE } =\frac { 1 }{ \sqrt { 2 } } \Rightarrow 30\sqrt { 2 } \) = BE
BE = 30 x 1.414 = 42.420 = 42.42 m

Given XY and X'Y' are two parallel tangents. Another tangent AB touches the circle at C and intersect XY at A and X'Y'at B.
To prove \(\angle\)AOB = 90°
Proof We know that, tangents drawn from an external point to a circle are equal in length.
\(\therefore\)   AP = AC   [\(\because\) A is an external point] ...(i)
Thus, in \(\Delta\)APO and \(\Delta\)ACO, AP = AC   [from Eq. (i)]
AO = AO  [common sides]
OP = OC [radii of circle]
\(\Delta\)APO \(\cong\)\(\Delta\)ACO  [by SSS congruence rule]
Then, \(\angle\)OAP = \(\angle\)OAC    [by CPCT]        ...(ii)
\(\Rightarrow\) \(\angle\)PAC = 2 \(\angle\)CAO                             ....(iii)
Similarly, we can prove that \(\angle\)CBO = \(\angle\)OBQ
\(\Rightarrow\)   \(\angle\)CBQ = 2 \(\angle\)CBO                           ...(iv)
since, XY || X'Y'      [given]
\(\therefore\) \(\angle\)PAC +  \(\angle\)QBC = 180°
[\(\because\) sum of interior angles on the same side of transversal is 180°]
\(\Rightarrow\) 2 \(\angle\)CAO + 2 \(\angle\)CBO = 180°    [from Eqs. (iii) and (iv)]
\(\Rightarrow\)   \(\angle\)CAO + \(\angle\)CBO = 90°                    ...(v)
Now, in \(\Delta\)AOB, \(\angle\)CAO + \(\angle\) CBO + \(\angle\)AOB = 180°
[by angle sum property of triangle]
\(\Rightarrow\) \(\angle\)CAO + \(\angle\)CBO = 180° - \(\angle\)AOB    ...(vi)
From Eqs. (v) and (vi), we get
180° - \(\angle\)AOB = 90° \(\Rightarrow\) \(\angle\)AOB = 90° Hence proved.

7.75 cm, square

Here AC is the diameter of the circle.
∴  ∠ABC = 90°     [ Angle in a semi-circle]
Now, in ΔACB, ∠A + ∠B + ∠C = 180°      [Sum of all interior angles of a triangle is 180°]
⟹     ∠A + 90° + 50° = 180°
⟹     ∠A + 140 = 180
⟹    ∠A = 180^o - 140° = 40°
Or     ∠OAB = 40°            …(i)
Here, OA ⏊ AT
⟹    ∠OAT = 90°          
⟹    ∠OAB + ∠BAT = 90°
⟹    ∠BAT = 90° - 40° = 50°    [Using (i)]
Hence, the value of ∠BAT is 50°.