The steps of construction are as follows :

1. Construct the triangle ABC in which AB = 6 cm, BC = 8 cm and \(\angle B=90^{ 0 }\)
2. Draw a perpendicular BD from B to AC.
3. Bisect BC at E. 
4. With E as centre and radius EC, draw a circle which will pass through B, C and O.
5. Join AE.
6. Bisect AE at F.
7. With F as centre and radius FE, draw a circle which intersects the circle (with centre E) at P and B.
8. Join AP and AB. and AP and AB are the two required tangents.

Given, AD = 8 cm, AC = 6 cm and BT = 4 cm
⇒ \(\angle\)CAD = 90° [angle in a semi-circle]
So, in \(\angle\)ACD,
CD² = AC² + AD² = 36 +64 = 100
[by Pythagoras rheorerrf]
⇒ CD =10cm
Therefore, OC = OD = OB =\({10\over 2}cm\) = 5 cm
Now, ㄥOBT = 90°
[angle between radius and tangent]
So, in ΔOBT, OT² = OB² + BT² = 25 + 16 = 41
[by Pythagoras theorem]
=> OT = \(\sqrt{41}cm\)

Given,   \(\angle PCA={ 110 }^{ \circ }\)
Join CO and a tangent PC.
   \(\because\)         \(\angle PCA={ 110 }^{ \circ }\)
Then  \(\angle ACT={ 180 }^{ \circ }-{ 110 }^{ \circ }={ 70 }^{ \circ }\)  [linear pair]
But  \(\angle OCT={ 90 }^{ \circ }\)  [radius and tangent]
So,  \(\angle OCA={ 90 }^{ \circ }-{ 70 }^{ \circ }={ 20 }^{ \circ }\) \(\left[ \because OA=OC \right] \)
Hence,  ​​\(\angle OAC=\angle OAC={ 20 }^{ \circ }\)
   \(\because\)  \(\angle CBA={ 90 }^{ \circ }\angle OAC\)
\(={ 90 }^{ \circ }-{ 20 }^{ \circ }={ 70 }^{ \circ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \because\angle ACB={ 90 }^{ \circ } \right] \)     
Again,  \(\angle AOC={ 180 }^{ \circ }\angle OCA-\angle OAC \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[angle\ \ sum \ \ property \ \ of \ \ a \ \ triangle \right] \)
\(\Rightarrow\)        ​​​​​​ ​\(\angle AOC={ 180 }^{ \circ }-{ 20 }^{ \circ }-{ 20 }^{ \circ }={ 140 }^{ \circ }\)
Now,     \(\angle BCO=\angle OBC=\angle CBA\)  \(\left[ \because OB=OC \right] \)
So,   \(\angle BCO={ 70 }^{ \circ }\) 

Let O be the centre of a circle of radius 10 cm.
Here, chord AB = 16 cm
Since OP bisects chord AB
ஃ AL = LB = 8 cm
Consider rt, \(\angle \)ed \(\triangle\)OLB, by using Pythagoras Theorem,
we have OL² = OB² - LB²
= 10² - 8² = 100 - 64 = 36
OL = 6 cm
Let PL be x cm
ஃ   OP = OL + PL = (6+x)cm
In rt ed BLP, we have
PB² = BL² + PL²
PB² = 8² + x²       ........(i)
⇒  x² = PB² - 6⁴
Again, in rt, \(\angle \)ed \(\triangle\)OBP, we have
PB² = OP² - OB²
PB² = (6 x)² - 10²   ......(ii)
From (i) and (ii), we obtain
(6+x)² - 100 = 64 + x² 
36 + x² + 12x - 100 - 64 - x² = 0
12x - 128 = 0
\(x={128\over12}={32\over3}cm\)
From (i), we have
PB² = 64 + \(({32\over3})^2\)
PB² = \(64+{1024\over9}={576+1024\over9}={1600\over9}\)
PB = \(40\over3\) cm

Since α and β are roots of x²-7x+12=0
∴ Sum of the roots, α+β=7 and 
Product of roots (αβ)=12
Sum of roots=(α+2)+(β+2)
=α+β+4
=7+4=11
Product of roots=(α+2)(β+2)
=αβ+2(α+β)+4
=12+2(7)+4
=12+14+4=30
Thus, the required equation in x is
x²-11x+30=0

52m

Let AB be the tower of height h m, P and Q be the two points in line with the foot of the tower AB, such that
ㄥAPB=∝ and ㄥAQB=β
Consider rt. ㄥed ΔQBA,
\(\frac { AB }{ QB } \)=tan β
QB=AB.\(\frac { 1 }{ tan\beta } \)
=AB cot β....(i)
Again consider rt ㄥed ΔPBA,
\(\frac { AB }{ PB } \)=tan α
\(\frac { AB }{ PQ+QB } \)=tan α
⇒ PQ+QB=AB.\(\frac { 1 }{ tan\alpha } \)
=AB cot α....(ii)
Subtracting (i) from (ii), we obtain
PQ=AB cot α - AB cot β
PQ=AB (cot α - cot β)
PQ=h(cot α - cot β)

Let a be the first term and d be the common difference og given A>P
a₅ = a + 4d = 19 ......... (i)
And a₁₃ - a₈ = 20
a + 12d - a - 7d = 20
5d = 20
d = 4
From (i), we have
a + 4(4)=19
a=19 - 6 = 3
Hence,the required A.P is 3,7,11,15,.....

\(x = {4 \over 3}\) or 3

Two-digit numbers divisible by 5 are 10, 15, 20,....., 95
     a_n = a + (n - 1)d
\(\Rightarrow\)  95 = 10 + (n - 1)5
\(\Rightarrow\)  95 = 10 + 5n - 5
\(\Rightarrow\)  5n = 90
\(\Rightarrow\)    n = 18

77.94 m

109.9 m

7 m

\(K = {-9\over 2}\)

Steps of Construction:
1. Draw a circle C' with the help ofa bangle, for finding the centre, take three non collinear points A, B and C, lying on the circle. Join AB and BC and draw perpendicular bisector of AB and BC, both intersect at a point O, 'O' is centre of the circle.
2. Take a point P outside the circle. Join OP.
3. Draw perpendicular bisector of OP, which intersects OP at point O'.
4. Take O' as the centre with OO' as radius draw a circle which passes through O and P, intersecting previous circle at points R and Q.
5. Join PQ and PR.
6. PQ and PR are the required pair of tangents.

Justification:

Join OQ and OR.
In ΔOQP and ΔOPR
OQ = OR   [Radii of the circle]
OP = OP   [Common]
ㄥQ = ㄥR = 90^o   [Radius is 丄 to tangent]
ΔOQP ≅ ΔORP   [by RHS]
PQ = PR
A pair Of tangents can be drawn to a circle from an externar point lying outside the circle.
These two tangents are equal in lengths.
∴ PQ = PR

x²+17x+c=0
Roots are -2 and -15
\(\Rightarrow \) (-2)²+17x(-2)+c=0
\(\Rightarrow \) 4-34+c=0 \(\Rightarrow \) c=30
Original equation, x²_{+13x+30=0
\(\Rightarrow \)} x²+10x+3x+30=0
_{\(\Rightarrow \)} x(x+10)+3(x+10) =0
_{\(\Rightarrow \)} x(x+10) + 3(x+10)=0
_{\(\Rightarrow \)} x=-10 or x=-3
Hence original roots are -10 and -3

Let breadth of a rectangular mango grove be x m
Then, length of a rectangular mango grove = 2x m
According to the question,
Area of rectangular mango grove = 800 m²
\(\Rightarrow\) 2x(x) = 800  [\(\because\) Area = length \(\times\) breadth]
\(\Rightarrow\) 2x² = 800 \(\Rightarrow\) x² = 400
\(\Rightarrow\) x = \(\pm\)20
But, x = -20 is not possible because breadth can never be negative. so, x = 20.
Thus, length = 2x = 40 m and breadth = 20 m.

a = 25, d = 50 - 25 = 25, x = 1000
A.T.Q.,         a_k = x
\(\Rightarrow\)    a + (k - 1)d = 1000
\(\Rightarrow\) 25 + (k - 1)25 = 1000
\(\Rightarrow\)  (k - 1)25 = 975
\(\Rightarrow\)   k - 1  = \(\frac{975}{25}\)
\(\Rightarrow\)   k - 1 = 39
\(\Rightarrow\)  k = 40

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 \(\sqrt { 6 } \)

12 cm

1268 m

36yr, 9yr

(i) 14.66 km
(ii) Presence of mind, ability to take promote decisions.

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. 

Given : A ΔABC, in which AB = 5 cm, BC = 6 cm and CA = 7 cm. 

Required: ΔA'BC' ∼ ΔABC with \(\frac { 4 }{ 5 } \) (reduced) scale-factor.
Steps of Construction : 
1. Construct ΔABC, such that AB=5cm, BC = 6 cm and CA = 7 cm. 
2. Through B, construct an acute ㄥCBX, 90^o).
3. Mark five points on BX such that BB₁ = B₁B₂ = B₂B₃ - B₃B₄ = B₄B₅.
4. Join B₅C. 
5. Through B₄, draw B₄C'||B₅C, intersecting BC in C'. 
6. Through C', draw C'A'||CA, intersecting BA in A'.
Hence, ΔA'BC' is the required triangle.

Here a=-2, d=-7-(-2)=-5 and a_n= -77
\(\therefore \) a+(n-1)d=a_n
\(\Rightarrow \quad -2+(n-1)\quad (-5)=-77\\ \Rightarrow \quad (n-1)(-5)=-75\\ \Rightarrow \quad n-1=15\\ \Rightarrow \quad n=16\)
Now,  \({ S }_{ 16 }=\frac { 16 }{ 2 } (a+l) \)
\(\\=8[-2+(-77)]\\ =8(-79)=-632\quad \)
Hence, -77 is the 16th term and sum of these 16 terms are -632.

25 terms

Let SQ be the height and PQ be the shadow of a man.
According to the question, SQ = PQ

Again, let the angle of elevation of the sun be \(\theta \) 
In right angled \(\\ \Delta PQS\)
\(tan\theta =\frac { perpendicular }{ Base } =\frac { QS }{ PQ }\)
\( \Rightarrow tan\theta =\frac { QS }{ QS } \left[ \because PQ = QS \right]\)
\( \\ \Rightarrow tan\theta =1=tan 45° \left[ \because tan 45°=1 \right] \)
\( \therefore \theta =45°\)
Hence, the angle of elevation of the sun is 45\(°\)

Since, AD is a diameter of a circle, so AD is perpendicular to the tangent AB.
ㄥDAB =90°
In ΔABD,
ㄥDAB + ㄥABD +ㄥADB = 180°
90° + 50° + ㄥADB = 180°
ㄥADB = 180° -140° = 40°
In ΔODC,
OD = OC [same radii of circle]
ㄥOCD = ㄥCDO = 40°
ㄥDOC + ㄥOCD + ㄥCDO = 180°
ㄥDOC + 40° + 40° = 180°
ㄥDOC = 1,00°
Since, AD is a straight line.
ㄥDOC +ㄥCOA = 180°
100° +ㄥCOA = 180°
ㄥCOA = 80°

Given AP is 34+32+30+...+10.
Here, first term, a = 34
Common difference, d = 32 - 34 = -2 and last term, \({ l }={ a }_{ n }=10\) 
\(\because\) \({ a } _ { n } = a + ( n-1 ) d\)
\(\therefore\) 10 = 34 + (n-2) (-2)​​​​​​​
\(\Rightarrow\) (-2) (n-1) = 10 - 34​​​​​​​
\(\Rightarrow\) (-2) (n-1) = -24​​​​​​​
\(\Rightarrow\)  n-1 = 12​​​​​​​
\(\Rightarrow\) n =  12 + 1 = 13​​​​​​​
\(\because\) Sum of n terms of an AP, \({ S } _ { n } = \frac { n } { 2 } { (a + 1) }\)
\(\therefore\)  \({ S } _ { 13 } = \frac { 13 } { 2 } (34 +10) = \frac { 13 }{ 2 } \times 44\)
​​​​​​​\(= 13 \times 22 = 286\)

kx(x - 2) + 6 = 0
or kx² - 2kx + 6 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = k, b = - 2k and c = 6
= ( - 2k)² - 4 (k) (6)
= 4k² - 24k
For equal roots,
b² - 4ac = 0
4k² - 24k = 0
4k (k - 6) = 0
Either 4k = 0 or
k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x²' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.

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.