Let AB be the tree whose part AC breaks and touches the ground at D
Then,  BD = \(2\sqrt { 3 } m\)
and AC = CD
In right angled \(\Delta \)CBD,
\(cos 60°=\frac { B }{ H } =\frac { BD }{ CD } \)
\(\Rightarrow \frac { 1 }{ 2 } =\frac { 2\sqrt { 3 } }{ CD } \)
\(\left[ \because cos 60°=\frac { 1 }{ 2 } and BD=2\sqrt { 3 } m \right] \)

\(\Rightarrow\) \(CD=2\times 2\sqrt { 3 } =4\sqrt { 3 } \)
\(=4\times 1.732=6.928\ m [\because \sqrt { 3 } =1.732]\)
\(\therefore\)  AC = CD = 6.928 m
Again, in right angled \(\Delta \)CBD,
\(tan\quad 60°=\frac { P }{ B } =\frac { BC }{ BD } \)
\(\Rightarrow \sqrt { 3 } =\frac { BC }{ 2\sqrt { 3 } } [\because tan60°=\sqrt { 3 } and BD=2\sqrt { 3 } m]\)
\(\Rightarrow BC=\sqrt { 3 } \times 2\sqrt { 3 } =6 m\)
Now, AB = AC + BC
= 6.928 + 6 = 12.928 m (approx)
Hence, the height of the tree is 12.928 m.

(i) Given, PA = PB = 8 m
We know that, tangents drawn from an external point to the circle are equal in length.

So, PA and PB are tangents.
Now, draw OP which bisects ㄥAPB and perpendicular to the chord AB.
ㄥAPC = ㄥBPC = 30°
and ㄥACP = ㄥBCP = 90°
In ΔACP, ㄥAPC + ㄥACP + ㄥPAC = 180°
30° + 90° + LPAC = 180°
ㄥPAC = 180° -120° = 60°
Similarly, ㄥPBC = 60°
Thus, ΔAPB is an equilateral triangle.
AB = AP = BP = 8 m
(ii) Given, OC = 3 m
We know that, if a perpendicular drawn from the centre of the circle to the chord of that circle, then it bisects the chord.
∴ \(AC=BC={AB\over 2}={8\over 2}=4\)
In right angled ΔACO,
OA²=AC²+OC²
OA²=(4)⁴+(3)²
⇒16+9=25
⇒ OA=5m
which is the radius of the circle.
(iii) Pythagoras theorem
(iv) Creating awareness and taking initiative.

Let AB=80 m be the width of the road. On both sides of the road, poles AE = BD = h m are standing. Let C be any point on AB such that from point C, angles of elevation are

\(\angle BCD=60°\), and \(\angle ACE=30°\).
Let BC = x m.
Then, AC = AB - BC = (80 - x)m
In right angled \(\Delta\)CAE,  \(\tan 30^{\circ}=\frac{P}{B}=\frac{A E}{A C}\)
\(\begin{aligned} & \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{80-x} \quad\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right] \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad 80-x=h \sqrt{3} \Rightarrow h \sqrt{3}+x=80 \end{aligned}\)  ...(i)
and in right angled \(\Delta\)CBD,
\(\tan 60^{\circ}=\frac{B D}{B C} \Rightarrow \sqrt{3}=\frac{h}{x} \quad\left[\because \tan 60^{\circ}=\sqrt{3}\right]\)
\(\Rightarrow \quad b=\sqrt{3 x}\)    ...(ii)
On putting h=\(\sqrt3\)x in Eq. (i), we get
\(\begin{aligned} &\sqrt{3} x(\sqrt{3})+x=80 \Rightarrow 3 x+x=80\\ \end{aligned}\)
\(\Rightarrow\) \(\begin{aligned} &4 x=80 \Rightarrow x=20 \mathrm{~m} \end{aligned}\)
On putting x = 20 m in Eq. (ii), we get
\(h=20 \sqrt{3} \mathrm{~m}\)
Now, AC = 80 - x = 80 - 20 = 60 m
Hence, height of the poles is 20\(\sqrt3\)m and the distances of the point C from the poles are 60 m and 20 m.

In the given figure, AB is the height of the pole and AC = 20 m is the length of rope which is tied from the top of the pole to the ground at point C.
In right angled \(\Delta ABC\) ,
\(sin30°=\frac { AB }{ AC } =\frac { AB }{ 20 } \)
⇒ \(\frac { 1 }{ 2 } =\frac { AB }{ 20 } \Rightarrow AB=\frac { 20 }{ 2 } =10\quad m\)
Hence, the height of the pole is 10 m.

- 5 + ( -8 ) + (-11) + .... + (-230)
Here,    a = -5, a_n = -230
d = - 8 - ( - 5) = -3
an = a + ( n - 1) d
\(\Rightarrow\)   -230 = -5 + ( n - 1 ) ( -3 ) \(\Rightarrow\) \(\frac{-225}{-3}\) = n -1
\(\Rightarrow\)  75 + 1 = n or n = 76
\(\therefore\)  S₇₆ = \(\frac{76}{2}[-5+(-230)]=38\times(-235)=-8930\) 

Let original duration of the flight be x hours.
 Distance = 2800 km 
\(\therefore \) Usual method = \(\frac { 2800 }{ x } km/h\)
When time = \(\left( x+\frac { 1 }{ 2 } \right) \) hrs
The new speed = \(\frac { 2800 }{ x+\frac { 1 }{ 2 } } =\frac { 5600 }{ 2x+1 } \)
ATQ \(\frac { 2800 }{ x } -\frac { 5600 }{ 2x+1 } =100\quad \Rightarrow \frac { 2800(2x+1)-5600x }{ (2x+1)x } =100\)
\(\Rightarrow 2800=100(2x^{ 2 }+x)\Rightarrow 2x^{ 2 }+x-28=0\)
\(\Rightarrow 2x^{ 2 }+8x-7x-28=0\quad \Rightarrow 2x(x+4)-7(x+4)=0\)
\(\Rightarrow (x+4)(2x-7)=0\quad \Rightarrow x=-4\) (rejected) or x=\(\frac { 7 }{ 2 } =3\frac { 1 }{ 2 } \)
\(\therefore \) Original duration = \(3\frac { 1 }{ 9 } \) hours

OP = Radius of the circle OA = 5 cm; AP = 4 cm
o [RAdius and tangent are ⊥ ar]
OA² = AP² + OP²   [By pythagoras theorem]
5² = 4² + OP²
⇒ 25 = 16 + OP²  ⇒ 25 - 16 = OP² ⇒ 9 = OP² ⇒ OP = \(\sqrt9\) = 3
Radius = 3 cm

111222333

Given: A right triangle with sides of length 5 cm and 4 cm making a right angle. 

Required: Triangle whose sides are \(\frac { 3 }{ 5 } \)  times the corresponding sides of the 1st triangle. 
Steps of Construction : 
1 . Construct a right triangle ABC right-angled at B with sides BC= 5 cm and AB=4 cm. 
2. Through B, construct an acute angle CBX, (<90^o). 
3. Mark Five points B₁, B₂, B₃,B₄,B₅ such that BB₁ =B₁B₂=B₂B₃=B₃B₄=B₄B₅.Join B₃C.
4. Through B₃, draw B₃C'||B₅C, intersecting BC in C'.
5. Through C', draw C'A'||CA, intersecting AB in A'.
Hence, ΔA 'BC' is the required triangle.

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.

\([-1, 5]\)

\(\left[{2\over 5 } , {2\over 5}\right]\)

Let C be the position of peacock and A be the position of serpent.

Given, ∠DCA = 30°
\(\Rightarrow \angle B A C=\angle D C A=30^{\circ}\) [alternate angles]
∵ Distance = Speed x Time
∴ \(A C=300 \ \frac{12}{60}\left[\because 12 \mathrm{~s}=\frac{12}{60} \mathrm{~min}\right]\)
\(\Rightarrow A C=60 \mathrm{~m}\)
In right angled ΔABC,
\(\sin 30^{\circ}=\frac{P}{H}=\frac{B C}{A C}=\frac{h}{60}\)
30 m.

87

In rt. ΔABC, tan60^o = \(\frac { AB }{ BC } =\frac { h }{ x } \)
⇒ \(\sqrt { 3 } =\frac { h }{ x } \Rightarrow h=\sqrt { 3 } x\)
In rt. ∆ABD, tanፀ = \(\frac { AB }{ BD } \)
⇒  tanፀ = \(\frac { h }{ 3x } \)
⇒  tanፀ = \(\frac { \sqrt { 3 } x }{ 3x } =\frac { 1 }{ \sqrt { 3 } } \)⇒ ፀ = 30^o​​​

Draw OC ⊥ AB and join OP, OT and AO. in right OCP.

OP² = PC² + OC²
OP²=[AP+AC]² + OC²
OP²=\({ \left[ 4+\frac { 1 }{ 2 } AB \right] }^{ 2 }\) + OC²
[∵ OC⊥AB,  ஃ AC=BC]
⇒ OP2=\({ \left[ 4+\frac { 1 }{ 2 } AB \right] }^{ 2 }\)+OC2
⇒ OP2 = \({ \left( \frac { 13 }{ 2 } \right) }^{ 2 }\)+OC2 ......(i)
In right OCA
OA² = OC² + AC²
OA² - AC² = OC²
OA² - AC² = OC²
OA² - \({ \left( \frac { 13 }{ 2 } \right) }^{ 2 }\) + OC²   .....(ii)
ஃ  eq (i) becomes.   ......(ii)
OP² = \({ \left( \frac { 13 }{ 2 } \right) }^{ 2 }\) + OA² - \({ \left( \frac { 5 }{ 2 } \right) }^{ 2 }\)
OP² = \(\frac { 169 }{ 4 } -\frac { 25 }{ 4 } \)+OA2
OP² = \(144\over4\)+OA²
⇒   OP² = 36 + OA²     ....(iii)
Also,   OP² = OT² + PT²  ......(iv)
From (iv) and (iii)
PT² + OT² = 36 + OA²
⇒   PT² = 36   [∵ OT = OA = radii]
PT = 6 cm

(i) Draw a line segment BC = 5.5 cm.
(ii) Perpendicular bisector XY of BC is drawn intersecting BC at L.
(iii) Point A is marked on XL such that AL=3 cm
(iv) AB and AC are joined.
(v) An acute ㄥCBP is drawn below BC.
(vi) On BP, points B₁ ,B₂ ,B₃ and B₄ are marked such that BB₁=B₁B₂=B₂B₃=B₃B₄
(vii) B₄C is joined
(viii) B₃C' is drawn parallel to B₄C interesting BC at C'
(ix) C'A' is drawn parallel to CA intersecting BA at A'

Here,    AB = AC (Given)      .......(i)
AF = AE (Tangent from A)    ......(ii)
AB - AF = AC - AE
⇒ BF = CE
Now,   BF = BD  (Tangent from B)
Also,  CE = CD  (Tangent from C)
⇒ BD = CD

Steps of construction:
1. Draw a line segment AB = 7
1. Draw a line segment AB = 7 cm.
2. Draw any ray AX making an acute angle with AB.
3. Draw the point A ll  such that
\(A_{ 1 }A_{ 2 },A_{ 3 }.....A_{ 6 }\)
4.Join BA_g
5.Through the point A31 draw a line parallel to BAs.

Then AP : PB = 3 : 5

\(4x^{ 2 }+4bx+b^{ 2 }-a^{ 2 }\)
\(\Rightarrow (2x+b)^{ 2 }-a^{ 2 }=0\)
\(\Rightarrow (2x+b+a)(2x+b-a)=0\)
\(\Rightarrow x=\frac { \left( a+b \right) }{ 2 } ,x=\frac { a-b }{ 2 } \)

Here a = 18, d = - 2, S_n = 0 
Therefore n/2 [36 + (n -1) (- 2)] = 0 
\(\Rightarrow\)n = 19

Let 1^st term be a and common difference be d
a₂ = a+d
a + d = 8 ...........(i)
a_s = a + 4d
a + 4d = 17 .......(ii)
From (i) and (ii),
a = 5, d = 3,  
a₁₉= a + 18d
= 5 + 54 = 59 

Here, 
\(a=\sqrt { 7 } ,a+d=\sqrt { 28 } \)
\(d=\sqrt { 28 } -\sqrt { 7 } =2\sqrt { 7 } -\sqrt { 7 } \)
\(=\sqrt { 7 } \)
\(\Rightarrow Next\quad term=\sqrt { 7 } +(4-1)\sqrt { 7 } \)
\(\Rightarrow =\sqrt { 7 } (4)\)
\(\Rightarrow =\sqrt { 7\times 16 } \)
\(=\sqrt { 112 } \)

Let Sanjay's present age = x
Sanjay's age 27 yr hence = (x + 27)
29yr ago = (x - 29)
37 yr.

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

We know that, length of the tangents from an external to a circle point are equal. 
⇒ BP = BQ = 27 cm and 
BC = 38
⇒ BQ + CQ = 38 
CQ = 38 - BQ = 38 -27 = 11 cm 
Now, CR = CQ = 11 cm 
Join OR ⇒ OS ⊥ AD and OR ⊥ DC 
⇒ DROS is a square of side 10 cm. 
therefore, DR = 10 cm 
Thus, CD = CR + DR = 11 + 10 = 21 cm 

\(10\sqrt{3}m\)

28 units, 7 units

\(x^{2} + 14x - 392 = 0\) ; Never leave your old parents alone.

Here, \(\frac { 3+5+7+...to\quad n\quad terms }{ 5+8+11+...to\quad 10\quad terms } =7\)
\(\Rightarrow \frac { \frac { n }{ 2 } \left\{ 2(3)+(n-1)2 \right\} }{ \frac { 10 }{ 2 } \left\{ 2(5)+(10-1)3 \right\} } =7\)
\(\Rightarrow \frac { n\left\{ 6+2n-2 \right\} }{ 10\left\{ 10+27 \right\} } =7\)
\(\Rightarrow \frac { 2n^{ 2 }+4n }{ 370 } =7\)
\(\Rightarrow \) 2n²+ 4n - 2590 = 0
\(\Rightarrow \) n² + 2n - 1295 = 0
\(\Rightarrow \) (n + 37) (n - 35) = 0
\(\Rightarrow \) n = 35 or n = 37
rejecting n = - 37
\(\therefore \) No of terms cannot be negative.
We have n = 35

\({15\over2}{\sqrt {3}} \ m \)

\({3\over16}\)

\(35\left( 1+\sqrt { 2 } \right) m\)

Steps of Construction: 
(i) A line segment BC = 9 cm is drawn. 
(ii) ㄥABC = 60^o is constructed at B.
(iii) An arc of 6 cm radius to be drawn with B as centre, cutting BA at A.
(iv) A and C are joined. Then triangle ABC is constructed.
(v) An acute angle CBX is drawn below BC.
(vi) Points B₁,B₂,B₃ are taken on BX, such that BB₁= B₁B₂ = B₂B₃.
(vii) B₃ and C are joined. 
(viii) B₂C' is drawn parallel to B₃C, meeting BC at C' 
(ix) C'A' is drawn parallel to CA, meeting BA at A'.
(x) A' BC' is the required triangle similar to ΔABC whose sides are \(\frac { 2 }{ 3 } \)  of the corresponding sides of Δ ABC.

Given: PAQ is a tangent to the circle with centre O at a point A as shown in figure \(\angle \)OBA = 35^o

To find: \(\angle \)BAQ and \(\angle \)ACB 
Proof: OA = OB  [Radii of the same circle]
⇒ \(\angle \)3 = 35^o      [Angles opposite to equal sides of a triangle are equal]
but,   \(\angle \)1 + \(\angle \)2 + \(\angle \)3 = 180^o [Angle sum property]
⇒ 35^o + 35^o + \(\angle \)2 = 180^o

Steps of Construction: 
1. Draw a line segment BC = 6 cm and at point B draw a ㄥABC = 60^o. 
2. Cut AB 5 cm. Join AC. We obtain ABC is triangle. 
3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A. 
4. Locate 4 points A₁,A₂,A₃ and A₄ on the ray BX so that BA₁=A₁A₂=A₂A₃=A₃A₄.
5. Join A₄ to C. 
6. At A₃ draw A₃C' || A₄C. Where C' is a point on the line segment BC.
7. At C' draw C'A' || CA, where A' is a point on the line segment BA.

Δ A'BC' is the required triangle.
Justification:
In Δ A'BC' and ΔABC A'C' || AC
∴ By BPT \(\frac { A'B }{ AB } =\frac { BC' }{ BC } \)    ...(i)
From (i) and (ii),
\(\frac { A'B' }{ AB } =\frac { 3 }{ 4 } \Rightarrow A'B=\frac { 3 }{ 4 } AB\)
In ΔBA₃C' and ΔBA₄C
\(\frac { BC' }{ BC } =\frac { { BA }_{ 3 } }{ { BA }_{ 4 } } =\frac { 3 }{ 4 } \)      ...(ii) 
∴ Sides of new triangle formed are \(\frac { 3 }{ 4 } \)  times the corresponding sides of first triangle.

mx(x-1)+49=0
\(\Rightarrow \) mx²-7mx+49=0
Here a=m,b=-7m,c=49
For equal roots ,D=0
\(\Rightarrow D=b^{ 2 }-4ac\)
\(\Rightarrow 0=(-7m)^{ 2 }-4\times m\times 49\)
\(\Rightarrow 0=49m^{ 2 }-196m\)
\(\Rightarrow 49m^{ 2 }-196m=0\)
\(\Rightarrow 7m(7m-28)=0\)
\(\Rightarrow 7m=0\) or 7m-28 =0
\(\Rightarrow m=0\) or \(m=\frac { 28 }{ 7 } =4\)
but \(m\neq 0\) [ \(\therefore \) In quadratic equation  \(a\neq 0\) ]

Given A right-angled triangle with sides of lengths 8 cm and 6 cm making a right angle. Required Triangle whose sides are \(\frac { 3 }{ 4 } \) times of the corresponding sides of given triangle.
Steps of Construction
1. Construct a right angled \(\angle ABC\)   right angle at B with sides BC = 8 cm and AB = 6 cm
2. Through B, construct an acute LCBX on the side \(\angle CBX\) opposite to the vertex A.
 
3. Mark four points B₁, B₂ , B₃, and B₄ on BX such that BB₁ = B₁B₂ = B₂B₃,= B₃B₄
4. Join B₄C
5. Through B₃, draw B₃C' II B₄C intersecting BC .at C'.
6. Through C', draw C'A' II CA intersecting AB at A'. Hence, \(\Delta A'BC'\)  is the required triangle.