Since tangent to a circle is perpendicular to the radius through the point of contact.

ஃ \(\angle \)OLP = \(\angle \)OMP = 90°
Now, in quad OLPM, we have
\(\angle \)OLP +\(\angle \)LPM + \(\angle \)OMP + \(\angle \)LOM = 360°
⇒  90° + LPM + 90° + LOM = 360°
⇒ LPM + LOM = 180°     ........(i)
In OLM, using angle sum property, we have
\(\angle \)OLM + \(\angle \)OML + \(\angle \)LOM = 180°
\(\angle \)OLM + \(\angle \)OLM + \(\angle \)LOM = 180°
[ ∵ OL = OM = r ⇒ \(\angle \)OLM = \(\angle \)OML, angles opp. to equal sides of a \(\triangle\)]
2\(\angle \)OLM + \(\angle \)LOM = 180°    .........(ii)
From (i) and (ii)
\(\angle \)LPM + \(\angle \)LOM = 2 \(\angle \)OLM + \(\angle \)LOM
⇒ \(\angle \)LPM = 2\(\angle \)OLM

Area of the circle having radius 8cm
= π x 8 x 8 = 64cm²
Area of the circle having radius 6 cm
=π x 6 x 6 = 36π cm²
Now, area of the required circle = sum of areas of two circles
⇒ πr² = 64π + 36π
⇒ πr² =100π
⇒ r² = 100
⇒ r = 10
Hence, the radius of the required circle is 10 cm.

5 cm

Let AB is pole and BC is its shadow
∴ AB = 6m, BC = 2\(\\ \\ \\ \\ \sqrt { 3 } \) m
In right ΔABC, \(\frac { AB }{ BC } \)=tanፀ
⇒ tanፀ = \(\frac { 6 }{ 2\sqrt { 3 } } \)
⇒ tanፀ = \(\sqrt { 3 } \)
⇒ ፀ = 60^o

Let AB is tower and BC is its shadow
∴ AB = 20 m and BC = 20\(\sqrt { 3 } \)
In right ΔABC, \(\frac { AB }{ BC } \) = tanፀ
⇒ tanፀ = \(\frac { 20 }{ 20\sqrt { 3 } } \)
⇒ tanፀ = \(\frac { 1 }{ \sqrt { 3 } } \)
⇒ ፀ = 30^o

Let AB be a diameter of a given circle and LM and PQ be the tangent lines drawn to the circle at points A and B, respectively.

To prove LM || PQ
Proof We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OA \(\perp\) PQ and OB \(\perp\) LM
\(\Rightarrow\) AB \(\perp\) PQ
and AB \(\perp\) LM
\(\Rightarrow\) \(\angle\)PAB = 90°
and  \(\angle\)ABM = 90°
\(\Rightarrow\) \(\angle\)PAB = \(\angle\)ABM
[each = 90°]
But these are alternate angles.
\(\therefore\) PQ || LM
Hence, the tangents drawn at the ends of a diameter of a circle are parallel.
Hence proved.

Radius of the well = \(\frac { 7 }{ 2 } m=3.5\quad m\)
Volume of the earth taken out \(=\pi { \left( \frac { 7 }{ 2 } \right) }^{ 2 }\times 10\)
\(=\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times 10\)
= 385 m³
Area of the rectangular field = 20 \(\times\) 14
= 280 m²
Area of the top of the well = \(\\ \pi { \left( \frac { 7 }{ 2 } \right) }^{ 2 }\)
\(=\frac { 77 }{ 2 } { m }^{ 2 }\)
Area of the remaining field = \(280-\frac { 77 }{ 2 } \)
\(=\frac { 483 }{ 2 } { m }^{ 2 }\)
Let 'h' is the rise in the level of the field 
\(h=\frac { 385 }{ \frac { 483 }{ 2 } } =1.6m\) (approx)

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.

Let BC be the width of the river and A, B be the ends of river such that
AB = 150 m = Length of the bridge

and  \(\angle\)BAC = 45 \(°\)
In right angled \(\Delta ACB,\)
\(sin\quad 45°=\frac { P }{ H } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 2 } } =\frac { BC }{ AB } =\frac { BC }{ 150 } \)
\(\therefore BC=\frac { 150 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } \) [by rationalising]
\(=\frac { 150 }{ 2 } \sqrt { 2 } =75\sqrt { 2 } \)
\(\\ =75\times 1.414\quad [\because \sqrt { 2 } =1.414]\)
=106.05 m (approx)
Hence, width of the river is 106.05 m.

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.

Radius (r₁) of 1^st circle = 19 cm
Radius (r₂) or 2^nd circle = 9 cm
Let the radius of 3^rd circle be r.
Circumference of 1^st circle = 2πr₁ = 2π (19) = 38π
Circumference of 2^nd circle = 2πr₂ = 2π (9) = 18π
Circumference of 3^rd circle = 2πr
Given that,
Circumference of 3^rd circle = Circumference of 1^st circle + Circumference of 2^nd circle
2πr = 38π + 18π = 56π
\(r=\frac{56 \pi}{2 \pi}=28\)
Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.

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

79.58

51.96m

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

Here OA = 6.5 cm and OC = 2.5 cm
\(\therefore \quad AC=\sqrt { { OA }^{ 2 }-{ OC }^{ 2 } } \)
\(=\sqrt { { \left( 6.5 \right) }^{ 2 }-{ \left( 2.5 \right) }^{ 2 } } \)
\(=\sqrt { 42.25-6.25}\) = 6 cm
\(\therefore\)  AB = 12 cm

Since the tangent is perpendicular to the radius,
\(\angle OAB=\angle OBA\)    (\(\because \)  OA = OB)
\(\angle OBA\) = 30°
 \(\angle AOB\) = 180° - (30° + 30°)
\(\angle AOB\)  = 120°
\(\angle ABP=\angle OBP-\angle OBA\)
= 90° - 30° = 60°

56 cm

10 m

Since OM bisects chord AE
\(\therefore \ A M=\frac{A B}{2}=5 \mathrm{~cm}\)
\(\text { In } \Delta O M A\)
\(\sin \theta=\frac{A M}{O A}=\frac{1}{\sqrt{2}} \quad\left[\because \sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}\right]\)
\(\Rightarrow \ \theta=45^{\circ}\)
\(\text { Area of segment }=\text { Area of sector } \Delta A B O-\text { Area of } \Delta O A B\)
= 14.28 cm²

Let P be the position of the aeroplane. Then, PM=200m and let A and B be two points on the two banks of a river such that the angles of depression at A and B are \({ 60 }^{ \circ }\)and \({ 45 }^{ \circ }\) , respectively. 
Let  Am=x m and BM= y m.
Then, \(\angle XPB=\angle MBP={ 45 }^{ \circ }\)  [altenatives angles]
and  \(\angle XPB=\angle MBP={ 60 }^{ \circ }\)  [altenatives angles]
In right angled  \(\Delta AMP,\) \(\tan { { 60 }^{ \circ } } =\frac { PM }{ AM } \)
\(\Rightarrow \sqrt { 3 } =\frac { 200 }{ x } \Rightarrow 200=\sqrt { 3x } \)
\(\Rightarrow x=\frac { 200 }{ \sqrt { 3 } } m \)
In right angled  \(\Delta BMP,\)
\(\tan { { 45 }^{ \circ } } =\frac { PM }{ BM } \Rightarrow 1=\frac { 200 }{ y }\)
\(\Rightarrow y=200m \)
Now, width of the river, AB=BM+MA
\(\Rightarrow AB=x+y=\frac { 200 }{ \sqrt { 3 } } +200\)
\( =200\left( \frac { 1 }{ \sqrt { 3 } } +1 \right)\)
\(=200(1.5773)=315.46m. [\because \sqrt { 3 } =1.732]\)
Hence, the width of the river is 315.46m.

60°

\({ 100 }^{ \circ }\)

Since, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
i.e  ㄥAOB + ㄥCOD = 180^o
⇒  ㄥCOD =180^o  - ㄥAOB
⇒ ㄥCOD =  = 180^o  - 125^o = 55^o