141.96 m

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

25^∘

20 m

CA ⊥ AP
CB ⊥ BP
PA ⊥ PB
Also AP = PB
ஃ BPAC is a square.
⇒ AP = PB = BC = 4 cm

∠BPA = 90° (Angle in semicircle)
In BPA
∠ABP + ∠BPA + ∠PAB = 180°
⇒  30° + 90 + PAB = 180°
⇒ ∠PAB = 60°
Also ∠POA = 2 ∠PBA
⇒ ∠POA = 2 x 30° = 60°
⇒ OP = AP
(sides opposite to equal angles)   .....(i)
In △OPT, △OPT = 90°
∠POT = 60° and ∠PTO = 30°     [angle sum property of a ]
Also  ∠APT + ∠ATP = PAO    (exterior angle property)
ஃ APT + 30 = 60
⇒ APT = 30

Since, Ajay stands midway between two trees
\(\therefore B O=O D=d \mathrm{~m} \text {. }\)
Now, in \(\triangle B O E\) right angle at O
\(B E^2  =B O^2+O E^2 \)
\(B E^2  =d^2+d^2=2 d^2\)
\(B E =\sqrt{2} d\) [by Pythagoras theorem]
Similarly, \(D E=\sqrt{2} d\)
Let \(D C=h \mathrm{~m}\),
then \(A B=3 h \mathrm{~m}\)
In \(\triangle E B A, \tan \theta=\frac{A B}{B E}\)
\(\tan \theta=\frac{3 h}{\sqrt{2} d}\) [from Eq. (i) and (iv)]
Similarly, in right \(\triangle E D C\),
\(\tan \left(90^{\circ}-\theta\right) =\frac{C D}{E D} \ {\left[\because \cdot \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]} \)
\(\Rightarrow \quad \cot \theta  =\frac{h}{\sqrt{2} d} \ {[\text { from Eqs. (ii) and (iii) }]}\)
On multiplying $\tan \theta$ and $\cot \theta$, we get
\(\tan \theta \times \cot \theta =\frac{3 h}{\sqrt{2} d} \times \frac{h}{\sqrt{2} d} \)
\(1  =\frac{3 h^2}{2 d^2}\)
\(\Rightarrow \quad \frac{2}{3} d^2  =h^2\) \([\because \tan \theta \times \cot \theta=1]\)
Thus, \(h=d \times \sqrt{\frac{2}{3}}\)

1268 m

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.

Here,   AC = \(\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } =\sqrt { { 8 }^{ 2 }+{ 15 }^{ 2 } } \)
= \(\sqrt { 64+225 } =\sqrt { 289 } \)
= 17 cm
Clearly, OPBQ is a square
ஃ BP = BQ = OP = OQ = r cm
AR = AP = AB - BP = (8 - r) cm

Also,  Cr = CQ = CB - BQ = 15 - r
⇒  CR + AR = AC
⇒ 15 - r + 8 - r = 17
⇒ 23 - 2r = 17
⇒ -2r = -6
⇒ r = 3 cm

Given: A circle with centre O. AB is the diameter of this circle. I is tangent to the circle. AH and BK are perpendicular to I from A and B at H  and K respectively. 
To prove: AH + BK = AB 
Proof: AH and HP are tangents from the external point H 
ஃ  AH = HP        .....(i)
and BK, KP are tangent from the external point K
ஃ   BK = KP      ......(ii)
Adding (i) and (ii) we get
AH + BK = HP + PK = HK ......(iii)
AB ⊥  AH
AB ⊥ BK
[Tangent makes 90° angle with radius at the point of contact]
⇒ \(\angle \)1 = \(\angle \)2 = 90°
Given that AH ⊥ i ⇒ \(\angle \)3 = 90°
and BK ⊥ i ⇒ \(\angle \)4 = 90°
∵ \(\angle \)1 = \(\angle \)2 = \(\angle \)3 = \(\angle \)4 = 90°​​​​​​​
⇒ AHKB is a rectangle
⇒  AB = HK        ......(iv)
[Opposite sides of a rectangle are equal]
From (iii) and (iv) ⇒ PH + PK = AB
AH + BK = AB from (i) and (ii) Hence proved.

Given ABCD be a parallelogram circumscribing a circle with centre O.
To prove ABCD is a rhombus.

Proof We know that the tangents drawn to a circle from an exterior point are of equal lengh.
\(\therefore\) AP = AS, BP = BQ, CR = CQ and DR = DS
Adding the above four equations
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
\(\therefore\)        AB + CD = AD + BC
or    AB = BC = DC = AD
Therefore, ABCD is a rhombus.
Join OC, OP, OQ, OR, OD, OS, OB and OA
Now, area of rhombus = Area of \(\Delta\)BOC + Area of \(\Delta\)AOB + Area of \(\Delta\)AOD + Area of \(\Delta\)COD
Area of \(\Delta\)BOC \(=\frac{1}{2} \times B C \times O Q\)
\(\left[\because \text { Area of triangle }=\frac{1}{2} \times \text { Base } \times \text { Height }\right]\)
\(=\frac{1}{2} \times 10 \times 3=15 \mathrm{~cm}^2\)
\(\therefore\) Area of \(\Delta\)BOC = Area of \(\Delta\)AOB
= Area of \(\Delta\)AOD
= Area of \(\Delta\)COD
= 15 cm²
Area of rhombus=4 \(\times\) Area of \(\Delta\)BOC
= 4 \(\times\) 15 cm² = 60 cm²
Therefore, area of rhombus is 60 cm².

16 minutes 24 seconds

Let AB = h km be the height of the hill and C, D be two consecutive stones such that CD = 1 km.
Let BC be x km , then BD = BC + CD = (x + 1) km

Now, ∠ADB = ∠XAD = 30° [alternate angles]
and ∠ACB = ∠XAC = 45° [alternate angles]
In right angled ΔABC,
\(\tan 45^{\circ}=\frac{P}{B}=\frac{A B}{B C} \Rightarrow 1=\frac{h}{x} \Rightarrow x=h\)
Now, in right angled ΔABD, tan\(30^{\circ}=\frac{A B}{B D}\)
\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+1} \left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]\)
\(\left(\frac{\sqrt{3+1}}{2}\right) \mathrm{km}\)

OP is tangent of the circle having centre O'
So \(\angle \)OPO' = 90°
[ ∵ Radius and tangent are to ⊥ each other at the point of contact]
In right-angled OPO'
OP = 4 cm
O'P = 3 cm  [Given]
In right-angled \(\triangle\)OPO'
OP = 4 cm
O'P = 3 cm   [Given]
OO'² = OP² + O'P²
= 4² + 3² = 16 + 9 = 25
OO' = 5 cm
If two circles intersect each other then line joining the two centre always ⊥ bisector of the common chord.
OO' ⊥ PQ and PT = TQ
Area of \(\triangle\) OO'P = \(1\over2\) x base x altitude
Here, base = 4 cm, attitude = 3 cm.
Area = \(1\over2\) x 4 x 3 = 6 cm²   ...(i)
But if base OO' = 5 cm altitude = PT
Area \(\triangle\)POO' = \(1\over2\) x 5 x altitude   .....(ii)
6 cm2 = \(1\over2\)x 5 x altitude  .....(ii)
Comparing (i) and (ii)
6 cm² = \(1\over2\) x 5 x altitude
⇒ \(\frac { 2\times 6 }{ 5 } \) = Altitude
⇒ \(\frac { 12 }{ 5 } \) = PT
⇒ PQ = 2PT = \(\frac { 2\times 12 }{ 5 } =\frac { 24 }{ 5 } \)cm
So, length of common chord = \(24\over5\) cm
= 4.8 cm

Let C₁ and C₂ be the two concentric circles, with centre at O and respective radii being r₁ = 4cm and r₂ = 5cm.
Draw a chord AC of circle C₂, to touch circle C₁ at B.
Join OB.

Here OB 丄 AC [As, tangent at any point of circle is perpendicular to radius through the point of contact]
Thus, in right angled OAB ,we have:
OA² = AB² + OB²     [By using phythagoras theorem]
⇒  52 =  AB² + 4²
⇒  AB² = 25 - 16  = 9
∴ Length of chord AC = 2 AB = 2 x 3 = 6 cm

Here in right angled triangle ABC, AB = 6 m and BC= 8 cm.
\(\therefore\) By Pythagoras theorem \(A C=\sqrt{(A B)^{2}+(B C)^{2}}\)
\(=\sqrt{(8)^{2}+(6)^{2}}=\sqrt{100}=10 \mathrm{~m}\)
Also, AP = x m.
(i) (c):AR=AP = xm      ..(1)
[Since, length of tangents drawn from an external point are equal]
(ii) (b): BQ = BR = AB - AR = (6 - x) m         (Using (1))
(iii) (d): CQ = CP = AC - AP = (10 - x) m
Also, CQ = BC - BQ = BC - BR = 8 - (6 - x) = 2 + x
(iv) (b): Since, CQ = 10 - x = 2 + x
\(\Rightarrow\) 8 = 2x \(\Rightarrow\) x = 4
\(\therefore\)  AR = AP = 4 m, BR = BQ = 2 m
and CP = CQ = 6 m
Also, OQ.\(\perp\)BQ and OR.l BR
\(\therefore\) BROQ is a square.
(v) (a): Radius of the circle, OR = BR = 2 cm

(i) (b): 

Here, OS the is radius of circle.
Since radius at the point of contact is perpendicularto tangent.
So, \(\angle\)OSA = 90°
(ii) (d): Since, length of tangents drawn from an external point to a circle are equal.
\(\therefore\) AS=AP,BP=BQ,
CQ = CR and DR = DS
(iii) (a): AP = AS = AD _ DS = AD _ DR (Using (1)
= 11 - 7 = 4 cm
(iv) (b): In quadrilateral OQCR,

\(\angle\)QCR = 60° (Given)
And \(\angle\)OQC = \(\angle\)ORC = 90° [Since, radius at the point of contact is perpendicular to tangent.]
\(\therefore\) \(\angle\)QOR = 360° - 90° - 90° - 60° = 120°
(v) (c): From (1), we have AS = AP, DS = DR,
BQ = BP and CQ = CR
Adding all above equations, we get
AS + DS + BQ + CQ = AP + DR + BP + CR
\(\Rightarrow\) AD + BC = AB + CD

(i) (c): \(\text { In } \Delta O P Q\), we have
\(\tan 60^{\circ}=\frac{P Q}{P O} \)
\(\Rightarrow \sqrt{3}=\frac{20}{P O} \)
\(\Rightarrow P O=\frac{20}{\sqrt{3}} \mathrm{~m}\)
(ii) (b): In \(\Delta\)ORS, we have
\(\tan 30^{\circ}=\frac{R S}{O R} \Rightarrow \frac{1}{\sqrt{3}}=\frac{20}{O R} \Rightarrow O R=20 \sqrt{3} \mathrm{~m}\)
(iii) (d): Clearly, width of the road = PR
\(\begin{array}{l} =P O+O R=\left(\frac{20}{\sqrt{3}}+20 \sqrt{3}\right) \mathrm{m} \\ =20\left(\frac{4}{\sqrt{3}}\right) \mathrm{m}=\frac{80}{\sqrt{3}} \mathrm{~m}=46.24 \mathrm{~m} \end{array}\)
(iv) (a): \(\text { In } \Delta O P Q \text { , if } \angle P O Q=45^{\circ} \text { , then }\)
\(\tan 45^{\circ}=\frac{P Q}{P O} \Rightarrow 1=\frac{20}{P O} \Rightarrow P O=20 \mathrm{~m}\)
(v) (b)