(i) (a): Speed = 1200 km/hr
\(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\)
\(\therefore\) Required distance = Speed x Time
\(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\)
(ii) (c): Speed = 1500 km/hr
Time = \(\frac{3}{2}\) hr.
\(\therefore\) Required distance = Speed x Time
\(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\)
(iii) (b): Clearly, directions are always perpendicular to each other.
\(\therefore \quad \angle P O Q=90^{\circ}\)
(iv) (a): Distance between aeroplanes after \(1\frac{1}{2}\) hour 
\(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\)
(v) (d): Area of \(\Delta\)POQ= \(\frac{1}{2}\)x base x height
\(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)

(i) (a): Since \(\Delta\)ABC - \(\Delta\)PQR

\(\therefore \quad \frac{A B}{P Q}=\frac{B C}{Q R} \Rightarrow \frac{x}{20}=\frac{50}{10} \Rightarrow x=100\)
Thus, height of tower is 100 m.
(ii) (c): Since \(\Delta\)ABC - \(\Delta\)PQR

\(\therefore \quad \frac{100}{20}=\frac{x}{15} \Rightarrow x=\frac{1500}{20}=75 \mathrm{~m}\)
(iii) (d): Since, the shapes are similar

\(\therefore \quad \frac{x}{20}=\frac{20}{10} \)
\(\Rightarrow \quad x=\frac{20 \times 20}{10}=40 \mathrm{~m}\)
(iv) (b): Since, the shapes are similar, so, \(\frac{40}{100}=\frac{x}{40}\)
\(\Rightarrow\) x = 16m

(v) (d): Since, the shapes are similar, so \(\frac{20}{100}=\frac{x}{40}\)
\(\Rightarrow \quad x=\frac{20 \times 40}{100}=8 \mathrm{~m}\)

(i) (b)
(ii) (c): Using Pythagoras theorem, we have PQ² = PR² + RQ²
\(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \)
\(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\)
\(\Rightarrow x^{2}+12 x-5 x-60=0 \)
\(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \)
\(\Rightarrow x=5, x=-12\)
\(\therefore \quad x=5\) [Since length can't be negative]
(iii) (a) : PR = 2x = 2 x 5 = 10 km
(iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km
(v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km

(i) (c)
(ii) (b): Since \(\Delta\)AOB ~ \(\Delta\)COD [ByAA similarity criterion]
\(\therefore \frac{A O}{O C}=\frac{B O}{O D} \Rightarrow \frac{3}{x-3}=\frac{x-5}{3 x-19}\)
\(\Rightarrow 3(3 x-19)=(x-5)(x-3) \)
\(\Rightarrow 9 x-57=x^{2}-3 x-5 x+15 \Rightarrow x^{2}-17 x+72=0 \)
\(\Rightarrow(x-8)(x-9)=0 \Rightarrow x=8 \text { or } 9\)
(iii) (c) : Since, \(\Delta\)AOB ~ \(\Delta\)COD [ByAA similarity criterion]
\(\therefore \frac{A O}{O C}=\frac{B O}{O D} \Rightarrow \frac{6 x-5}{2 x+1}=\frac{5 x-3}{3 x-1}\)
\(\Rightarrow(6 x-5)(3 x=1)=(5 x-3)(2 x+1) \)
\(\Rightarrow \quad 18 x^{2}-6 x-15 x+5=10 x^{2}+5 x-6 x-3 \)
\(\Rightarrow \quad 8 x^{2}-20 x+8=0 \Rightarrow 2 x^{2}-5 x+2=0\)
From options, x = 2 is the only value that satisfies this equation.
(iv) (d): Since, \(\Delta\)APQ ~ \(\Delta\)ABC [ByAA similarity criterion]
\(\therefore \quad \frac{A P^{\circ}}{A B}=\frac{A Q}{A C}=\frac{P Q}{B C} \Rightarrow \frac{2.4}{A B}=\frac{2}{5}=\frac{P Q}{6} \)
\(\therefore \quad A B=\frac{2.4 \times 5}{2}=6 \mathrm{~cm} \text { and } P Q=\frac{2 \times 6}{5}=2.4 \mathrm{~cm} \)
\(\therefore \quad A B+P Q=6+2.4=8.4 \mathrm{~cm}\)
(v) (a): Since,  \(\Delta\)DRS ~ \(\Delta\)DEF  [ByAA similarity criterion]
\(\therefore \quad \frac{D E}{D R}=\frac{D F}{D S} \Rightarrow \frac{D E}{D R}-1=\frac{D F}{D S}-1 \)
\(\Rightarrow \frac{D E-D R}{D R}=\frac{D F-D S}{D S} \Rightarrow \frac{E R}{D R}=\frac{F S}{D S} \)
\(\Rightarrow \quad \frac{D R}{E R}=\frac{D S}{F S} \Rightarrow \frac{4 x-3}{3 x-1}=\frac{8 x-7}{5 x-3} \)
\(\Rightarrow \quad 20 x^{2}-12 x-15 x+9=24 x^{2}-8 x-21 x+7 \)
\(\Rightarrow \quad 4 x^{2}-2 x-2=0 \Rightarrow 2 x^{2}-x-1=0\)
Only option (a) i.e., x = 1 satisfies this equation.

(i) (b): Let \(\Delta\)ABC is the triangle formed by both hotels and mountain top. \(\Delta\)CDE is the triangle formed by both huts and mountain top. Clearly, DE || AB and so 
\(\triangle A B C \sim \triangle D E C\)  [By AA-similarity criterion]

Now, required ratio = Ratio of their corresponding sides \(=\frac{B C}{E C}=\frac{10}{7}\) i.e., 10:7.
(ii) (c): Since, DE || AB, therefore
\(\frac{C D}{A D}=\frac{C E}{E B} \Rightarrow \frac{10}{A D}=\frac{7}{3} \Rightarrow A D=\frac{10 \times 3}{7}=4.29 \text { miles }\)
(iii) (b) : Since, \(\triangle A B C \sim \triangle D E C\)
\(\therefore \quad \frac{B C}{E C}=\frac{A B}{D E}\)  [ \(\because\) Corresponding sides of similar triangles are proportional]
\(\Rightarrow \frac{10}{7}=\frac{A B}{8} \Rightarrow A B=\frac{80}{7}=11.43 \text { miles }\)
(iv) (a) Given, DC= 5+ BC.
Clearly, BC = 10-5 = 5 miles
Now, CE = \(\frac{7}{10}\)x BC = \(\frac{7}{10}\) x 5 = 3.5 miles
(v) (d) :Clearly the radio of areas of two angles (i,e \(\triangle A B C \sim \triangle D E C\))
\(\begin{array}{l} =\left(\frac{B C}{E C}\right)^{2}=\left(\frac{10}{7}\right)^{2}=\frac{100}{49} \\ \therefore \quad \text { Required ratio }=\frac{\operatorname{ar}(\Delta C D E)}{a r(E B A D)}=\frac{49}{100-49}=\frac{49}{51} \end{array}\)