1. (a) Problem: Triangle ABC is an enlargement of triangle PQR. Given sides of PQR: 7 cm, 8 cm, 5 cm; and side QR of ABC is 12 cm. Find the surface area of triangle ABC. Step 1: Identify scale factor. Since QR corresponds to side 5 cm in PQR and 12 cm in ABC, scale factor $k = \frac{12}{5} = 2.4$. Step 2: Calculate area of triangle PQR using Heron's formula. Semi-perimeter $s = \frac{7 + 8 + 5}{2} = 10$ cm. Area $= \sqrt{s(s-7)(s-8)(s-5)} = \sqrt{10 \times 3 \times 2 \times 5} = \sqrt{300} = 10\sqrt{3}$ cm$^2$. Step 3: Area of ABC is $k^2$ times area of PQR. Area ABC $= (2.4)^2 \times 10\sqrt{3} = 5.76 \times 10\sqrt{3} = 57.6\sqrt{3} \approx 99.76$ cm$^2$. Answer: Surface area of triangle ABC is approximately 99.76 cm$^2$. 2. (b) Problem: In triangle with $|AB|=10$ cm, $|AC|=12$ cm, $|CE|=5$ cm, and DE parallel to AB, find (i) $|DE|$ and (ii) area of triangle DCE. Step 1: Since DE is parallel to AB, triangles DCE and CAB are similar. Step 2: Ratio of sides $\frac{|DE|}{|AB|} = \frac{|CE|}{|CB|}$. Given $|CE|=5$ cm, $|AC|=12$ cm, so $|CB| = |AB| - |DE|$ is unknown, but since DE is parallel to AB, $\frac{|DE|}{10} = \frac{5}{12}$. Step 3: Solve for $|DE|$: $|DE| = \frac{5}{12} \times 10 = \frac{50}{12} = 4.17$ cm. Step 4: Area of triangle DCE is proportional to square of similarity ratio. Area of triangle ABC is $\frac{1}{2} \times 10 \times 12 = 60$ cm$^2$. Similarity ratio $r = \frac{5}{12}$. Area DCE $= r^2 \times 60 = \left(\frac{5}{12}\right)^2 \times 60 = \frac{25}{144} \times 60 = \frac{1500}{144} = 10.42$ cm$^2$. Answer: (i) $|DE| = 4.17$ cm, (ii) area of triangle DCE is approximately 10.42 cm$^2$. 3. (a) Problem: Time $t$ varies partly as distance $d$ and partly as inverse of speed $s$: $t = ad + \frac{b}{s}$. Given $t=2$ hrs for $d=100$ km, $s=50$ km/h and $t=1.5$ hrs for $d=90$ km, $s=60$ km/h. Find $t$ for $d=120$ km, $s=40$ km/h. Step 1: Write equations: $2 = 100a + \frac{b}{50}$ $1.5 = 90a + \frac{b}{60}$ Step 2: Multiply first by 60 and second by 50 to eliminate denominators: $120 = 6000a + 1.2b$ $75 = 4500a + 0.8333b$ Step 3: Multiply second by 1.44 to align $b$ coefficients: $75 \times 1.44 = 108 = 4500 \times 1.44 a + 0.8333 \times 1.44 b$ $108 = 6480a + 1.2b$ Step 4: Subtract first from this: $108 - 120 = 6480a - 6000a + 1.2b - 1.2b$ $-12 = 480a$ $a = -\frac{12}{480} = -0.025$ Step 5: Substitute $a$ into first equation: $2 = 100(-0.025) + \frac{b}{50} = -2.5 + \frac{b}{50}$ $2 + 2.5 = \frac{b}{50} \Rightarrow 4.5 = \frac{b}{50}$ $b = 225$ Step 6: Calculate $t$ for $d=120$, $s=40$: $t = -0.025 \times 120 + \frac{225}{40} = -3 + 5.625 = 2.625$ hours. Answer: Time taken is 2.625 hours. 4. (b) Problem: Given $\tan x = 1$ and $180^\circ \leq x \leq 270^\circ$, find (i) $\sin x$ and $\cos x$, (ii) evaluate $\frac{2\sin x - 3\cos x}{2x \tan x}$. Step 1: Since $\tan x = 1$ in third quadrant, $x = 225^\circ$. Step 2: $\sin 225^\circ = -\frac{\sqrt{2}}{2}$, $\cos 225^\circ = -\frac{\sqrt{2}}{2}$. Step 3: Evaluate numerator: $2\sin x - 3\cos x = 2 \times \left(-\frac{\sqrt{2}}{2}\right) - 3 \times \left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2} + \frac{3\sqrt{2}}{2} = \frac{-2\sqrt{2} + 3\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$. Step 4: Evaluate denominator: $2x \tan x = 2 \times 225 \times 1 = 450$ (degrees treated as number here). Step 5: Final value: $\frac{\frac{\sqrt{2}}{2}}{450} = \frac{\sqrt{2}}{900}$. Answer: (i) $\sin x = -\frac{\sqrt{2}}{2}$, $\cos x = -\frac{\sqrt{2}}{2}$; (ii) value is $\frac{\sqrt{2}}{900}$. Final answers summarized: 1. (a) Surface area of ABC $\approx 99.76$ cm$^2$. 2. (b) (i) $|DE| = 4.17$ cm, (ii) area DCE $\approx 10.42$ cm$^2$. 3. (a) Time for 120 km at 40 km/h is 2.625 hours. 4. (b) (i) $\sin x = -\frac{\sqrt{2}}{2}$, $\cos x = -\frac{\sqrt{2}}{2}$, (ii) value $= \frac{\sqrt{2}}{900}$.