Here are the solutions to the problems: 1.a. Calculate the area of the shaded region. Step 1: Identify the properties of the equilateral triangle PQR. The side length of the equilateral triangle PQR is $s = 18$ cm. All angles in an equilateral triangle are $60^\circ$, so $\angle P = \angle Q = \angle R = 60^\circ$. Step 2: Calculate the area of the equilateral triangle PQR. The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4} s^2$. $$ \text{Area of } \triangle PQR = \frac{\sqrt{3}}{4} (18)^2 = \frac{\sqrt{3}}{4} \times 324 = 81\sqrt{3} \text{ cm}^2 $$ Using $\sqrt{3} \approx 1.73205$: $$ \text{Area of } \triangle PQR \approx 81 \times 1.73205 = 140.29605 \text{ cm}^2 $$ Step 3: Determine the radius of the sector PAB. The arc is centered at P and touches QR at M. This means PM is the radius of the sector. In an equilateral triangle, the line segment from a vertex to the midpoint of the opposite side (PM) is also an altitude. Consider the right-angled triangle PQM, where $PQ = 18$ cm and $QM = \frac{1}{2} QR = \frac{1}{2} \times 18 = 9$ cm. Using the Pythagorean theorem: $PM^2 + QM^2 = PQ^2$ $$ PM^2 + 9^2 = 18^2 $$ $$ PM^2 + 81 = 324 $$ $$ PM^2 = 324 - 81 = 243 $$ $$ PM = \sqrt{243} = \sqrt{81 \times 3} = 9\sqrt{3} \text{ cm} $$ So, the radius of the sector is $r = 9\sqrt{3}$ cm. Step 4: Calculate the area of the sector PAB. The angle of the sector is $\angle QPR = 60^\circ$. The formula for the area of a sector is $\frac{\theta}{360^\circ} \pi r^2$. Using $\pi = \frac{22}{7}$: $$ \text{Area of sector PAB} = \frac{60}{360} \times \frac{22}{7} \times (9\sqrt{3})^2 $$ $$ = \frac{1}{6} \times \frac{22}{7} \times (81 \times 3) $$ $$ = \frac{1}{6} \times \frac{22}{7} \times 243 $$ $$ = \frac{11}{3 \times 7} \times 243 $$ $$ = \frac{11}{21} \times 243 $$ $$ = \frac{11 \times 81}{7} = \frac{891}{7} \text{ cm}^2 $$ $$ \text{Area of sector PAB} \approx 127.28571 \text{ cm}^2 $$ Step 5: Calculate the area of the shaded region. The shaded region is the area of the triangle PQR minus the area of the sector PAB. $$ \text{Area of shaded region} = \text{Area of } \triangle PQR - \text{Area of sector PAB} $$ $$ \text{Area of shaded region} \approx 140.29605 - 127.28571 $$ $$ \text{Area of shaded region} \approx 13.01034 \text{ cm}^2 $$ Correct to two decimal places: The area of the shaded region is $\boxed{13.01 \text{ cm}^2}$. 1.b. Solve the equation $a - \sqrt{a^2 + 12} + 2 = 0$ Step 1: Isolate the square root term. $$ a + 2 = \sqrt{a^2 + 12} $$ Step 2: Square both sides of the equation to eliminate the square root. $$ (a+2)^2 = (\sqrt{a^2 + 12})^2 $$ $$ a^2 + 4a + 4 = a^2 + 12 $$ Step 3: Solve for $a$. Subtract $a^2$ from both sides: $$ 4a + 4 = 12 $$ Subtract 4 from both sides: $$ 4a = 12 - 4 $$ $$ 4a = 8 $$ Divide by 4: $$ a = \frac{8}{4} $$ $$ a = 2 $$ Step 4: Check the solution by substituting $a=2$ back into the original equation. $$ 2 - \sqrt{2^2 + 12} + 2 = 0 $$ $$ 2 - \sqrt{4 + 12} + 2 = 0 $$ $$ 2 - \sqrt{16} + 2 = 0 $$ $$ 2 - 4 + 2 = 0 $$ $$ 0 = 0 $$ The solution is correct. The value of a is $\boxed{2}$. 1.c. The ratio of the age of a mother to that of the daughter is 11:3. If the mother is 55 years old, what will be the ratio of their ages in 5 years time? Step 1: Determine the current ages of the mother and daughter. Let the mother's age be $M$ and the daughter's age be $D$. Given ratio $M:D = 11:3$. Given mother's current age $M = 55$ years. We can write the ratio as $\frac{M}{D} = \frac{11}{3}$. Substitute $M=55$: $$ \frac{55}{D} = \frac{11}{3} $$ To find $D$, cross-multiply: $$ 11D = 55 \times 3