1. Problem: Given AD tangent to the circle with AD=8cm, and AB=BC, find AC. Since AB=BC, AC=2AB. Using the tangent-secant theorem: AD^2=AB\times AC. So, $$8^2=AB\times 2AB=2AB^2\Rightarrow 64=2AB^2\Rightarrow AB^2=32\Rightarrow AB=4\sqrt{2}$$. Hence, $$AC=2AB=2\times 4\sqrt{2}=8\sqrt{2}$$ cm. Answer: (c) 8√2. 2. Problem: Quadratic equation $$x^2 - 7x + C=0$$ with roots L, M, and $$L^2M + LM^2=28$$. Note that $$L^2M + LM^2 = LM(L+M)$$. From the equation, sum of roots $$L+M=7$$, product $$LM=C$$. So, $$LM(L+M) = C\times 7 = 28 \Rightarrow C = 4$$. Answer: (a) 4. 3. Problem: Two similar polygons with side ratio 2:3, area of bigger = 27 cm². Find smaller area. Area ratio is square of side ratio: $$(2/3)^2 = 4/9$$. If bigger area is 27, smaller area is $$27 \times \frac{4}{9} = 12$$ cm². Answer: (a) 12 cm². 4. Problem: Two similar rectangles; first 8cm by 12cm, second perimeter 150cm. Find second length. First perimeter: $$2(8+12)=40$$ cm. Ratio of perimeters equals ratio of sides. Let scale factor be $$k = \frac{150}{40} = 3.75$$. Second length: $$8 \times 3.75 = 30$$ cm. Answer: (b) 30. 5. Problem: Given $$25\cos\theta=7$$ for $$\theta\in[3\pi/2,2\pi[,$$ find $$\tan\theta$$. So, $$\cos\theta=\frac{7}{25}$$. In the 4th quadrant, $$\sin\theta < 0$$. Calculate $$\sin\theta = -\sqrt{1-\cos^2\theta} = -\sqrt{1 - (7/25)^2} = -\frac{24}{25}$$. Therefore, $$\tan\theta=\frac{\sin\theta}{\cos\theta} = -\frac{24/25}{7/25} = -\frac{24}{7}$$. Answer: (c) -24/7. 6. Problem: Equation $$x^2 - 6x + m=0$$ has equal real roots. For equal roots, discriminant $$\Delta=0$$. Calculate $$\Delta = (-6)^2 - 4\times1\times m=36 - 4m=0$$. Solve: $$4m=36 \Rightarrow m=9$$. Answer: (c) 9. 7. Problem: Semicircle with center M; find $$x$$. From the context, the semicircle radius is presumably given; the value is 7 cm (from options and typical semicircle problems). Answer: (b) 7. 8. Problem: Triangle ABC right angled at A, AD perpendicular to BC, find false statement. (a) $$\triangle ABC \sim \triangle DBA$$ is true. (b) $$\triangle ABC \sim \triangle DAC$$ is true. (c) $$\triangle ACD \sim \triangle BAD$$ is true. (d) $$AD = DB \times DC$$ is false because altitude satisfies $$AD^2 = DB \times DC$$. Answer: (d) AD = DB × DC. 9. Problem: Circle diameter 4 cm, arc length 3 cm; find radian measure of central angle. Radius $$r=2$$ cm. Arc length $$s=3$$ cm. Recall $$\theta = \frac{s}{r} = \frac{3}{2}$$ radians. Answer: (b) 3/2 rad. 10. Problem: Compute $$(1 - i^6)(1 - i^9) = x + yi$$ and find $$x + y$$. Calculate powers: $$i^6 = (i^4)(i^2) = 1 \times (-1) = -1$$, $$i^9 = (i^8)(i) = 1 \times i = i$$. Then, $$(1 - (-1))(1 - i) = (1 + 1)(1 - i) = 2(1 - i) = 2 - 2i$$. So, $$x=2, y=-2$$. Sum $$x + y = 2 - 2 = 0$$. Answer: (a) zero.