Item 2: Determine the shape and area of a ceiling given coordinates. Item 4: Find the HCF of two numbers related to a party.

Here are the solutions for Item 2 and Item 4. Item 2: Ceiling Design i) To help the engineer, we plot the given coordinates on a Cartesian plane. The coordinates are (1,-1), (-2,-1), (-2,2), and (1,2). Let's label the points: A = (1,-1) B = (-2,-1) C = (-2,2) D = (1,2) Plotting these points reveals a rectangle. The side AB is horizontal, from x=-2 to x=1 at y=-1. The side BC is vertical, from y=-1 to y=2 at x=-2. The side CD is horizontal, from x=-2 to x=1 at y=2. The side DA is vertical, from y=-1 to y=2 at x=1. The shape formed by these coordinates is a rectangle. More specifically, since the lengths of adjacent sides are equal, it is a square. ii) To determine the area of the ceiling, we find the lengths of the sides of the square. Length of horizontal side (e.g., AB): |1 - (-2)| = |1 + 2| = 3 units Length of vertical side (e.g., BC): |2 - (-1)| = |2 + 1| = 3 units Since it's a square with side length 3 units, the area is: Area = side × side = 3 × 3 = 9 square units The area of the ceiling is 9 square units. Item 4: Graduation Party 1. To find the greatest number of guests Stephen must invite so that each guest receives an equal number of items with no leftovers, we need to find the Highest Common Factor (HCF) of the number of cup cakes, apples, and chocolate bars. The quantities are: Cup cakes: 72 Apples: 144 Chocolate bars: 288 Step 1: Find the prime factorization of each number. 72 = 2 × 36 = 2 × 2 × 18 = 2 × 2 × 2 × 9 = 2^3 × 3^2 144 = 2 × 72 = 2 × (2^3 × 3^2) = 2^4 × 3^2 288 = 2 × 144 = 2 × (2^4 × 3^2) = 2^5 × 3^2 Step 2: Identify the common prime factors and their lowest powers. The common prime factors are 2 and 3. The lowest power of 2 is 2^3. The lowest power of 3 is 3^2. Step 3: Calculate the HCF. HCF = 2^3 × 3^2 = 8 × 9 = 72 The greatest number of guests Stephen must invite is 72. 2. The number of guests in words is Seventy-two. 3 done, 2 left today. You're making progress.