(a) 
The area of the rectangle is 29 cm² greater than the area of the square.
The difference between the perimeters of the two shapes is k cm.
Find the value of k. You must show all your working.
(b) 
The volume of the larger cube is 5 cm³ greater than the volume of the smaller cube.
(i) Show that 3y² + 3y – 4 = 0.
(ii) Find the volume of the smaller cube. Show all your working and give your answer correct to 2 decimal places.
▶️ Answer/Explanation
(a) k = 12
First, set up the area equation: (2x+1)(x-1) – x² = 29.
Expand and simplify to get x² – x – 30 = 0.
Factorize to (x-6)(x+5)=0, so x=6 (as length can’t be negative).
Calculate perimeters: Rectangle=2(2x+1 + x-1)=6x=36cm, Square=4x=24cm.
Difference k=36-24=12cm.
(b)(i)
Volume difference: (y+1)³ – y³ = 5.
Expand (y+1)³ = y³ + 3y² + 3y + 1.
Subtract y³: 3y² + 3y + 1 = 5.
Simplify to required form: 3y² + 3y – 4 = 0.
(b)(ii) 0.09 cm³
Solve quadratic: y = [-3 ± √(9+48)]/6 = [-3 ± √57]/6.
Take positive solution: y ≈ 0.7583.
Volume = y³ ≈ (0.7583)³ ≈ 0.44 cm³.
(a) 
The total of the areas of rectangles A and B is 20 cm².
(i) Show that \( 3x^2 + 6x – 22 = 0 \).
(ii) Solve the equation \( 3x^2 + 6x – 22 = 0 \), giving your answers correct to 4 significant figures. You must show all your working.
(iii) Find the perimeter of rectangle B.
(b) 
The diagram shows two rectangles where \( H – h = 1 \).
By forming a quadratic equation and factorising, find the value of \( y \).
▶️ Answer/Explanation
(a)(i)
Area of A = x(3x + 4)
Area of B = 2(x – 1)
Total area: x(3x + 4) + 2(x – 1) = 20
Expand: 3x² + 4x + 2x – 2 = 20
Simplify: 3x² + 6x – 22 = 0
(a)(ii) -3.887 and 1.887
Using quadratic formula: x = [-6 ± √(36 + 264)]/6
x = [-6 ± √300]/6 = [-6 ± 10√3]/6
Simplify: x = -1 ± (5√3)/3
Calculate: x ≈ -3.887 or x ≈ 1.887
(a)(iii) 5.77 cm
Using positive x value (1.887):
Dimensions of B: (1.887 – 1) = 0.887 cm and 2 cm
Perimeter = 2(0.887 + 2) ≈ 5.77 cm
(b) 5
From areas: H = 15/(y-2) and h = 20/y
Given H – h = 1: 15/(y-2) – 20/y = 1
Multiply through by y(y-2): 15y – 20(y-2) = y(y-2)
Simplify: -5y + 40 = y² – 2y → y² + 3y – 40 = 0
Factor: (y+8)(y-5) = 0 → y = 5 (positive solution)