Foundation November 2021 Paper 1 Q26
26 The diagram shows a cube with edges of length \(x\) cm and a sphere of radius 3 cm.
The surface area of the cube is equal to the surface area of the sphere.
Show that \(x = \sqrt{k\pi}\) where \(k\) is an integer. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| Shown | M1 | for a correct expression for the area of one face of the cube, eg \(x^2\) or a correct expression for the surface area of the cube, eg \(6 \times x^2\) |
| M1 | for a correct expression for the surface area of the sphere, eg \(4 \times \pi \times 3^2\ (= 36\pi)\) | |
| M1 | for forming a suitable equation, eg \(6 \times x^2 = 4 \times \pi \times 3^2\) or \(6x^2 = \text{``}36\pi\text{''}\) | |
| A1 | for completing the method to \(x = \sqrt{6\pi}\) or \(k = 6\) |
Additional guidance
No marks for \(x = \sqrt{6\pi}\) without any working.
\(6 \times x^2 = 4 \times \pi \times 3^2\)
\(x^2 = 36\pi \div 6\)
\(x = \sqrt{6\pi}\)