Higher June 2022 Paper 2R Q24

EdexcelHigherCurrent spec6 marksSimilar Shapes Volume & Surface Area

24 The diagram shows a frustum of a cone, and a sphere.

The frustum, shown shaded in the diagram, is made by removing the small cone from the large cone.
The small cone and the large cone are similar.

A shaded frustum with top radius r cm and base radius kr cm; the removed small cone of height h cm is shown dotted on top, and the large cone has height kh cm. Beside it, a sphere of radius r cm

Diagram NOT accurately drawn

The height of the small cone is \(h\) cm and the radius of the base of the small cone is \(r\) cm.
The height of the large cone is \(kh\) cm and the radius of the base of the large cone is \(kr\) cm.
The radius of the sphere is \(r\) cm.

The sphere is divided into two hemispheres, each of radius \(r\) cm.

Solid A is formed by joining one of the hemispheres to the frustum.
The plane face of the hemisphere coincides with the upper plane face of the frustum, as shown in the diagram below.

Solid B is formed by joining the other hemisphere to the small cone that was removed from the large cone.
The plane face of the hemisphere coincides with the plane face of the base of the small cone, as shown in the diagram below.

Solid A: the frustum with a hemisphere of radius r cm on its upper face; base radius kr cm. Solid B: the small cone with a hemisphere joined to its base

Diagram NOT accurately drawn

The volume of solid A is 6 times the volume of solid B.

Given that \(k \gt \sqrt[3]{7}\)
find an expression for \(h\) in terms of \(k\) and \(r\)

(6)

Mark scheme

Scheme	Marks
eg \(\dfrac{4}{3}\pi r^3 \div 2\;\left(= \dfrac{2}{3}\pi r^3\right)\) oe	M1
eg \(\dfrac{1}{3}\pi(kr)^2kh - \dfrac{1}{3}\pi r^2h\;\left(= \dfrac{1}{3}\pi r^2h(k^3 - 1)\right)\) oe	M1
eg \(\dfrac{1}{3}\pi r^2h(k^3 - 1) + \dfrac{2}{3}\pi r^3\) or \(\dfrac{1}{3}\pi r^2h + \dfrac{2}{3}\pi r^3\) oe	M1
eg \(\dfrac{1}{3}\pi r^2h(k^3 - 1) + \dfrac{2}{3}\pi r^3 = 6\left(\dfrac{1}{3}\pi r^2h + \dfrac{2}{3}\pi r^3\right)\) oe	M1
eg \(h(k^3 - 1) - 6h = 12r - 2r\) oe	M1
NB: note that simplest form was not required Answer: \(\dfrac{10r}{k^3 - 7}\)	A1
	(6)
	(6 marks)

Notes

M1: for finding the volume of hemisphere

M1: for finding the volume of the frustum

M1: for a correct expression for the volume of Solid A or Solid B

M1: for a correct equation using the volumes of Solid A and Solid B (\(\pi\) could be cancelled out)

M1: for simplifying to a point where the \(h\) terms are on one side of an equation and other terms the other side – must be correct

A1: oe eg \(\dfrac{4r - \frac{2}{3}r}{\frac{1}{3}k^3 - 2\frac{1}{3}}\)