Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 6

Table of contents
- a 6
  - Solution a
- b 6
  - Solution b

a

Solution

The sphere has surface area

4 π R^{2} = 400 π m^{2}

. Dividing the given surface-patch area by that total gives the dimensionless fraction

\frac{10}{400 π} = \frac{1}{40 π} \approx 0.00796,

or about $0.796%$ of the surface. The patch area is already measured on the surface, so no planar projection or spherical-cap conversion is needed.

b

Solution

Interpret the stated side lengths as all eight edges of the square pyramid. The equal lateral edges put its apex above the base centre. The base-centre-to-vertex distance is half the diagonal,

1/ 2 m

. The Pythagorean theorem therefore gives

h = 1^{2} - (1/ 2)^{2} = \frac{1}{2} m .

Using the pyramid volume formula with a

1 m^{2}

base,

V = \frac{1}{3} (1) \frac{1}{2} = \frac{1}{3 2} m^{3} \approx 0.2357 m^{3} .

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version