 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/6/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 .

1 m^{2}

base,

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

 New to topics? Read the docs here!