+ = 6 7n M1 g(3 n) – 3g + = 6 7n gn – + 7n = 6 3g M1 gn – + 7) = 6 3g M1 n(g – = A1 n (Total 4 marks) Question 20. The diagram shows a pyramid with base ABC. is perpendicular to both and CD CA CB.
g(3 + n) = 6 – 7n M1 3g + gn = 6 – 7n gn + 7n = 6 – 3g M1 n(g + 7) = 6 – 3g M1 n = A1 (Total 4 marks) Question 20. The diagram shows a pyramid with base ABC. CD is perpendicular to both CA and CB.