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Question: The median m divides the ABC triangle in 2 triangles, ABM and ACM. What is the ratio between the areas of the 2 triangles, Area_ABM/Area_ACM ?

(a) 1

(b) 1/2

(c) 2

(d) It cannot be determined from the information given

(e) 3

• Answer: The areas of the 2 triangles will be equal because they are both (1/2) · h ·(BC/2), where h is the altitude. The correct answer is (a).
  

Question: In the figure below, angle AMB is 100^o, angle BMD is 40^o and angle CME is 60^o. What is the value of angle CMD?
Figure not drawn to scale

(a) 20^o
(b) 30^o
(c) 40^o
(d) 50^o
(e) 80^o

• Answer: If angle AMB is 100^o, then angle BME is 180^o - 100^o = 80^o
  BMC + CMD = 40^o
  BMC + CMD + DME = 80^o
  If we subtract the 2 equations, DME = 40^o
  CMD + DME = 60^o
  BMC + CMD + DME = 80^o
  If we subtract the 2 equations, BMC = 20^o
  CMD = BME - BMC - DME = 80^o - 40^o - 20^o = 20^o