Question 137
If the lengths of the sides AB, BC and CA of a triangle ABC are 10 cm, 8 cm and 6 cm respectively and if M is the mid - point of BC and MN II AB to cut AC at N, then the area of the trapezium ABMN is equal to
Solution
ABIIMN, $$\triangle$$ABC$$\sim$$ $$\triangle$$NMC
$$\frac{AB}{MN}$$ = $$\frac{BC}{CM}$$
$$\frac{10}{MN}$$ = $$\frac{8}{4}$$
MN=5cm
Area of trapezium ABMN= $$\frac{AB+MN}{2} \times 4$$
Area= 30 square cm
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