SNAP 2013 Question Paper

Question 29

In the triangle ABC, MN is parallel to AB. Area of trapezium ABNM is twice the area of triangle CMN. What is ratio of CM : AM ?

Solution

We have MN || AB 
Now let area of CMN be A 
so area of ABMN will be 2A 
Now Area of triangle ABC will be 3A

Now since MN || AB 
so we can say CMN is similar to CAB
So we can say
(CM/CA)^2 = Area of CMN : Area of CAB
so we can say CM :CA =$$1:\sqrt{\ 3}$$
Now therefore CM :AM = $$1:\sqrt{\ 3}-1\ =\ \sqrt{\ 3}+1:2$$

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