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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 ?
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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 = $$1:\sqrt{\ 3}-1\ =\ \frac{\ 1\ \left(\sqrt{\ 3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{\ 3}+1\right)}=\ \sqrt{\ 3}+1:2$$
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