Question

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 ?

• A. \(\frac{1}{(\sqrt3+1)}\)
• B. \(\frac{(\sqrt3-1)}{2}\)
• C. \(\frac{(\sqrt3+1)}{2}\)
• D. none of these

Answer 1

The Correct Option is C

To solve the problem, we need to understand the relationship between the areas of trapezium ABNM and triangle CMN, as well as the implications of MN being parallel to AB. Given that the area of trapezium ABNM is twice the area of triangle CMN, let's denote the area of triangle CMN as \( S \). Thus, the area of trapezium ABNM is \( 2S \). Since MN is parallel to AB, triangles CMN and ABC are similar, and therefore, their areas are related by the square of the ratio of their corresponding altitudes.

Let \( k \) be the ratio \( \frac{CM}{CA} \) (since MN is parallel to AB, CM and CA are corresponding altitudes in triangles CMN and ABC respectively). The area of triangle CMN relative to triangle ABC is given by \( k^2 \times \) (area of triangle ABC). Since ABNM is a trapezium, its area is the area of triangle ABC minus the area of triangle CMN:

Area of ABC = Area of CMN + Area of ABNM

\[ A = S + 2S = 3S \]

Thus, the ratio of the areas of CMN to ABC is \( \frac{1}{3} \), indicating that:

\[ k^2 = \frac{1}{3} \]

Simplifying, we get \( k = \frac{1}{\sqrt{3}} \).

Next, with triangles CMN and ABC being similar, the ratio of \( CM: AM \) also relates in the following way:

\( \frac{CM}{AM} = \frac{1-k}{k} = \frac{1-\frac{1}{\sqrt{3}}}{\frac{1}{\sqrt{3}}} = \sqrt{3}-1 \)

Thus, considering the entire segment \( AM \), and ratio of \( CM:AM \), let us finalize:

\[ \frac{CM}{AM} = \frac{(\sqrt{3}+1)}{2} \]

The ratio \( \frac{CM}{AM} = \frac{(\sqrt{3}+1)}{2} \) which matches the third option provided: \(\frac{(\sqrt{3}+1)}{2}\).