Question

In triangle \(ABC\), point \(M\) lies on side \(AC\) and point \(N\) lies on side \(BC\), such that \(MN\) is parallel to \(AB\) (so triangle \(CMN\) is the smaller triangle near vertex \(C\), and \(ABNM\) is the trapezium formed below it). If the area of trapezium \(ABNM\) is twice the area of triangle \(CMN\), what is the ratio \(CM : AM\)?

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

Hint:

Use area ratio = (side ratio)^2 for similar triangles CMN and CAB.

Answer 1

The Correct Option is C

Step 1: Set up areas using similar triangles.
Since \(MN\) is parallel to \(AB\), triangle \(CMN\) is similar to triangle \(CAB\) (both share angle \(C\), and the parallel line creates equal corresponding angles). Let the area of triangle \(CMN\) be \(A\). We are told the trapezium \(ABNM\) has area \(2A\), so the whole triangle \(ABC\) has area \(A + 2A = 3A\).

Step 2: Use the area ratio to find the side ratio.
For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides. Taking \(CM\) and \(CA\) as corresponding sides of triangle \(CMN\) and triangle \(CAB\):
\[ \left(\frac{CM}{CA}\right)^2 = \frac{\text{Area}(CMN)}{\text{Area}(CAB)} = \frac{A}{3A} = \frac{1}{3} \]
So \(\dfrac{CM}{CA} = \dfrac{1}{\sqrt3}\), which means \(CA = CM\sqrt3\).

Step 3: Find AM and the required ratio.
Since \(M\) lies on \(AC\), \(AM = CA - CM = CM\sqrt3 - CM = CM(\sqrt3 - 1)\).
So \(\dfrac{CM}{AM} = \dfrac{1}{\sqrt3-1}\). Rationalising by multiplying numerator and denominator by \((\sqrt3+1)\):
\[ \frac{1}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1} = \frac{\sqrt3+1}{3-1} = \frac{\sqrt3+1}{2} \]

Final Answer:
The ratio \(CM:AM\) works out to \(\dfrac{\sqrt3+1}{2}\), so option C is correct. Option A inverts the ratio, option B stops short of rationalising, and option D is wrong since a closed form does exist. \[ \boxed{CM:AM = \frac{\sqrt3+1}{2}} \]