Question

In the triangle \(A B C, M N\) is parallel to \(A B\). Area of trapezium \(A B N M\) is twice the area of triangle \(C M N\). What is ratio of \(C M: A M ?\) (a) \(\frac{1}{\sqrt{3}+1}\) (b) \(\frac{\sqrt{3}-1}{2}\) (c) \(\frac{\sqrt{3}+1}{2}\) (d) none of these

Short answer

a) 1/3 b) 2/3 c) 4/3 d) none of these Answer: d) none of these Explanation: We found that the ratio of CM to AM in triangle CMA is 1/2, which does not match any of the given options. Therefore, the correct answer is (d) none of these.

Step-by-step solution

Understand the relationship between triangle areas and their bases

Since MN is parallel to AB, triangles CMN and CMA are similar. As the areas are related, we are primarily interested in the heights of triangles CMN and CMA. The area of a triangle is given by the formula $$ Area = \frac{1}{2}bh $$ where \(b\) is the base of the triangle, and \(h\) is its height. We will consider the heights of triangles CMN and CMA, denoted by h₁ and h₂, respectively.

Set up the area relationship between the two triangles

According to the problem, the area of trapezium ABNM is twice the area of triangle CMN. As triangle CMN and trapezium ABNM are side by side, the distance from C to MN, which we called h₁, must be half the distance from A to MN (called h₂). So we have h₂ = 2h₁

Translate the problem into a ratio involving side lengths of CMA and CMN

Since triangles CMN and CMA are similar, their sides are proportional. So the ratio of CM to AM can be expressed as $$ \frac{CM}{AM} = \frac{h_1}{h_2} $$

Plug in the relationship between the heights of the two triangles into the ratio

From step 2, we know h₂ = 2h₁. Plugging this into the ratio we have $$ \frac{CM}{AM} = \frac{h_1}{2h_1} $$

Calculate the ratio

$$ \frac{CM}{AM} = \frac{h_1}{2h_1} = \frac{1}{2} $$ The ratio of CM to AM is \(\frac{1}{2}\). However, none of the given options match this value, so the correct answer is (d) none of these.

Key concepts

Similar Triangles

Understanding the concept of similar triangles is crucial for students to solve problems related to geometric proportions and ratios. In the context of our example, similar triangles are those that have the same shape but different sizes. This is a consequence of having corresponding angles that are equal and the sides that are proportionate to each other.

When two triangles are similar, it means that one can be scaled up or down to match the other. The scaling factor is the ratio of any pair of corresponding sides. In our exercise, since line segment MN is parallel to AB, triangles CMN and CMA are similar by AA (Angle-Angle) similarity postulate. This setup is instrumental in solving the problem because it establishes a proportional relationship between the sides of the triangles and ultimately the heights of the triangles.

Once we recognize that the two triangles are similar, we can confidently say that corresponding sides and heights will be in the same ratio. Hence, the ratio of CM to AM can be directly inferred from the ratio of their corresponding heights.

Trapezium Area

Determining the area of a trapezium is also an important aspect of the problem. A trapezium, or trapezoid as it is known in some regions, is a quadrilateral with at least one pair of parallel sides. The formula for the area of a trapezium involves the lengths of the parallel sides (a and b) and the distance (height) between them (h), expressed as: $$\text{Area} = \frac{1}{2}(a+b)h$$
In this specific exercise, the trapezium ABNM's area is stated to be twice that of triangle CMN. The parallel sides for the trapezium are AB and MN, and the height is the distance between these sides, which corresponds to the heights of the triangles involved. An understanding of how the area of a trapezium relates to its components allows students to see the connection between the areas of the trapezium and the triangle within it, which is pivotal to solve the problem.

Triangle Height Proportions

When we talk about heights in triangles, especially in similar triangles, we are referring to the length of the line segment drawn from a vertex perpendicular to the line containing the opposite side, often called the altitude. In similar triangles, the ratio of the triangles' heights is equal to the ratio of corresponding sides. This property is useful in our exercise to find the ratio of CM to AM.

Based on the relationship between the area of the trapezium and the triangle, we concluded that the height of the trapezium (from A to MN) is twice the height of triangle CMN (h₁). If we denote the height of triangle CMA as h₂, the proportion between the heights h₁ and h₂ directly reflects the proportion between other corresponding lengths in triangles CMN and CMA, such as CM and AM. The interplay of triangle height proportions is therefore essential to determine the lengths' ratio simply by understanding the areas and heights' relationship.