Question

What is the ratio of the area of triangle ABC to the area of square ADFC if CB=(CF)/4? 

 

• A. 1/4
• B. 1/8
• C. 1/16
• D. 2/5
• E. 3/8

Hint:

In geometry problems, assign a variable (like 's') to a key length. Express all other lengths and areas in terms of this variable. The variable will cancel out when you calculate the ratio.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks for the ratio of the area of a triangle to the area of a square. We are given a geometric relationship between the base of the triangle and the side of the square.
Step 2: Key Formula or Approach:
- Area of a square with side \(s\): \(A_{square} = s^2\)
- Area of a triangle: \(A_{triangle} = \frac{1}{2} \times \text{base} \times \text{height}\)
Step 3: Detailed Explanation:
Let the side length of the square ADFC be \(s\).
Therefore, \(AD = DF = FC = CA = s\).
The area of the square ADFC is \(s^2\).
We are given the relationship \(CB = \frac{CF}{4}\). Since \(CF = s\), we have:
\[ CB = \frac{s}{4} \] Now, let's find the area of triangle ABC.
We can consider CB as the base of the triangle. The height of the triangle corresponding to this base is the perpendicular distance from vertex A to the line containing the base CF.
Since ADFC is a square, the side AC is perpendicular to the side CF.
Therefore, the height of \(\triangle ABC\) is the length of the side AC, which is \(s\).
Area of \(\triangle ABC\) = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\[ \text{Area}(\triangle ABC) = \frac{1}{2} \times CB \times AC = \frac{1}{2} \times \left(\frac{s}{4}\right) \times s = \frac{s^2}{8} \] The question asks for the ratio of the area of triangle ABC to the area of square ADFC.
\[ \text{Ratio} = \frac{\text{Area}(\triangle ABC)}{\text{Area}(ADFC)} = \frac{s^2/8}{s^2} = \frac{1}{8} \] Step 4: Final Answer:
The ratio of the area of triangle ABC to the area of square ADFC is 1/8.