Consider a square ABCD of side length 16. Let E, F be points on CD such that CE = EF = FD. Let the line BF and AE meet in M. The area of triangleMAR is? Answer is integer between 0-99

We are given a square \( ABCD \) with side length 16. Points \( E \) and \( F \) divide side \( CD \) into three equal parts, so that \( CE = EF = FD = \frac{16}{3} \). The lines \( BF \) and \( AE \) meet at point \( M \), and we are asked to find the area of triangle \( \triangle MAR \), where \( R \) is yet to be defined, but we can assume we will need the geometry of the triangle formed by points \( M \), \( A \), and some other important point on the figure.

Let’s break down the problem step-by-step:

### Step 1: Coordinates of points
We can assign coordinates to the vertices of the square based on a Cartesian coordinate plane.

- Let \( A = (0, 16) \)
- \( B = (16, 16) \)
- \( C = (16, 0) \)
- \( D = (0, 0) \)

Since \( E \) and \( F \) divide \( CD \) into three equal parts:
- \( E = (16 - \frac{16}{3}, 0) = (\frac{32}{3}, 0) \)
- \( F = (16 - 2 \times \frac{16}{3}, 0) = (\frac{16}{3}, 0) \)

### Step 2: Equations of lines
Now, we need to find the equations of lines \( BF \) and \( AE \).

#### Line \( BF \):
Using the slope formula, the slope of \( BF \) is:
\[
\text{slope of } BF = \frac{0 - 16}{\frac{16}{3} - 16} = \frac{-16}{\frac{16}{3} - \frac{48}{3}} = \frac{-16}{\frac{-32}{3}} = \frac{16 \times 3}{32} = \frac{3}{2}
\]
The equation of the line \( BF \) (in slope-intercept form) is:
\[
y - 16 = \frac{3}{2}(x - 16)
\]
Simplifying:
\[
y - 16 = \frac{3}{2}x - 24 \quad \Rightarrow \quad y = \frac{3}{2}x - 8
\]

#### Line \( AE \):
The slope of \( AE \) is:
\[
\text{slope of } AE = \frac{0 - 16}{\frac{32}{3} - 0} = \frac{-16}{\frac{32}{3}} = \frac{-16 \times 3}{32} = -\frac{3}{2}
\]
The equation of the line \( AE \) is:
\[
y - 16 = -\frac{3}{2}(x - 0) \quad \Rightarrow \quad y = -\frac{3}{2}x + 16
\]

### Step 3: Finding point \( M \)
To find the coordinates of \( M \), the point of intersection of lines \( BF \) and \( AE \), we set the equations of the two lines equal:
\[
\frac{3}{2}x - 8 = -\frac{3}{2}x + 16
\]
Solving for \( x \):
\[
\frac{3}{2}x + \frac{3}{2}x = 16 + 8 \quad \Rightarrow \quad 3x = 24 \quad \Rightarrow \quad x = 8
\]
Substitute \( x = 8 \) into either equation to find \( y \). Using \( y = \frac{3}{2}x - 8 \):
\[
y = \frac{3}{2}(8) - 8 = 12 - 8 = 4
\]
Thus, \( M = (8, 4) \).

### Step 4: Area of triangle \( \triangle AMR \)
Assuming \( R \) is some significant point (possibly one of the square's vertices), but if we interpret this geometrically based on the given coordinates, a common interpretation is focusing on \( A, M, D \), since \( D \) often appears in such problems for the area of a triangle relative to the geometry of a square.

Using \( A(0,16) \), \( M(8,4) \), and \( D(0,0) \), we apply the formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substitute \( A(0,16) \), \( M(8,4) \), and \( D(0,0) \):
\[
\text{Area} = \frac{1}{2} \left| 0(4 - 0) + 8(0 - 16) + 0(16 - 4) \right| = \frac{1}{2} \left| 8(-16) \right| = \frac{1}{2} \times 128 = 64
\]

Thus, the area of triangle \( \triangle AMD \) (or \( \triangle MAR \)) is \( \boxed{64} \).

Would you like further details or clarification? 

Here are five related questions:
1. How do you find the area of a triangle using vertices?
2. What is the significance of using coordinates in geometry problems?
3. How do you solve for the point of intersection between two lines?
4. Why is the slope important when calculating line equations?
5. How can dividing a square into equal parts assist in problem-solving?

**Tip:** Always verify your geometric interpretations to ensure the correct triangle is analyzed!

Consider a square ABCD of side length 16. Let E, F be points on CD such that CE = EF = FD. Let the line BF and AE meet in M. The area of triangle MAR is? Answer is integer between 0-99

Solve the problem of finding the area of triangle MAR in a square with side length 16, divided into three equal parts. Learn how to apply coordinate geometry, slope, and the area formula for triangles with a step-by-step solution. Suitable for Grades 10-12.

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