Question

What is the ratio of the area of the smaller triangle in relation to the area of the larger triangle?
Larger
A (5,9)
B (5,3)
C (2,3)
Smaller
A’ (12,6)
B’ (12,2)
C’ (10,2)

Answer 1

To find the ratio of the areas of the triangles, we can use the formula for the area of a triangle given its vertices:

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

### Area of the Larger Triangle (A, B, C)
Vertices:
- A(5, 9)
- B(5, 3)
- C(2, 3)

Let's plug in the coordinates into the area formula:

\[
\text{Area}_\text{larger} = \frac{1}{2} \left| 5(3 - 3) + 5(3 - 9) + 2(9 - 3) \right|
\]

Calculating each term:
1. \( 5(3 - 3) = 5 \cdot 0 = 0 \)
2. \( 5(3 - 9) = 5 \cdot (-6) = -30 \)
3. \( 2(9 - 3) = 2 \cdot 6 = 12 \)

Combining these:
\[
\text{Area}_\text{larger} = \frac{1}{2} \left| 0 - 30 + 12 \right| = \frac{1}{2} \left| -18 \right| = \frac{18}{2} = 9
\]

### Area of the Smaller Triangle (A’, B’, C’)
Vertices:
- A’(12, 6)
- B’(12, 2)
- C’(10, 2)

Now, we calculate the area:

\[
\text{Area}_\text{smaller} = \frac{1}{2} \left| 12(2 - 2) + 12(2 - 6) + 10(6 - 2) \right|
\]

Calculating each term:
1. \( 12(2 - 2) = 12 \cdot 0 = 0 \)
2. \( 12(2 - 6) = 12 \cdot (-4) = -48 \)
3. \( 10(6 - 2) = 10 \cdot 4 = 40 \)

Combining these:
\[
\text{Area}_\text{smaller} = \frac{1}{2} \left| 0 - 48 + 40 \right| = \frac{1}{2} \left| -8 \right| = \frac{8}{2} = 4
\]

### Ratio of Areas
Now we can find the ratio of the area of the smaller triangle to the area of the larger triangle:

\[
\text{Ratio} = \frac{\text{Area}_\text{smaller}}{\text{Area}_\text{larger}} = \frac{4}{9}
\]

Thus, the ratio of the area of the smaller triangle to the area of the larger triangle is \(\frac{4}{9}\).