Question

A big triangle with vertices (5,3) , (5,9) and (2,3)

A smaller triangle with vertices (12,2) , (12,6) and (10,2)

What is the ratio of the area of the smaller triangle in relation to the area of the larger triangle

Answer 1

To find the area of both triangles, we can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

### 1. Area of the larger triangle with vertices \((5, 3)\), \((5, 9)\), and \((2, 3)\):

Let:
- \( (x_1, y_1) = (5, 3) \)
- \( (x_2, y_2) = (5, 9) \)
- \( (x_3, y_3) = (2, 3) \)

Plugging into the area formula:

\[
\text{Area} = \frac{1}{2} \left| 5(9-3) + 5(3-3) + 2(3-9) \right|
\]
\[
= \frac{1}{2} \left| 5 \cdot 6 + 0 + 2(-6) \right|
\]
\[
= \frac{1}{2} \left| 30 - 12 \right|
\]
\[
= \frac{1}{2} \left| 18 \right| = 9
\]

### 2. Area of the smaller triangle with vertices \((12, 2)\), \((12, 6)\), and \((10, 2)\):

Let:
- \( (x_1, y_1) = (12, 2) \)
- \( (x_2, y_2) = (12, 6) \)
- \( (x_3, y_3) = (10, 2) \)

Plugging into the area formula:

\[
\text{Area} = \frac{1}{2} \left| 12(6-2) + 12(2-2) + 10(2-6) \right|
\]
\[
= \frac{1}{2} \left| 12 \cdot 4 + 0 + 10(-4) \right|
\]
\[
= \frac{1}{2} \left| 48 - 40 \right|
\]
\[
= \frac{1}{2} \left| 8 \right| = 4
\]

### 3. Ratio of the area of the smaller triangle to the area of the larger triangle:

Now that we have the areas:
- Area of the larger triangle = \(9\)
- Area of the smaller triangle = \(4\)

The ratio is:

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

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