Step 1: Draw the triangle

To draw the triangle $\triangle ABC$, we plot the given vertices on a Cartesian plane:

• $A(1, 3)$
• $B(0, 5)$
• $C(0, 0)$

Let's plot these points:

1. $A$ is at $(1, 3)$, so it is 1 unit to the right and 3 units up from the origin.
2. $B$ is at $(0, 5)$, so it is directly above the origin by 5 units.
3. $C$ is at $(0, 0)$, which is the origin.

Now, we draw the lines connecting these points to form the triangle.

Step 2: Find the area of $\triangle ABC$

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

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

Substitute the coordinates of $A(1, 3)$, $B(0, 5)$, and $C(0, 0)$:

$\text{Area} = \frac{1}{2} \left| 1(5 - 0) + 0(0 - 3) + 0(3 - 5) \right|$ $= \frac{1}{2} \left| 1(5) + 0 + 0 \right|$ $= \frac{1}{2} \times 5 = 2.5$

So, the area of $\triangle ABC$ is 2.5 square units.

Step 3: Find the length of the side $AB$

The length of a line segment between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

For $AB$, with $A(1, 3)$ and $B(0, 5)$, we calculate:

$AB = \sqrt{(0 - 1)^2 + (5 - 3)^2}$ $AB = \sqrt{(-1)^2 + (2)^2}$ $AB = \sqrt{1 + 4} = \sqrt{5}$

So, the length of side $AB$ is $\sqrt{5} \approx 2.236$ units.

---

Would you like further details or clarifications on any of the steps?

5 Related Questions:

1. How can you apply the area formula to other polygons?
2. What is the formula for the area of a right triangle, and does it apply here?
3. How can you calculate the area of a triangle using base and height?
4. What is the formula to calculate the perimeter of a triangle?
5. How do you calculate the lengths of the other sides of the triangle (i.e., $BC$ and $AC$)?

Tip: Remember that the area of a triangle can also be calculated using the determinant method when given the vertices, and it can be a quick way to check your results!