Question

4
8
6
13
First, find the area of the triangle.
6
Find the Area.
8
8
Triangle
[?] +
13
Remember:
A₁ = bh
Rectangle Total Area
=
units²

Answer 1

Answer:

128cm²

Step-by-step explanation:

1. Solve for the area of the triangle:

To find the area of a triangle, you can use the formula:

$\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]$

Given that the base is 6 units and the height is 8 units, you can plug these values into the formula:

$\[ \text{Area} = \frac{1}{2} \times 6 \times 8 \]

\ \\ \text{Area} = \frac{1}{2} \times 48 \] \\

\[ \text{Area} = 24 \]$

So, the area of the triangle is 24 units².

2. Solve for the area of the rectangle:

To find the area of a rectangle, you multiply its length by its width.

For a rectangle with sides of length 8 units and 13 units:

$\[ \text{Area} = \text{length} \times \text{width} \] \\

\[ \text{Area} = 8 \text{unit} \times 13 \text{unit} \] \\

\ \text{Area} = 104 \text{ units}^2 \]$

So, the area of the rectangle is 104 units².

3. Add both areas = 104 + 24 = 128 units²

Answer 2

Answer:

$\begin{aligned} \textsf{ Traingle} & \quad \textsf{ Rectangle} & \sf \quad \textsf{ Total Area } \\\\ \boxed{\,24 \, } &\quad \boxed{ \, 104 \, } & \sf = \boxed{ \, 128 \, } \textsf{ units}^2 \end{aligned}$

Step-by-step explanation:

To find the area of the figure, which is a combination of a triangle and a rectangle, we need to find the area of each component and then add them together.

Area of the Triangle ($A_{\textsf{triangle}})$:

$A_{\textsf{triangle}} = \dfrac{1}{2} \times \textsf{base} \times \textsf{height}$

$A_{\textsf{triangle}} = \dfrac{1}{2} \times 6 \times 8$

$A_{\textsf{triangle}} = 24$

Area of the Rectangle ($A_{\textsf{rectangle}})$:

$A_{\textsf{rectangle}} = \textsf{length} \times \textsf{width}$

$A_{\textsf{rectangle}} = 13 \times 8$

$A_{\textsf{rectangle}} = 104$

Total Area ($A_{\textsf{total}})$:

$A_{\textsf{total}} = A_{\textsf{triangle}} + A_{\textsf{rectangle}}$

$A_{\textsf{total}} = 24 + 104$

$A_{\textsf{total}} = 128$

Therefore, the total area of the figure is $128$ units²