The problem asks for the area of the figure, which is a combination of a rectangle and two triangles. The rectangle has a height of 7 inches and a width of 16 inches. The two triangles are congruent and their combined base is 8 inches + 8 inches = 16 inches, and the height is 4 inches (7 inches - 3 inches, where 3 inches is the remaining height on the sides of the triangle after subtracting 4 inches from the total height of 7 inches).

1. Problem Description

The problem asks for the area of the figure, which is a combination of a rectangle and two triangles. The rectangle has a height of 7 inches and a width of 16 inches. The two triangles are congruent and their combined base is 8 inches + 8 inches = 16 inches, and the height is 4 inches (7 inches - 3 inches, where 3 inches is the remaining height on the sides of the triangle after subtracting 4 inches from the total height of 7 inches).

2. Solution Steps

First, we find the area of the rectangle.

$Area_{rectangle} = length \times width = 16 \times 7 = 112$ square inches.

Next, we find the area of the two triangles. The two triangles can be viewed as one triangle with a base of 8 in + 8 in = 16 in and height of 7-4 =3 in. Since the base is $b = 8 + 8 = 16$ and the height is $h = 7 - 4 = 3$, the total height of the two rectangles on top of it is 7 in. Therefore, the difference gives the height of 3 in. The area is given as:

$Area_{triangles} = \frac{1}{2} \times base \times height$

The base of each individual triangle cannot be added. We can consider the top triangles to be triangles which have sides of 8 inches and a height of

3. In doing so, the combined area is $2 \times (\frac{1}{2} \times 8 \times 3) = 24$. This approach is wrong.

The entire bottom length is 16 inches. The rectangular part has a length of 16 inches and the top side of rectangle has a total horizontal length of $8 + 8 = 16$, each sides are of equal lengths so this gives an isosceles triangle. The height is equal to $7 - 4 = 3$ inches. This means that the total area of the two triangles is:

$Area_{triangles} = \frac{1}{2} \times 16 \times 3 = \frac{48}{2} = 24$ square inches.

Finally, we add the area of the rectangle and the area of the two triangles.

$Total Area = Area_{rectangle} + Area_{triangles} = 112 - 24 = 88$ wrong!

$Total Area = Area_{rectangle} - Area_{triangles} = 112 - 16 \times 3 / 2= 112 + 24 =88$

Let the length of the rectangle is 16 and height is 4

Area of rectangle = $16 \times 4= 64$

Then, area of two equal sized rectangles with width 8 and height of 3 on the sides, with area $8 \times 3 = 24$. Add these two sides

Total = $64+24=88$

3. Final Answer

B. 88 in.$^2$