Area of a Triangle: How many times does the shape fit inside of another shape?

To solve this problem, we will find how many times a triangle can fit inside a square based on given dimensions:

• Step 1: Compute the area of the larger square.
• Step 2: Compute the area of the smaller square (side: $3.5$) which forms the base and height of the triangle.
• Step 3: Compute the area of the triangle using the area of the smaller square.
• Step 4: Divide the area of the larger square by the area of the triangle to find how many triangles fit.

Now, let's proceed with these calculations:
Step 1: Area of the large square = $7 \times 7 = 49$.
Step 2: Area of the smaller square = $3.5 \times 3.5 = 12.25$.
Step 3: Since the triangle fits perfectly within this square, and is a right isosceles triangle, its area = $\frac{1}{2} \times 3.5 \times 3.5 = 6.125$.
Step 4: Dividing the area of the large square by the area of the triangle: $\frac{49}{6.125} \approx 8$.

Therefore, the large square can completely fit exactly 8 triangles inside it.

To solve this problem, we'll follow these steps:

• Step 1: Calculate the area of triangle KBC.
• Step 2: Determine the relevant dimensions for trapezoid AKCD and calculate its area.
• Step 3: Divide the area of the trapezoid by the area of the triangle to determine how many triangles fit into the trapezoid.

Now, let's work through each step:

Step 1: Calculate the area of triangle KBC.
KBC is a right triangle where $KB = 4 \, \text{cm}$ and $BC = 5 \, \text{cm}$ (since $AD = 5 \, \text{cm}$ is a vertical line segment in the rectangle, $BC$ must be equal to $AD$).
The area of triangle KBC is given by:

$\text{Area}_{KBC} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 5 = 10 \, \text{cm}^2$

Step 2: Calculate the area of trapezoid AKCD.
For trapezoid AKCD, $AK$ is the shorter parallel side, and $DC = 14 \, \text{cm}$ is the longer parallel side. The height is the same as the height of rectangle AD or BC, $h = 5 \, \text{cm}$.
To find $AK$, since $KB = 4 \, \text{cm}$ and total $AB = 14\, \text{cm}$, thus $AK = AB - KB = 14 - 4 = 10\, \text{cm}$.
The area of trapezoid AKCD is given by:

$\text{Area}_{AKCD} = \frac{1}{2} \times (AK + DC) \times h = \frac{1}{2} \times (10 + 14) \times 5 = \frac{1}{2} \times 24 \times 5 = 60 \, \text{cm}^2$

Step 3: Calculate how many triangles KBC fit into trapezoid AKCD.
To find how many triangles fit into the trapezoid, divide the area of trapezoid by the area of the triangle:

$\frac{\text{Area}_{AKCD}}{\text{Area}_{KBC}} = \frac{60}{10} = 6$

Therefore, the solution to the problem is that 6 triangles identical to triangle KBC are needed to create the trapezoid AKCD.