Question

Find the answer please

Answer 1

For this question, we are going to be dealing with 3 right triangles.

The square on the left side has dimensions of 7m x 7m (Due to it being a square, it must have equal side lengths.

The rectangle on the right then has dimensions of 3m x 8m (Due to the total length being 16m and subtracting 7m from the square = 8m)

So the rightmost shaded right triangle we solve for readily. We know that the area of a triangle is represented by: $\frac{1}{2}*bh$. We already know the b and h of this triangle:

$\frac{1}{2}*(3m)*(8m)=12 m^{2}$

For the left side, it is a tad more complex. We are going to find the area of the two right triangles that are unshaded and subtract that area from the total area of the square.

For one of the triangles, the base is 4m (Found by subtracting 3m from 7m) and the height is 7m. So the area is:

$\frac{1}{2}*(4m)*(7m)=14 m^{2}$

For the second triangle, the base is 7m and the height is 7m. So the area is:

$\frac{1}{2} *(7m)*(7m)=24.5 m^{2}$

So the total of the two unshaded triangles in the square totals: $14 m^{2} +24.5 m^{2}=38.5 m^{2}$

We know that the total area in the square is: $7m*7m=49 m^{2}$

The total area of the shaded region will be the total area of the square minus the area of the unshaded right triangles. So let's solve:

$49 m^{2}-38.5 m^{2}=10.5 m^{2}$

We now know that the shaded area inside the square is equal to $10.5 m^{2}$.

Let's add this shaded area to the shaded area inside the rectangle:

$10.5 m^{2}+12 m^{2}=22.5 m^{2}$

So now we know that the area of the total shaded area in the figure is equal to: $22.5 m^{2}$