From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(18 \times 9)=81\)

Next, recall how to find the area of a circle.

\(\displaystyle \text{Area of Circle}=\pi\times\text{radius}^2\)

\(\displaystyle \text{Area of Circle}=\pi\times 9^2= 81\pi\)

To find the area of the shaded region, subtract the two areas.

\(\displaystyle \text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}\)

\(\displaystyle \text{Area of Shaded Region}=81\pi-81=173.47\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(26 \times 13)=169\)

Next, recall how to find the area of a circle.

\(\displaystyle \text{Area of Circle}=\pi\times\text{radius}^2\)

\(\displaystyle \text{Area of Circle}=\pi\times 13^2= 169\pi\)

To find the area of the shaded region, subtract the two areas.

\(\displaystyle \text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}\)

\(\displaystyle \text{Area of Shaded Region}=169\pi-169=361.93\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(44 \times 22)=484\)

Next, recall how to find the area of a circle.

\(\displaystyle \text{Area of Circle}=\pi\times\text{radius}^2\)

\(\displaystyle \text{Area of Circle}=\pi\times 22^2= 484\pi\)

To find the area of the shaded region, subtract the two areas.

\(\displaystyle \text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}\)

\(\displaystyle \text{Area of Shaded Region}=484\pi-484=1036.53\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(46 \times 23)=529\)

Next, recall how to find the area of a circle.

\(\displaystyle \text{Area of Circle}=\pi\times\text{radius}^2\)

\(\displaystyle \text{Area of Circle}=\pi\times 23^2= 529\pi\)

To find the area of the shaded region, subtract the two areas.

\(\displaystyle \text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}\)

\(\displaystyle \text{Area of Shaded Region}=529\pi-529=1132.90\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(30 \times 15)=225\)

Next, recall how to find the area of a circle.

\(\displaystyle \text{Area of Circle}=\pi\times\text{radius}^2\)

\(\displaystyle \text{Area of Circle}=\pi\times 15^2= 225\pi\)

To find the area of the shaded region, subtract the two areas.

\(\displaystyle \text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}\)

\(\displaystyle \text{Area of Shaded Region}=225\pi-225=481.86\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the given image, you should notice that the base of the triangle is also the diameter of the circle. In addition, the height of the triangle is also the radius of the circle.

Thus, we can find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(52 \times 26)=676\)

Next, recall how to find the area of a circle.

\(\displaystyle \text{Area of Circle}=\pi\times\text{radius}^2\)

\(\displaystyle \text{Area of Circle}=\pi\times 26^2= 676\pi\)

To find the area of the shaded region, subtract the two areas.

\(\displaystyle \text{Area of Shaded Region}=\text{Area of Circle}-\text{Area of Triangle}\)

\(\displaystyle \text{Area of Shaded Region}=676\pi-676=1447.72\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

In order to find the area of the shaded region, we will need to find the areas of the triangle and of the parallelogram.

First, recall how to find the area of a parallelogram.

\(\displaystyle \text{Area of Parallelogram}=\text{base}\times\text{height}\)

\(\displaystyle \text{Area of Parallelogram}=2\times 1 = 2\)

Next, recall how to find the area of a triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base}\times\text{height})\)

Now, find the height of the triangle.

\(\displaystyle \text{Height of Triangle}=\frac{\text{Height of Parallelogram}}{2}\)

\(\displaystyle \text{Height of Triangle}=\frac{1}{2}\)

Plug this value in to find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(2\times \frac{1}{2})=\frac{1}{2}=0.5\)

Subtract the two areas to find the area of the shaded region.

\(\displaystyle \text{Area of Shaded Region}=\text{Area of Parallelogram}-\text{Area of Triangle}\)

\(\displaystyle \text{Area of Shaded Region}=2-0.5=1.5\)

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar. 

\(\displaystyle \angle AVB\) and \(\displaystyle \angle CVD\) are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, \(\displaystyle \angle AVB \cong \angle CVD\).

\(\displaystyle \angle ABV\) and \(\displaystyle \angle CDV\) are alternating interior angles formed by two parallel lines \(\displaystyle m\) and \(\displaystyle n\) cut by a transversal \(\displaystyle t\). As a consequence, \(\displaystyle \angle ABV \cong \angle CDV\).

The conditions of the Angle-Angle Similarity Postulate are satisfied, and it holds that \(\displaystyle \bigtriangleup AVB \sim \bigtriangleup CVD\).

The given information is actually inconclusive.

By the Angle-Angle Similarity Postulate, if two pairs of corresponding angles of a triangle are congruent, the triangles themselves are similar. Therefore, we seek to prove two of the following three angle congruence statements:

\(\displaystyle \angle AVB \cong \angle CVD\)

\(\displaystyle \angle ABV \cong \angle CDV\)

\(\displaystyle \angle BAV \cong \angle DCV\)

\(\displaystyle \angle AVB\) and \(\displaystyle \angle CVD\) are a pair of vertical angles, having the same vertex and having sides opposite each other. As such, \(\displaystyle \angle AVB \cong \angle CVD\). 

\(\displaystyle \angle ABV \cong \angle DCV\), but this is not one of the statements we need to prove. Also, without further information - for example, whether \(\displaystyle m\) and \(\displaystyle n\) are parallel, which is not given to us - we have no way to prove either of the other two necessary statements. 

The correct response is "false".

As we are establishing whether or not \(\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF\), then \(\displaystyle A\), \(\displaystyle B\), and \(\displaystyle C\) correspond respectively to \(\displaystyle D\), \(\displaystyle E\), and \(\displaystyle F\).

\(\displaystyle \bigtriangleup DEF\) is an isosceles triangle, so it must have two congruent angles. \(\displaystyle \angle F\) has measure \(\displaystyle 30^{\circ }\), so either \(\displaystyle \angle D\) has this measure, \(\displaystyle \angle E\) has this measure, or \(\displaystyle \angle D \cong \angle E\). If we examine the second case, it immediately follows that \(\displaystyle \angle B \ncong \angle E\). One condition of the similarity of triangles is that all pairs of corresponding angles be congruent; since there is at least one case that violates this condition, it does not necessarily follow that  \(\displaystyle \bigtriangleup ABC \sim \bigtriangleup DEF\). This makes the correct response "false".