If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total .

 and  are two such angles, so 

Setting  and , and solving for :

,

The statement turns out to be true regardless of the value of . Therefore, without further information, the value of  cannot be determined.

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total .

 and  are two such angles, so 

Setting  and , and solving for :

,

the correct response.

  is a diameter, so  is a semicircle - therefore, . By the Arc Addition Principle,

If we let , then

,

and

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since  and  are such segments intercepting  and , it holds that

Setting , , and :

The inscribed angle that intercepts this arc, , has half this measure:

.

This is the correct response.

 is a diameter, so  is a semicircle, which has measure . By the Arc Addition Principle,

If we let , then, substituting:

,

and

the ratio of the length of  to that of  is 7 to 5; this is also the ratio of their degree measures; that is,

Setting  and :

Cross-multiply, then solve for :

, and 

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since  and  are such segments whose angle  intercepts  and , it holds that:

Set the area of the circle equal to four times the circumference πr² = 4(2πr). 

Cross out both π symbols and one r on each side leaves you with r = 4(2) so r = 8 and therefore d = 16.

The perimeter of a circle = 2 πr = πd

Therefore d = 36

Obtain the radius of the circle from the area.

Split the square up into 4 triangles by connecting opposite corners. These triangles will have a right angle at the center of the square, formed by two radii of the circle, and two 45-degree angles at the square's corners. Because you have a 45-45-90 triangle, you can calculate the sides of the triangles to be , , and . The radii of the circle (from the center to the corners of the square) will be 9. The hypotenuse (side of the square) must be .

The area of the square is then .

For the circle to contain all 3 vertices, the hypotenuse must be the diameter of the circle. The hypotenuse, and therefore the diameter, is 5, since this must be a 3-4-5 right triangle.

The equation for the area of a circle is A = πr².

Write the formula for the circumference.

Substitute the circumference.

To solve, simply use the formula for the area of a circle to find the radius, and then multiply it by 2 to find the diameter. Thus,