Given:In ⊙O, OA = 15, BC = 18, and CD = 12. Find:AABCD in square units

Question

Consider the following figure.

A circle contains a quadrilateral with four labeled vertices, a labeled center point, and a dashed line segment.

1. Each vertex of the quadrilateral A B C D is on the circle.
2. The first side of the quadrilateral starts at vertex A, goes up and to the left, and ends at vertex B.
3. The second side of the quadrilateral starts at vertex B, goes down and to the left, and ends at vertex C.
4. The third side of the quadrilateral starts at vertex C, goes down and to the left, and ends at vertex D.
5. The fourth side of the quadrilateral starts at vertex D, goes down and to the right, and ends at vertex A.
6. The center of the circle is point O.
7. A dashed line segment starts at vertex C, goes through the center point O, and ends at vertex A.

Given:In ⊙O, OA = 15, BC = 18, and CD = 12.

Find:A_ABCD  in square units

AABCD = .  units2

 

 

 

 

Answer 1

Because angles B and D intercept semi-circles (180 degrees), their measure is 90 degrees and we have two right triangles. We just need to compute the area of each triangle and add them up.

The hypotenuse of both right triangles is AC whose length is 2 times the radius OA (15) = 30

By the Pythagorean Theorem, AC**2 = CD**12 + AD**2 and since CD = 12 and AC = 30:

30**2 = 12**2 + AD**2

AD = 27.5

Area of right triangle = 1/2 the product of the two legs.

Area of ACD= 1/2 * AD * CD = 1/2 * 27.5 * 12 = 165

By the Pythagorean Theorem, BC**2 + AB**2 = AC**2 and since BC is 18 and AC = 30:

30**2 = 18**2 + AB**2

AB = 24

Area of ABC = 1/2 * 18 * 24 = 216

Sum of areas of triangles is 216 + 165 = 381

Area of ABCD is thus 381