Question
Angle D is a circumscribed angle of circle O.
Circle O is shown. Triangle A C B is inscribed within circle O. Side A B goes through point O. The length of A C is 15 and the length of C B is 8. Angle A C B is a right angle. Line segment O E is a radius. Tangents B D and E D intersect at point D outside of the circle to form kite O B D E. Angles O B D and D E O are right angles. The length of E D is 5.
What is the perimeter of kite OBDE?
17 units
23 units
27 units
60 units
Circle O is shown. Triangle A C B is inscribed within circle O. Side A B goes through point O. The length of A C is 15 and the length of C B is 8. Angle A C B is a right angle. Line segment O E is a radius. Tangents B D and E D intersect at point D outside of the circle to form kite O B D E. Angles O B D and D E O are right angles. The length of E D is 5.
What is the perimeter of kite OBDE?
17 units
23 units
27 units
60 units
Answers
GPT-4o mini
To find the perimeter of kite \( OBDE \), we first need to determine the lengths of its sides.
Since \( OBDE \) is a kite, we know that \( OB = OD \) and \( BE = DE \).
1. **Finding Length \( OB \)**:
- Since \( ACB \) is a right triangle and point \( O \) lies on line \( AB \), we can use the Pythagorean theorem.
- We have:
\[
AC^2 + CB^2 = AB^2
\]
- Plugging in the known lengths:
\[
15^2 + 8^2 = AB^2 \\
225 + 64 = AB^2 \\
AB^2 = 289 \\
AB = \sqrt{289} = 17
\]
- Since point \( O \) bisects \( AB \), we have:
\[
OA = \frac{AB}{2} = \frac{17}{2} = 8.5
\]
- Therefore, \( OB = OA = 8.5 \).
2. **Finding Length \( OD \)**:
- Since \( O \) is the center and \( DE \) is a radius, we have:
\[
OD = OE = 5 \text{ (since } ED \text{ is tangent and a radius)}
\]
3. **Finding Length \( BE \)**:
- Since \( DE \) is given as length \( 5 \) and by the properties of the kite:
\[
BE = DE = 5
\]
4. **Perimeter Calculation**:
- The perimeter \( P \) of kite \( OBDE \) can now be calculated as:
\[
P = OB + OD + BE + DE
\]
- Substituting the values we found:
\[
P = 8.5 + 5 + 5 + 8.5 = 27
\]
Thus, the perimeter of kite \( OBDE \) is \( \boxed{27} \) units.
Since \( OBDE \) is a kite, we know that \( OB = OD \) and \( BE = DE \).
1. **Finding Length \( OB \)**:
- Since \( ACB \) is a right triangle and point \( O \) lies on line \( AB \), we can use the Pythagorean theorem.
- We have:
\[
AC^2 + CB^2 = AB^2
\]
- Plugging in the known lengths:
\[
15^2 + 8^2 = AB^2 \\
225 + 64 = AB^2 \\
AB^2 = 289 \\
AB = \sqrt{289} = 17
\]
- Since point \( O \) bisects \( AB \), we have:
\[
OA = \frac{AB}{2} = \frac{17}{2} = 8.5
\]
- Therefore, \( OB = OA = 8.5 \).
2. **Finding Length \( OD \)**:
- Since \( O \) is the center and \( DE \) is a radius, we have:
\[
OD = OE = 5 \text{ (since } ED \text{ is tangent and a radius)}
\]
3. **Finding Length \( BE \)**:
- Since \( DE \) is given as length \( 5 \) and by the properties of the kite:
\[
BE = DE = 5
\]
4. **Perimeter Calculation**:
- The perimeter \( P \) of kite \( OBDE \) can now be calculated as:
\[
P = OB + OD + BE + DE
\]
- Substituting the values we found:
\[
P = 8.5 + 5 + 5 + 8.5 = 27
\]
Thus, the perimeter of kite \( OBDE \) is \( \boxed{27} \) units.