Engineering Board Exam Reviewer · Ptolemy's Theorem and Cyclic Quadrilaterals · hard
A cyclic quadrilateral has sides 9, 10, 17 and 6. Find its area.
- 88.99
- 90
- 7920
- 105
Answer
A. 88.99
Solution
Brahmagupta's formula with s=21: A=sqrt((21-9)(21-10)(21-17)(21-6))=sqrt(12(11)(4)(15))=sqrt(7920) 88.99.
More Ptolemy's Theorem and Cyclic Quadrilaterals questions
- A cyclic quadrilateral ABCD has AB=CD=5, BC=6 and AD=14. Find the length of diagonal AC.
- The angle A of a cyclic quadrilateral ABCD is 70°. Find the opposite angle C.
- A cyclic quadrilateral ABCD has sides AB=6, BC=7, CD=8, and DA=5. If diagonal AC=10, find diagonal BD using Ptolemy's theorem.
- A cyclic quadrilateral ABCD has consecutive sides AB=4, BC=5, CD=6, DA=7. If diagonal AC=7.5, find diagonal BD using Ptolemy's theorem.
- A rectangle (a special cyclic quadrilateral) has sides 6 and 8. Find the length of its diagonal.
- A cyclic quadrilateral ABCD has AB=7, BC=24, CD=15, DA=20 and diagonal AC=25. Find the radius of the circumscribed circle.