the sides of a cyclic quadrilateral measures 8cm, 9cm, 12cm and 7cm. Find the area of the circumscribing circle
Ans. 134.37 cm^2
Solution:
Area of cyclic Quadrilateral
\(\displaystyle A_{cyc} = \sqrt{(s-a)(s-b)(s-c)(s-d)}\)
where
\(\displaystyle s = \frac{a + b + c + d}{2} \)
and the radius of the circumscribing circel
\(\displaystyle r = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4A_{cyc}}\)
\(\displaystyle s = \frac{9 + 8 + 7 + 12}{2} \)
\(\displaystyle s = 18 \)
\(\displaystyle A_{cyc} = \sqrt{ (18-9)(18-8)(18-7)(18-12) }\)
\(\displaystyle A_{cyc} = 69.714\)
\(\displaystyle r = \frac{\sqrt{((9)(8)+(7)(12))(9(7)+8(12))(9(12)+8(7))}}{4(69.714)}\)
\(\displaystyle r = 7.233\)
\(\displaystyle A_{circ} = \pi{7.233}^2\)
Ans. \(\displaystyle A_{circ} = 164.37\)
WHere did i get wrong?
Ans. 134.37 cm^2
Solution:
Area of cyclic Quadrilateral
\(\displaystyle A_{cyc} = \sqrt{(s-a)(s-b)(s-c)(s-d)}\)
where
\(\displaystyle s = \frac{a + b + c + d}{2} \)
and the radius of the circumscribing circel
\(\displaystyle r = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4A_{cyc}}\)
\(\displaystyle s = \frac{9 + 8 + 7 + 12}{2} \)
\(\displaystyle s = 18 \)
\(\displaystyle A_{cyc} = \sqrt{ (18-9)(18-8)(18-7)(18-12) }\)
\(\displaystyle A_{cyc} = 69.714\)
\(\displaystyle r = \frac{\sqrt{((9)(8)+(7)(12))(9(7)+8(12))(9(12)+8(7))}}{4(69.714)}\)
\(\displaystyle r = 7.233\)
\(\displaystyle A_{circ} = \pi{7.233}^2\)
Ans. \(\displaystyle A_{circ} = 164.37\)
WHere did i get wrong?