Quadrilateral circle radius

Given is a quadrilateral ABCD inscribed in a circle, with the diagonal AC being the circle's diameter. The distance between point B and the diameter is 15 cm, and between point D and the diameter is 18 cm. Calculate the radius of the circle and the perimeter of the quadrilateral ABCD.

Final Answer:

r =  18 cm
o =  99.7 cm

Step-by-step explanation:

$$\begin{gathered} h_1=15\text{cm} \\ {h}_2=18\text{cm} \\ AC=2r \\ x+y=r \\ z+w=r \\ {h}_1^2=x\cdot (y+z+w) \\ {h}_2^2=w\cdot (x+y+z) \\ {h}_1^2=x\cdot (y+r) \\ {h}_2^2=w\cdot (r+z) \\ {h}_1^2=(r-y)\cdot (y+r) \\ {h}_2^2=(r-z)\cdot (r+z) \\ {h}_1^2=r^2-y^2 \\ {h}_2^2=r^2-z^2 \\ z=0 \\ r=\sqrt{h_2^2+z^2}=\sqrt{18^2+0^2}=18=18\text{cm} \\ w=r-z=18-0=18\text{cm} \\ {h}_1^2=r^2-y^2 \\ 15^2=18^2-y^2 \\ {y}^2-99=0 \\ {y}_{1,2}=\pm \sqrt{99}=\pm 9.949874371 \\ {y}_1=9.949874371 \\ {y}_2=-9.949874371 \\ y=y_1=9.9499=3\sqrt{11}\text{cm}\doteq 9.9499\text{cm} \\ x=r-y=18-9.9499\doteq 8.0501\text{cm} \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} a=\sqrt{h_1^2+x^2}=\sqrt{15^2+8.0501^2}\doteq 17.0236\text{cm} \\ b=\sqrt{h_1^2+(r+y{)}^2}=\sqrt{15^2+(18+9.9499{)}^2}\doteq 31.7206\text{cm} \\ c=\sqrt{h_2^2+w^2}=\sqrt{18^2+18^2}=18\sqrt{2}\text{cm}\doteq 25.4558\text{cm} \\ d=\sqrt{h_2^2+(r+z{)}^2}=\sqrt{18^2+(18+0{)}^2}=18\sqrt{2}\text{cm}\doteq 25.4558\text{cm} \\ o=a+b+c+d=17.0236+31.7206+25.4558+25.4558=99.7\text{cm} \end{gathered}$$

Try another example