Answer to Question #334537 in Geometry for Tc1foster

Question

Tammy has a circular garden with a diameter of 36 feet. She will create 4 stone

pathways in the garden that will form an inscribed quadrilateral in her circular

garden. Tammy wants the straight-line distance between one pair of opposite

vertices to be 36 feet. She also wants two consecutive pathways to have

lengths of 24 feet and 12 feet. Tammy will plant marigolds along the shorter edge of

the garden that is cutoff by the pathways of lengths 24 feet and 12 feet.

What length, to the nearest tenth of a foot, of the edge of the garden will be planted

with marigolds?

Answer 1

Consider the quadrilateral ABCD: let's draw AC - it will be perpendicular to the side AB, due to the property of an inscribed angle that relies on the diameter. Proceeding from this, the triangle ABD is right-angled, then:

$\cos \angle BAD=\dfrac{AB}{AD}$ ;

$\cos \angle BAD=\dfrac{12}{36}=0.(3)$

$\angle BAD \thickapprox 71^o$

$\angle ADB \thickapprox 19^o$

$\cup AB =38^o$

Behind the property of the angles of a quadrilateral inscribed in a circle:

$\angle A+\angle С = 180$

$\angle С = 180-142 = 38^o$

$BD^2=AD^2-AB^2$

$BD=\sqrt{36^2-12^2}\thickapprox 33.94$

$\dfrac{BD}{\cos \angle C } = \dfrac{BC}{\cos \angle BDC}$

$\cos \angle BDC = \dfrac{BC* \cos \angle C}{BD}$

$\cos \angle BDC = \dfrac{24* 0.778}{33.94}=0.55$

$\angle BDC \thickapprox 56^o$

$\cup BC = 112^o$

$L_{\cup AC}= \dfrac{\pi R \angle AOC}{180} = \dfrac{\pi * 36*(112+38)}{180}=94.25 foot$