Bunuel wrote:
Points A, B, and C lie on a circle centered on point O. If line ZY is tangent to the circle at point B, angle CYB measures 45 degrees, and the circle has an area of 16π, what is the area of triangle ZOY?
A. 8
B. \(8\sqrt{2}\)
C. 16
D. \(16\sqrt{2}\)
E. 32
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The area of ∆ ZOY can be found in many ways. Here: ∆ ZOY consists of two congruent right isosceles triangles.
Their summed bases equal the base of ∆ ZOY, and their shared side, OB, with a known length, is the height of ∆ ZOY
For large ∆ ZOY, the third angle, at vertex Z, must = 45°
∆ ZOY already has one 90 and one 45-degree angle (given)
Angle at vertex Z: (180 - 135) = 45
Large ∆ ZOY consists of two congruent right isosceles triangles, ∆ BOY and ∆ BOZ
-- Tangent line YZ is perpendicular to radius OB, creating two right angles, one in each small triangle
-- Each small triangle has a vertex outside the circle, at Z (derived) and at Y (given), that = 45°
-- Their vertices at the center of the circle, O, both = 45° (for both small triangles, 90 + 45 + third vertex at O = 180, third vertex = 45)
Sides opposite equal angles are equal
Each small triangle has two equal sides, and both pairs (OB/BZ and OB/BY) lie opposite equal 45° angles.
Hence the sides are equal: Radius / side OB (shared) = side BZ = side BY
Length of those sides = \(4\)
Circle area = \(πr^2 = 16π , r^2 = 16 , r = 4 =\) OB, BZ, and BY
Base of ∆ ZOY = \((4 + 4) = 8\)
Area of ∆ ZOY = \(\frac{b*h}{2} = \frac{8*4}{2} = 16\)
Answer C
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63% (02:09) correct 
