Explanation:
2) ∠AMB is 90°, because the angle in a semicircle is always 90° (Thales's Theorem).
∠CMA is 90°, because the measure of arc AC is 90° (Angles Subtended by Same Arc Theorem).
The measure of arc CMA is 270°.
Triangle ΔAMB is a right triangle with a 30° angle, so it must be a 30-60-90 triangle.
3) We can use properties of 30-60-90 triangles to find AM and MB.
MB is half the hypotenuse, so MB = 4 cm.
AM is MB times √3, so AM = 4√3 cm.
OB and MB are both 4 cm, and angle ∠OBM is 60°, so triangle ΔOMB must be an equilateral triangle.
4) Triangle ΔAOI is a 30-60-90 triangle. AO is 4 cm, and OI is 4/√3 cm.
The area is therefore:
A = ½ bh
A = ½ (AO) (OI)
A = ½ (4 cm) (4/√3 cm)
A = 8/√3 cm²
5) "Orthogonal projection" simply means "perpendicular line". If we draw a line perpendicular to AM that passes through O, the intersection of that line with AM is point E.
a) If OE is the height of the triangle and AI is the base, then we can use the area from question 4 to solve for OE.
But first, we need to solve for AI using properties of 30-60-90 triangles.
AI = 2×OI = 8/√3 cm
A = ½ bh
A = ½ (AI) (OE)
8/√3 cm² = ½ (8/√3 cm) (OE)
OE = 2 cm
b) Triangle ΔAOE is a 30-60-90 triangle, and triangle ΔABM is a 30-60-90 triangle. Therefore, both triangles are similar, and OE is parallel to BM.