3D Geometry & Coordinates
Point A is at (13, 1, 0) and point B is at (13, 25, 0). What is the length of AB?
AB = √[(13−13)² + (25−1)² + (0−0)²] = √576 = 24 cm
What is a Voronoi diagram used for in the context of fire stations?
A Voronoi diagram divides an area so that each region contains all points closest to one specific site (fire station). It shows which station is responsible for each location.
In triangle BCD, ∠CBD = 41°, ∠BCD = 120°. What is ∠BDC?
Angles in a triangle sum to 180°. ∠BDC = 180° − 41° − 120° = 19°
A sector has radius 5 and central angle 120°. Convert 120° to radians.
120° × (π/180) = 2π/3 ≈ 2.094 radians
A point A is 3 km due North of O. What are the coordinates of A if O is at the origin (East = x-axis, North = y-axis)?
Bearing 000° = due North. A = (0, 3)
A triangular prism has vertices A(13,1,0), B(13,25,0), C(13,25,7). Calculate the length of AC.
AC = √[(0)² + (24)² + (7)²] = √(0 + 576 + 49) = √625 = 25 cm
Find the midpoint of BC where B(16, 12) and C(8, 13).
Midpoint = ((16+8)/2, (12+13)/2) = (12, 12.5)
In triangle BCD, BC = 40 m, ∠BCD = 120°, ∠BDC = 19°. Use the sine rule to find BD.
BD/sin(120°) = 40/sin(19°). BD = 40 × sin(120°)/sin(19°) = 40 × 0.8660/0.3256 ≈ 106.4 m
A circle has centre O and radius r = 5. The central angle is 120° (2π/3 rad). Find the arc length of the minor arc.
Arc length = rθ = 5 × (2π/3) = 10π/3 ≈ 10.47 units
Point C is 5 km from O at bearing 090°. Point A is at (0, 3). What are the coordinates of C?
Bearing 090° = due East. C = (5, 0)
In triangle ABC with AB = 24, BC = 7, AC = 25 — verify it is right-angled and state which angle is 90°.
AB² + BC² = 576 + 49 = 625 = AC² ✓ — The right angle is at vertex B.
B is at (16, 12) and C is at (8, 13). Find the equation of the perpendicular bisector of BC (the Voronoi edge between B and C).
Gradient BC = (13−12)/(8−16) = −1/8. Perpendicular gradient = 8. Midpoint = (12, 12.5). Equation: y − 12.5 = 8(x − 12) → y = 8x − 83.5
A flagpole AD stands vertically. AB = 10 m horizontally. The angle of elevation from B to D is 30°. Find the height AD.
tan(30°) = AD/AB → AD = 10 × tan(30°) = 10 × (1/√3) ≈ 5.77 m
A sector has radius r = 5 and central angle θ = 2π/3. Find the area of the minor sector.
Area = ½r²θ = ½ × 25 × (2π/3) = 25π/3 ≈ 26.18 units²
A is at (0, 3). B is 4 km from A at bearing 090°. Find the coordinates of B and the distance BC where C = (5, 0).
B = (0+4, 3) = (4, 3). BC = √[(5−4)²+(0−3)²] = √(1+9) = √10 ≈ 3.16 km
A prism ABCDEF has A(13,1,0), B(13,25,0), C(13,25,7), D(4,25,7). The face ADFC is a rectangle. State the coordinates of vertex F.
F is translated from A by the same vector as D from C. Vector D−C = (−9, 0, 0). So F = A + (−9,0,0) = (4, 1, 0)
Cell towers are at A(1,5), B(7,8), C(4,12). Find the equation of the perpendicular bisector of AB.
Midpoint = (4, 6.5). Gradient AB = 3/6 = 1/2. Perpendicular gradient = −2. Equation: y − 6.5 = −2(x − 4) → y = −2x + 14.5
In triangle ABD, AB = AD = 85 m and BD ≈ 106.4 m. Use the cosine rule to find angle BAD.
cos(∠BAD) = (85² + 85² − 106.4²)/(2×85×85) = (7225+7225−11321)/14450 = 3129/14450 ≈ 0.2166. ∠BAD = cos⁻¹(0.2166) ≈ 77.5°
A sector of a circle has radius 4.5 m. The perpendicular distance from the centre to chord AB is 4 m. Find the central angle AÔB.
cos(θ/2) = 4/4.5 = 0.8889. θ/2 = cos⁻¹(0.8889) ≈ 27.27°. θ ≈ 54.5° (0.952 radians)
At 4:00 pm, the minute hand of a clock (length 10 cm) points to 12 and the hour hand (length 6 cm) points to 4. Find the angle AÔB and the distance AB.
∠AOB = 120° (hour hand at 4 = 4/12 × 360°). AB² = 10² + 6² − 2(10)(6)cos(120°) = 100+36+60 = 196. AB = 14 cm
Points P(2, −1, 3) and Q(5, 3, −1) are in 3D space. Find the midpoint M and the length PQ.
Midpoint M = ((2+5)/2, (−1+3)/2, (3−1)/2) = (3.5, 1, 1). PQ = √[(3)²+(4)²+(−4)²] = √(9+16+16) = √41 ≈ 6.40 units
Fire stations are at A(18,8), B(16,12), C(8,13) on a grid where 1 unit = 2.5 km. A fire is reported at (14,10). Which station responds, and why?
Distance to A: √[(4)²+(2)²] = √20 ≈ 4.47. Distance to B: √[(2)²+(2)²] = √8 ≈ 2.83. Distance to C: √[(6)²+(3)²] = √45 ≈ 6.71. Station B responds — it is closest to (14,10).
Using ∠BAD = 77°, AB = AD = 85 m, find the area bounded by path BD and fences AB and AD.
Area = ½ × AB × AD × sin(∠BAD) = ½ × 85 × 85 × sin(77°) = ½ × 7225 × 0.9744 ≈ 3520 m²
A circle has radius r = 5. The area of the shaded region (segment) = 12. Find θ, given: ½r²(θ − sinθ) = 12.
½(25)(θ − sinθ) = 12 → θ − sinθ = 0.96. Solving numerically: θ ≈ 1.91 → 1.91 − sin(1.91) ≈ 0.97. θ ≈ 1.92 → 0.984. By iteration: θ ≈ 1.92 radians
B is at (4, 3) and C is at (5, 0). Find the bearing of C from B.
Vector B→C = (1, −3). This points East and South. Angle from South toward East: tan(α) = 1/3, α ≈ 18.43°. Bearing = 180° − 18.43° ≈ 161.6° (bearing 162°)