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100+ Free Balearic PAU Technical Drawing II Practice Questions
Balearic Islands PAU Technical Drawing II Examination — UIB (Dibuix Tècnic II 2º Bachillerato UIB 2026) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: Balearic PAU Technical Drawing II Exam
90 Minutes
Exam time duration
UIB PAU Commission
EUR 69.21
Ordinary registration fee for Access Phase
UIB 2026 Fees
0–10 Scale
Grading scale for Bachillerato and PAU exams
Spanish Ministry of Education
4 Core Blocks
Plane Geometry, Dihedral System, Axonometry/Perspective, and Technical Standardization
UIB Syllabus
100 Questions
Practice bank size in OpenExamPrep
OpenExamPrep
Balearic Islands PAU Technical Drawing II (UIB 2026) is a 90-minute university entrance exam assessing 2º Bachillerato Technical Drawing II across plane geometry, dihedral system, axonometry, and ISO standardization.
Sample Balearic PAU Technical Drawing II Practice Questions
Try these sample questions to test your Balearic PAU Technical Drawing II exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1What is the measure of each interior angle in a regular convex hexagon?
A.108°
B.120°
C.135°
D.140°
Explanation: The sum of interior angles of an n-sided convex polygon is given by (n - 2) × 180°. For a hexagon (n = 6), the total sum is (6 - 2) × 180° = 720°. Dividing by 6 equal interior angles yields 720° / 6 = 120° per interior angle.
2In an equilateral triangle circumscribed about a circle of radius r, what is the side length s of the triangle in terms of r?
A.s = r√3
B.s = 2r√3
C.s = 3r
D.s = 2r
Explanation: The inradius r of an equilateral triangle with side s is related to its altitude h by r = h/3, where h = (s√3)/2. Substituting h gives r = (s√3)/6, which rearranges to s = 6r/√3 = 2r√3.
3How many non-isomorphic star octagons (represented by Schläfli symbols {8/k}) can be constructed from eight equally spaced points on a circle?
A.1
B.2
C.3
D.4
Explanation: A star polygon {n/k} requires k to be coprime to n and 1 < k < n/2. For n = 8, the integers strictly between 1 and 4 are 2 and 3. The integer 2 shares a common factor with 8 (gcd(8,2)=2, forming two overlapping squares), while 3 is coprime to 8 (gcd(8,3)=1). Thus, only k = 3 forms a single regular star octagon {8/3}.
4Which of the following regular n-sided polygons CANNOT be constructed using only a straightedge and compass according to the Gauss-Wantzel theorem?
A.15-gon (Pentadecagon)
B.17-gon (Heptadecagon)
C.9-gon (Enneagon / Nonagon)
D.20-gon (Icosagon)
Explanation: According to the Gauss-Wantzel theorem, a regular n-gon is constructible with straightedge and compass if and only if the odd prime factors of n are distinct Fermat primes (3, 5, 17, 257, 65537). For n = 9, the prime factorization is 3², which contains a repeated odd prime factor (3²), making the regular 9-gon non-constructible.
5What is the exact algebraic value of the golden ratio φ (phi)?
A.(1 + √3) / 2
B.(1 + √5) / 2
C.(1 + √5) / 3
D.(1 + √2) / 2
Explanation: The golden ratio φ is defined by the proportion (a+b)/a = a/b = φ, which yields the quadratic equation φ² - φ - 1 = 0. Solving this equation via the quadratic formula gives the positive root φ = (1 + √5) / 2 ≈ 1.618033...
6What are the interior angle measures of a golden isosceles triangle (an isosceles triangle whose sides are in the golden ratio φ : 1 : φ)?
A.36°, 72°, 72°
B.45°, 45°, 90°
C.30°, 75°, 75°
D.54°, 54°, 72°
Explanation: A golden triangle is an isosceles triangle where the ratio of a leg to the base is φ. Bisecting a base angle creates a smaller similar golden triangle, implying the base angles are double the vertex angle (2θ + 2θ + θ = 180° => 5θ = 180° => θ = 36°). Thus, the angle measures are 36°, 72°, and 72°.
7Given a segment AB of length L, which geometric construction correctly locates the golden section point C on AB such that AC / CB = φ?
A.Erect a perpendicular BD of length L/2 at B, draw hypotenuse AD, draw an arc from D with radius DB intersecting AD at E, then draw an arc from A with radius AE to cut AB at C.
B.Erect a perpendicular BD of length L at B, bisect hypotenuse AD at M, then draw an arc from A with radius AM to cut AB at C.
C.Construct an equilateral triangle ABD on AB, bisect side BD at M, then drop a perpendicular from M to cut AB at C.
D.Draw a semicircle on diameter AB, erect a perpendicular at mid-point O of length L/2 to intersect the arc at P, then drop P onto AB.
Explanation: In right triangle ABD with legs AB = L and BD = L/2, hypotenuse AD = √(L² + (L/2)²) = (L√5)/2. Subtracting DE = BD = L/2 gives AE = (L√5)/2 - L/2 = L(√5 - 1)/2. Transferring AE onto AB yields AC = L(√5 - 1)/2, which divides AB in the golden ratio because AC / CB = φ.
8What is the angle between a circle's radius and a line tangent to the circle at the point of contact?
A.45°
B.60°
C.90°
D.180°
Explanation: By fundamental circle geometry, a line is tangent to a circle if and only if it is perpendicular (90°) to the radius drawn to the point of tangency.
9Two circles with radii R1 = 9 cm and R2 = 4 cm are externally tangent to each other. What is the length of their common external tangent segment between the points of contact?
A.12 cm
B.13 cm
C.10 cm
D.6 cm
Explanation: The length t of the common external tangent segment between two externally tangent circles with radii R1 and R2 is given by t = 2√(R1 × R2). Substituting R1 = 9 and R2 = 4 yields t = 2√(9 × 4) = 2√(36) = 2 × 6 = 12 cm.
10In the classical Apollonius problem of finding circles tangent to three given geometric entities, how many solutions generally exist for the Case PPP (three non-collinear points)?
A.1 solution
B.2 solutions
C.4 solutions
D.8 solutions
Explanation: A circle passing through three non-collinear points (PPP) is uniquely defined as the circumcircle of the triangle formed by those three points. Its center is the unique circumcenter (intersection of the perpendicular bisectors), giving exactly 1 solution.
About the Balearic PAU Technical Drawing II Practice Questions
Verified exam format metadata for Balearic Islands PAU Technical Drawing II Examination — UIB (Dibuix Tècnic II 2º Bachillerato UIB 2026) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.