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100+ Free PAU Technical Drawing II (Galicia) Practice Questions

Galicia PAU Technical Drawing II (Dibujo Técnico II) Exam practice questions are available now; exam metadata is being verified.

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Key Facts: PAU Technical Drawing II (Galicia) Exam

90 min

Exam Duration

CIUG Galicia

0–10

Grading Scale (Min 4.0 in Access Phase)

Comisión Interuniversitaria de Galicia (CIUG)

EUR 63.67

Ordinary Registration Fee (50% Large Family Discount)

CIUG Galicia Fee Structure 2026

5 Core Blocks

Plane Geometry, Transformations, Dihedral System, Axonometry, ISO Normalization

2º Bachillerato LOMLOE Curriculum (Galicia)

60% / 40%

Weighting Ratio (Bachillerato GPA / PAU Access Mark)

CIUG University Admission Rules

The Technical Drawing II exam for the Galicia PAU (Prueba de Acceso a la Universidad) features a 90-minute university entrance test administered by the Comisión Interuniversitaria de Galicia (CIUG) for public universities (USC, UDC, UVigo). Graded on a 0–10 scale, a minimum mark of 4.0 is required in the Access Phase to combine with Bachillerato GPA (60% Bachillerato + 40% PAU >= 5.0 to pass). The official exam consists of practical graphic drawing tasks; this dataset provides an English-language MCQ study adaptation designed for core concept practice, rule verification, and theoretical mastery.

Sample PAU Technical Drawing II (Galicia) Practice Questions

Try these sample questions to test your PAU Technical Drawing II (Galicia) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1In plane geometry, what is the geometric relationship between a tangent line and the radius of a circle drawn to the point of tangency?
A.They are parallel to each other.
B.They intersect at an angle of 45 degrees.
C.They are perpendicular to each other.
D.They intersect at an angle of 60 degrees.
Explanation: A line tangent to a circle is strictly perpendicular (90 degrees) to the radius drawn to the point of contact. This perpendicularity is the fundamental geometric property used in constructing tangent lines and circles.
2In any triangle, the center of the inscribed circle (incircle) is determined by the intersection of which set of lines?
A.The three perpendicular bisectors of the sides
B.The three internal angle bisectors
C.The three altitudes from the vertices
D.The three medians connecting vertices to midpoints
Explanation: The incenter (incentro) of a triangle is the point where the three internal angle bisectors intersect. Because every point on an angle bisector is equidistant from the two adjacent sides, the incenter is equidistant from all three sides.
3Which triangle center is located at the intersection of the perpendicular bisectors (mediatrices) of its sides?
A.Orthocenter
B.Incenter
C.Circumcenter
D.Centroid
Explanation: The circumcenter (circuncentro) is the point of intersection of the three perpendicular bisectors of a triangle's sides. It is equidistant from all three vertices and serves as the center of the circumscribed circle.
4What is the relationship between the side length (s) of a regular hexagon inscribed in a circle and the circle's radius (R)?
A.s = R
B.s = R * sqrt(2)
C.s = R * sqrt(3)
D.s = 2 * R
Explanation: A regular hexagon can be partitioned from its center into six congruent equilateral triangles. Consequently, the length of each side of an inscribed regular hexagon is equal to the radius of the circumscribed circle (s = R).
5How is the power of a point P with respect to a circle of radius R and center O defined, where d is the distance OP?
A.Power = d + R
B.Power = d^2 - R^2
C.Power = d^2 + R^2
D.Power = (d - R)^2
Explanation: The power of a point P with respect to a circle (Power = d^2 - R^2) is a constant scalar value equal to the product of segment lengths along any secant line passing through P. For a point outside the circle, it equals the square of the tangent segment length from P to the circle.
6What is the locus of points that have equal power with respect to two non-concentric circles?
A.A line perpendicular to the segment joining the centers of the two circles
B.A line parallel to the segment joining the centers of the two circles
C.A circle concentric with the larger circle
D.An ellipse with foci at the centers of the two circles
Explanation: The radical axis (eje radical) of two non-concentric circles is the straight line containing all points that have equal power with respect to both circles. Geometrically, this radical axis is always perpendicular to the line connecting the centers of the two circles.
7The radical center (centro radical) of three circles with non-collinear centers is defined as:
A.The centroid of the triangle formed by the three circle centers
B.The common point of intersection of the three pairwise radical axes
C.The center of the smallest circle tangent to all three circles
D.The midpoint of the segment connecting the two outermost circle centers
Explanation: The radical center of three circles with non-collinear centers is the unique point where the three pairwise radical axes intersect. This point possesses equal geometric power with respect to all three circles.
8To construct the tangent lines from an external point P to a circle with center O, which auxiliary construction is performed using Thales' theorem?
A.Draw an auxiliary circle centered at P with radius equal to PO
B.Draw an auxiliary circle with diameter PO (centered at the midpoint of PO)
C.Draw a line parallel to PO at distance R from P
D.Construct an equilateral triangle on segment PO
Explanation: By constructing an auxiliary circle with diameter PO (using the midpoint of PO as center), the points where this auxiliary circle intersects the given circle subtend a 90-degree angle from segment PO (Thales' theorem). These intersection points are the exact points of tangency.
9In the classical Apollonius problem designated as 'PPP' (Point-Point-Point), what is being constructed?
A.A circle passing through three given non-collinear points
B.A circle tangent to three given lines
C.A line tangent to three given circles
D.A point equidistant from three given circles
Explanation: The PPP Apollonius problem requires constructing a circle that passes through three given non-collinear points. The center of this circle is found at the intersection of the perpendicular bisectors of the segments joining the points.
10How many theoretical solution circles exist for the general Apollonius problem 'CCC' (constructing circles tangent to three given mutually exterior circles of different radii)?
A.2 solutions
B.4 solutions
C.8 solutions
D.16 solutions
Explanation: The general CCC Apollonius problem (tangent to three non-intersecting, mutually exterior circles of different sizes) has up to 8 distinct solution circles, corresponding to the different combinations of internal and external tangencies ($2^3 = 8$).

About the PAU Technical Drawing II (Galicia) Practice Questions

Verified exam format metadata for Galicia PAU Technical Drawing II (Dibujo Técnico II) Exam is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.