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100+ Free Valencian Community PAU Technical Drawing II Practice Questions
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Key Facts: Valencian Community PAU Technical Drawing II Exam
Comprehensive preparation for the Valencian Community PAU Technical Drawing II (2026 LOMLOE syllabus) featuring 100 practice questions, detailed mathematical explanations, and step-by-step calculations across plane geometry, dihedric projections, axonometry, perspective systems, and UNE-EN ISO technical standards.
Sample Valencian Community PAU Technical Drawing II Practice Questions
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1In a direct homothety centered at point $O$ with scale ratio $k = -1.5$, a segment $AB$ of length $40\text{ mm}$ is transformed into segment $A'B'$. What is the length of $A'B'$ and its geometric orientation relative to $AB$?
A.Length $60\text{ mm}$, parallel to $AB$ but oriented in the opposite direction relative to center $O$
B.Length $60\text{ mm}$, perpendicular to $AB$ centered at $O$
C.Length $26.67\text{ mm}$, parallel and in the same direction as $AB$
D.Length $40\text{ mm}$, rotated $180^\circ$ about point $A$
Explanation: In homothety, the length of any segment is scaled by $|k|$, giving $|A'B'| = |k| \cdot |AB| = 1.5 \times 40\text{ mm} = 60\text{ mm}$. Because $k < 0$, corresponding points lie on opposite sides of the homothety center $O$, producing a parallel segment oriented in the opposite direction.
2In a plane inversion centered at $O$ with positive power $K = 900\text{ mm}^2$, what is the inverse of a straight line $r$ that does NOT pass through the center of inversion $O$?
A.A straight line parallel to $r$ passing through $O$
B.A circle passing through the center of inversion $O$
C.A circle centered at $O$ with radius $30\text{ mm}$
D.A parabola with focus at $O$
Explanation: According to the fundamental properties of plane inversion, the inverse of any straight line not passing through the center of inversion $O$ is a circle passing through $O$. The diameter of this circle lies along the perpendicular dropped from $O$ onto line $r$.
3Two non-concentric circles $C_1$ (radius $r_1 = 30\text{ mm}$) and $C_2$ (radius $r_2 = 20\text{ mm}$) have centers separated by distance $O_1O_2 = 60\text{ mm}$. What is the geometric locus of points having equal power with respect to both circles?
A.The line segment connecting $O_1$ and $O_2$
B.A circle concentric with $C_1$ passing through $O_2$
C.A straight line perpendicular to the line of centers $O_1O_2$
D.An ellipse with foci at $O_1$ and $O_2$
Explanation: The geometric locus of points having equal power relative to two non-concentric circles is their radical axis. The radical axis is always a straight line perpendicular to the line joining the centers $O_1O_2$ of the two circles.
4A point $P$ is located at distance $d = 50\text{ mm}$ from the center $O$ of a circle of radius $R = 30\text{ mm}$. What is the power of point $P$ relative to the circle, and what is the length of the tangent segment from $P$ to the circle?
A.Power = $1600\text{ mm}^2$, Tangent length = $40\text{ mm}$
B.Power = $3400\text{ mm}^2$, Tangent length = $58.3\text{ mm}$
C.Power = $2000\text{ mm}^2$, Tangent length = $44.7\text{ mm}$
D.Power = $800\text{ mm}^2$, Tangent length = $28.3\text{ mm}$
Explanation: The power of point $P$ relative to a circle of radius $R$ is $\mathcal{P} = d^2 - R^2 = 50^2 - 30^2 = 2500 - 900 = 1600\text{ mm}^2$. The tangent segment length $t$ from $P$ to the point of tangency is $t = \sqrt{\mathcal{P}} = \sqrt{1600} = 40\text{ mm}$.
5In solving Apollonius' tangency problem for two given circles $C_1$ (radius $R_1$) and $C_2$ (radius $R_2$) and a given point $P$, what auxiliary reduction simplifies the problem to finding a circle tangent to one circle passing through two points?
A.Shrinking or expanding both circles by radius $R_2$, reducing circle $C_2$ to a point
B.Rotating both circles by $90^\circ$ around point $P$
C.Projecting the circles onto an axonometric plane
D.Inverting both circles with respect to an arbitrary line
Explanation: By applying the dilation method (dilatación), the radius of circle $C_2$ is reduced to zero (turning $C_2$ into a point $O_2$), while the radius of circle $C_1$ is adjusted to $R_1 \pm R_2$ and point $P$ is shifted accordingly. This transforms the CCP problem into a PPC problem (circle passing through two points and tangent to one circle).
6On an architectural blueprint drawn at scale $1:250$, a building wall measures $d = 48\text{ mm}$ on the paper. What is the actual real-world length of the wall in meters?
A.$12.0\text{ meters}$
B.$5.2\text{ meters}$
C.$120.0\text{ meters}$
D.$1.92\text{ meters}$
Explanation: The scale formula is $S = \text{drawing length} / \text{real length} = d / D$. Here $1/250 = 48\text{ mm} / D \implies D = 48 \times 250 = 12000\text{ mm} = 12.0\text{ meters}$.
7According to the focal definition of an ellipse, what is the constant sum of the distances from any point $P$ on the ellipse to its two foci $F_1$ and $F_2$?
A.Equal to the length of the major axis $2a$
B.Equal to the length of the minor axis $2b$
C.Equal to the focal distance $2c$
D.Equal to the semi-major axis $a$
Explanation: An ellipse is defined as the locus of points $P$ in a plane such that the sum of distances to two fixed points (foci $F_1, F_2$) is constant: $PF_1 + PF_2 = 2a$, where $2a$ is the length of the major axis.
8An ellipse has a major axis length $2a = 100\text{ mm}$ and a focal distance $2c = 60\text{ mm}$. What is the total length of its minor axis $2b$?
A.$80\text{ mm}$
B.$40\text{ mm}$
C.$70\text{ mm}$
D.$90\text{ mm}$
Explanation: In an ellipse, the semi-axes and focal distance satisfy $a^2 = b^2 + c^2$. Here $a = 50\text{ mm}$ and $c = 30\text{ mm}$, so $b = \sqrt{a^2 - c^2} = \sqrt{50^2 - 30^2} = \sqrt{2500 - 900} = \sqrt{1600} = 40\text{ mm}$. Therefore, the total minor axis is $2b = 2 \times 40 = 80\text{ mm}$.
9A hyperbola has transverse (major) axis length $2a = 80\text{ mm}$ and focal distance $2c = 100\text{ mm}$. What is the length of its semi-conjugate (minor) axis $b$?
A.$30\text{ mm}$
B.$60\text{ mm}$
C.$40\text{ mm}$
D.$50\text{ mm}$
Explanation: For a hyperbola, the relationship between semi-axes and focal distance is $c^2 = a^2 + b^2$. Here $a = 40\text{ mm}$ and $c = 50\text{ mm}$, so $b = \sqrt{c^2 - a^2} = \sqrt{50^2 - 40^2} = \sqrt{2500 - 1600} = \sqrt{900} = 30\text{ mm}$.
10What is the fundamental geometric definition of a parabola?
A.The locus of points equidistant from a fixed point (focus) and a fixed line (directrix)
B.The locus of points whose difference of distances to two foci is constant
C.The locus of points whose product of distances to two fixed axes is constant
D.The locus of points whose sum of distances to two fixed points is constant
Explanation: A parabola is defined as the locus of points $P$ in a plane equidistant from a fixed point called the focus $F$ and a fixed straight line called the directrix $d$, meaning $PF = d(P, d)$.
About the Valencian Community PAU Technical Drawing II Practice Questions
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