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Balearic PAU Technical Drawing II study materials
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Balearic PAU Technical Drawing II practice questions

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Question 1Technical drawing: plane and spatial geometry, projections and dimensions

What is the measure of each interior angle in a regular convex hexagon?

  1. 108°
  2. 120° Correct answer.
  3. 135°
  4. 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.

Question 2Technical drawing: plane and spatial geometry, projections and dimensions

In an equilateral triangle circumscribed about a circle of radius r, what is the side length s of the triangle in terms of r?

  1. s = r√3
  2. s = 2r√3 Correct answer.
  3. s = 3r
  4. 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.

Question 3Technical drawing: plane and spatial geometry, projections and dimensions

How many non-isomorphic star octagons (represented by Schläfli symbols {8/k}) can be constructed from eight equally spaced points on a circle?

  1. 1 Correct answer.
  2. 2
  3. 3
  4. 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}.

Question 4Technical drawing: plane and spatial geometry, projections and dimensions

Which of the following regular n-sided polygons CANNOT be constructed using only a straightedge and compass according to the Gauss-Wantzel theorem?

  1. 15-gon (Pentadecagon)
  2. 17-gon (Heptadecagon)
  3. 9-gon (Enneagon / Nonagon) Correct answer.
  4. 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.

Question 5Technical drawing: plane and spatial geometry, projections and dimensions

What is the exact algebraic value of the golden ratio φ (phi)?

  1. (1 + √3) / 2
  2. (1 + √5) / 2 Correct answer.
  3. (1 + √5) / 3
  4. (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...

Question 6Technical drawing: plane and spatial geometry, projections and dimensions

What are the interior angle measures of a golden isosceles triangle (an isosceles triangle whose sides are in the golden ratio φ : 1 : φ)?

  1. 36°, 72°, 72° Correct answer.
  2. 45°, 45°, 90°
  3. 30°, 75°, 75°
  4. 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°.

Question 7Technical drawing: plane and spatial geometry, projections and dimensions

Given a segment AB of length L, which geometric construction correctly locates the golden section point C on AB such that AC / CB = φ?

  1. 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. Correct answer.
  2. 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.
  3. Construct an equilateral triangle ABD on AB, bisect side BD at M, then drop a perpendicular from M to cut AB at C.
  4. 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.

Question 8Technical drawing: plane and spatial geometry, projections and dimensions

What is the angle between a circle's radius and a line tangent to the circle at the point of contact?

  1. 45°
  2. 60°
  3. 90° Correct answer.
  4. 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.

Question 9Technical drawing: plane and spatial geometry, projections and dimensions

Two 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?

  1. 12 cm Correct answer.
  2. 13 cm
  3. 10 cm
  4. 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.

Question 10Technical drawing: plane and spatial geometry, projections and dimensions

In 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)?

  1. 1 solution Correct answer.
  2. 2 solutions
  3. 4 solutions
  4. 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.

Question 11Technical drawing: plane and spatial geometry, projections and dimensions

What is the locus of points having equal power with respect to two non-concentric circles?

  1. A circle concentric with the larger circle
  2. A straight line perpendicular to the line joining the centers of the two circles Correct answer.
  3. An ellipse with foci at the centers of the two circles
  4. A hyperbola passing through the points of contact

Explanation

The radical axis of two non-concentric circles is defined precisely as the locus of points P whose power with respect to both circles is equal (d1² - R1² = d2² - R2²). Geometrically, this locus is always a straight line perpendicular to the line connecting the centers of the two circles.

Question 12Technical drawing: plane and spatial geometry, projections and dimensions

How many distinct solution circles exist for the general Apollonius problem of finding circles tangent to three mutually non-intersecting, non-concentric circles (Case CCC)?

  1. Up to 2
  2. Up to 4
  3. Up to 8 Correct answer.
  4. Up to 16

Explanation

In the general Apollonius problem (CCC), each of the three given circles can be touched either externally or internally by a solution circle. This leads to 2 × 2 × 2 = 8 possible tangency combinations, yielding up to 8 distinct solution circles.

Question 13Technical drawing: plane and spatial geometry, projections and dimensions

Which of the following plane transformations is NOT a direct or indirect isometry (rigid transformation that preserves distances)?

  1. Translation
  2. Rotation
  3. Reflection (Axial symmetry)
  4. Homothety (Dilation with scale factor k = 2) Correct answer.

Explanation

An isometry preserves all distances between points. Homothety with scale factor k = 2 multiplies all distances by 2, altering lengths while preserving angles, which makes it a similarity transformation rather than an isometry.

Question 14Technical drawing: plane and spatial geometry, projections and dimensions

In a homothety with center O and scale factor k = -0.5, what happens to an original triangle of area A?

  1. It is inverted and its area becomes 0.25A Correct answer.
  2. It is inverted and its area becomes 0.5A
  3. It stays upright and its area becomes 0.25A
  4. It is inverted and its area becomes -0.5A

Explanation

A negative scale factor k = -0.5 produces an inverted figure on the opposite side of center O. Area scales by the square of the linear scale factor: A' = |k|² × A = (-0.5)² × A = 0.25A.

Question 15Technical drawing: plane and spatial geometry, projections and dimensions

In plane homology (central collinearity), what is the key characteristic of the axis of homology?

  1. It is a line of invariant points where every line intersects its corresponding transformed line Correct answer.
  2. It is a point through which all lines connecting corresponding points pass
  3. It is a line parallel to all direction vectors of transformation
  4. It is the circle along which shapes retain their original perimeter

Explanation

In plane homology, the axis of homology is a line of invariant points (fixed points). Any line L in the original figure and its homological image L' always intersect at a point lying on the axis of homology.

Question 16Technical drawing: plane and spatial geometry, projections and dimensions

Under a plane affinity transformation (homología afín), what geometric property is preserved for any plane figure?

  1. Lengths of segments
  2. Angles between intersecting lines
  3. Parallelism between lines Correct answer.
  4. Radii of circles

Explanation

Affinity is a special case of homology where the center of homology is at infinity. A fundamental property of affine transformations is that parallel lines always transform into parallel lines, and ratios of parallel segment lengths are preserved.

Question 17Technical drawing: plane and spatial geometry, projections and dimensions

An affinity transformation with axis e transforms a circle of radius R into an ellipse. If the direction of affinity is perpendicular to axis e and the scale ratio is k = 0.6, what are the semi-major axis a and semi-minor axis b of the resulting ellipse?

  1. a = R, b = 0.6R Correct answer.
  2. a = 0.6R, b = R
  3. a = R, b = R / 0.6
  4. a = 1.6R, b = 0.6R

Explanation

Because the affinity is perpendicular to axis e, lengths parallel to axis e remain unchanged, while lengths perpendicular to axis e are scaled by k = 0.6. Therefore, the diameter along the axis remains 2R (semi-major axis a = R), and the diameter perpendicular to the axis becomes 2 × 0.6R (semi-minor axis b = 0.6R).

Question 18Technical drawing: plane and spatial geometry, projections and dimensions

In an ellipse with semi-major axis a, semi-minor axis b, and focal distance 2c, what is the relation between a, b, and c?

  1. a² = b² + c² Correct answer.
  2. c² = a² + b²
  3. b² = a² + c²
  4. a = b + c

Explanation

By the focal definition of an ellipse, the distance from a co-vertex (end of minor axis) to either focus is equal to semi-major axis a. Right triangle geometry formed by the center, focus, and co-vertex yields the Pythagorean relationship a² = b² + c².

Question 19Technical drawing: plane and spatial geometry, projections and dimensions

An ellipse has major axis length 2a = 10 cm and minor axis length 2b = 6 cm. What is its focal distance 2c?

  1. 4 cm
  2. 8 cm Correct answer.
  3. 16 cm
  4. 2√34 cm

Explanation

Given 2a = 10 => a = 5 cm, and 2b = 6 => b = 3 cm. Using the ellipse relation a² = b² + c² gives 5² = 3² + c² => 25 = 9 + c² => c² = 16 => c = 4 cm. The total focal distance between the two foci is 2c = 2 × 4 = 8 cm.

Question 20Technical drawing: plane and spatial geometry, projections and dimensions

Which statement correctly defines a parabola in terms of distances from a point P on the curve?

  1. The sum of distances from P to two fixed points is constant.
  2. The difference of distances from P to two fixed points is constant.
  3. The distance from P to a fixed point (focus) equals the perpendicular distance from P to a fixed line (directrix). Correct answer.
  4. The product of distances from P to two fixed lines is constant.

Explanation

A parabola is defined as the locus of points P in a plane that are equidistant from a fixed point called the focus F and a fixed line called the directrix d (i.e., PF = PD).

Question 21Technical drawing: plane and spatial geometry, projections and dimensions

For a hyperbola with semi-transverse axis a = 4 cm and semi-conjugate axis b = 3 cm, what are the equations of its asymptotes relative to its center at the origin (0,0)?

  1. y = ± (3/4)x Correct answer.
  2. y = ± (4/3)x
  3. y = ± (5/4)x
  4. y = ± (3/5)x

Explanation

For a hyperbola centered at the origin with transverse axis along the x-axis, the asymptotes pass through the origin and the vertices of the central rectangle of dimensions 2a × 2b. The slopes are ± (b/a) = ± (3/4). Thus, the asymptote equations are y = ± (3/4)x.

Question 22Technical drawing: plane and spatial geometry, projections and dimensions

What is the locus of the focal projections (perpendicular feet dropped from a focus onto any tangent line) of an ellipse?

  1. The focal circle (director circle centered at a focus with radius 2a)
  2. The principal circle (auxiliary circle centered at the ellipse center with radius a) Correct answer.
  3. A parabola with the same focus
  4. A straight line parallel to the major axis

Explanation

By the property of tangents to an ellipse, the projection of a focus onto any tangent line lies on the principal circle (auxiliary circle centered at the midpoint of the major axis with radius a). The reflection of a focus across a tangent line lies on the focal circle (radius 2a centered at the other focus).

Question 23Technical drawing: plane and spatial geometry, projections and dimensions

How is a standard cycloid generated?

  1. By a point on the circumference of a circle rolling without slipping along a straight line Correct answer.
  2. By unwinding a taut string from a fixed circle
  3. By a point on the circumference of a circle rolling inside another fixed circle
  4. By the locus of points at constant distance from a sine wave

Explanation

A cycloid is the curve traced by a point attached to a circle as it rolls along a straight line without slipping.

Question 24Technical drawing: plane and spatial geometry, projections and dimensions

For a circle of radius R, what is the length of one full turn (revolution) of its involute starting from the point of contact on the circle?

  1. 2π R
  2. 2π² R Correct answer.
  3. 4π R
  4. π² R

Explanation

The arc length s of the involute of a circle of radius R as a function of the unwinding angle θ (in radians) is given by s = (1/2) R θ². For one complete revolution, θ = 2π radians. Substituting gives s = (1/2) R (2π)² = (1/2) R (4π²) = 2π² R.

Question 25Technical drawing: plane and spatial geometry, projections and dimensions

What is the total length L of one complete arch of a cycloid generated by a circle of radius R?

  1. 2π R
  2. 4 R
  3. 8 R Correct answer.
  4. 2π² R

Explanation

The parametric equations of a cycloid generated by a circle of radius R are x = R(θ - sin θ) and y = R(1 - cos θ). Integrating the arc length element ds = √(dx² + dy²) over θ from 0 to 2π yields L = 8R.

Question 26Technical drawing: plane and spatial geometry, projections and dimensions

In the Monge dihedral projection system, a point P has a positive elevation (z > 0, above the horizontal plane) and a negative plan/departure (y < 0, behind the frontal plane). In which quadrant is point P located?

  1. 1st Quadrant
  2. 2nd Quadrant Correct answer.
  3. 3rd Quadrant
  4. 4th Quadrant

Explanation

The 4 quadrants in the dihedral system are defined by signs of departure y and elevation z: 1st Q (y>0, z>0), 2nd Q (y<0, z>0), 3rd Q (y<0, z<0), and 4th Q (y>0, z<0). Point P with z>0 and y<0 lies in the 2nd Quadrant.

Question 27Technical drawing: plane and spatial geometry, projections and dimensions

In dihedral projections, what is a horizontal line (recta horitzontal)?

  1. A line parallel to the horizontal projection plane, whose elevation projection r' is parallel to the ground line Correct answer.
  2. A line parallel to the frontal projection plane, whose plan projection r is parallel to the ground line
  3. A line perpendicular to the ground line in both projections
  4. A line inclined at 45° to both projection planes

Explanation

A horizontal line is parallel to the horizontal projection plane (HP). All points on it have the same elevation z, meaning its frontal projection (elevation r') is a straight line parallel to the ground line (LT), while its horizontal projection r shows its true direction and length.

Question 28Technical drawing: plane and spatial geometry, projections and dimensions

What are the projection characteristics of a vertical projecting plane (pla projectant horitzontal)?

  1. Its horizontal trace α1 is oblique to the ground line, and its frontal trace α2 is perpendicular to the ground line. Correct answer.
  2. Its frontal trace α2 is parallel to the ground line, and its horizontal trace α1 does not exist.
  3. Both of its traces α1 and α2 are parallel to the ground line.
  4. Its frontal trace α2 is oblique to the ground line, and its horizontal trace α1 is parallel to the ground line.

Explanation

A vertical projecting plane is perpendicular to the horizontal projection plane and oblique to the frontal plane. Consequently, its frontal trace α2 is perpendicular to the ground line (LT), while all points/lines in the plane project onto its oblique horizontal trace α1.

Question 29Technical drawing: plane and spatial geometry, projections and dimensions

A line of maximum slope (recta de màxima pendent) of an oblique plane α has which characteristic relative to the plane's horizontal trace α1?

  1. Its horizontal projection r is perpendicular to the horizontal trace α1 of the plane. Correct answer.
  2. Its frontal projection r' is perpendicular to the frontal trace α2 of the plane.
  3. Its horizontal projection r is parallel to the ground line.
  4. Its frontal projection r' is parallel to the horizontal trace α1.

Explanation

A line of maximum slope of a plane is the line in the plane that forms the steepest angle with the horizontal plane of projection. By definition, its horizontal projection r is strictly perpendicular to the horizontal trace α1 of the plane.

Question 30Technical drawing: plane and spatial geometry, projections and dimensions

How can a plane passing through the ground line (pla que passa per la línia de terra) be uniquely defined in dihedral projection?

  1. By its single merged trace on the ground line and an auxiliary point P not on the ground line Correct answer.
  2. By two traces α1 and α2 that intersect at 90° away from the ground line
  3. By its horizontal trace alone, as its frontal trace is at infinity
  4. By a single angle specified relative to the 1st bisector plane

Explanation

Because a plane passing through the ground line contains the ground line itself, both of its traces coincide with the ground line (α1 = α2 = LT). To uniquely define its position in space, an additional auxiliary point P (or profile view) outside the ground line must be given.

Question 31Technical drawing: plane and spatial geometry, projections and dimensions

What is the standard procedure to find the intersection point I between an oblique line r and an oblique plane α in dihedral projection?

  1. Pass an auxiliary projecting plane β through line r, find the intersection line i between planes α and β, then locate the intersection I = r ∩ i. Correct answer.
  2. Rotate line r until it is parallel to the ground line and read the z-coordinate.
  3. Abat plane α onto the horizontal projection plane along trace α1 and project r directly onto it.
  4. Draw perpendiculars from the traces of line r to the traces of plane α and connect their midpoints.

Explanation

The universal method for line-plane intersection is: (1) containing the given line r in a suitable auxiliary plane β (usually a vertical or frontal projecting plane), (2) determining the intersection line i = α ∩ β between the two planes, and (3) finding the point I where line r intersects line i.

Question 32Technical drawing: plane and spatial geometry, projections and dimensions

Where do the horizontal trace i1 and frontal trace i2 of the line of intersection i between two planes α and β lie?

  1. i1 = α1 ∩ β1 and i2 = α2 ∩ β2 Correct answer.
  2. i1 = α1 ∩ β2 and i2 = α2 ∩ β1
  3. i1 = α1 ∩ LT and i2 = β2 ∩ LT
  4. i1 and i2 are always parallel to the ground line

Explanation

The line of intersection i between two planes α and β belongs to both planes. Therefore, its horizontal trace H (where i pierces HP) must lie on the intersection of the two horizontal traces α1 and β1 (i1 = α1 ∩ β1). Similarly, its frontal trace V lies on i2 = α2 ∩ β2.

Question 33Technical drawing: plane and spatial geometry, projections and dimensions

If two planes α and β have horizontal traces α1 and β1 that are parallel to each other (α1 || β1), what can be concluded about their line of intersection i?

  1. Line i is a horizontal line parallel to traces α1 and β1. Correct answer.
  2. Line i is a vertical line perpendicular to the horizontal plane.
  3. Line i is a profile line.
  4. Planes α and β do not intersect.

Explanation

When the horizontal traces α1 and β1 of two planes are parallel, their horizontal piercing point H = α1 ∩ β1 is at infinity. This means the intersection line i cannot pierce the horizontal projection plane, so line i must be parallel to the horizontal plane-that is, i is a horizontal line whose horizontal projection i is parallel to α1 and β1.

Question 34Technical drawing: plane and spatial geometry, projections and dimensions

To find the visibility (solid vs. hidden line) of an oblique line r piercing a opaque polyhedral solid in dihedral projection, what principle is applied at apparent crossing points of projections?

  1. Compare the coordinates (departure for plan view, elevation for frontal view) of the two overlapping points on the ray. Correct answer.
  2. Hidden segments always lie to the left of the piercing point.
  3. The line segment inside the solid is always drawn as continuous thick.
  4. Visibility depends entirely on whether the line is in the 1st or 4th quadrant.

Explanation

At an apparent crossing point between line r and a polyhedron edge in projection, two distinct 3D points overlap. In the horizontal projection (plan view), the point with greater elevation (z) is visible. In the frontal projection (elevation view), the point with greater departure (y) is visible.

Question 35Technical drawing: plane and spatial geometry, projections and dimensions

How is the common intersection point I of three oblique planes α, β, and γ determined?

  1. Find the intersection line i = α ∩ β, then find the point I where line i pierces plane γ. Correct answer.
  2. Take the average of the coordinates where α1, β1, and γ1 intersect.
  3. Abat all three planes onto HP and find their centroid.
  4. Draw perpendicular lines from LT to all three frontal traces.

Explanation

Three non-parallel planes intersect at a single point I. The standard method is to determine the intersection line i between any two planes (e.g. i = α ∩ β), and then find the point I where this line i intersects the third plane γ (I = i ∩ γ).

Question 36Technical drawing: plane and spatial geometry, projections and dimensions

What is the formula for the true length d of a line segment AB with coordinate differences Δx, Δy (departure difference), and Δz (elevation difference)?

  1. d = √((Δx)² + (Δy)² + (Δz)²) Correct answer.
  2. d = Δx + Δy + Δz
  3. d = √((Δx)² + (Δy)²)
  4. d = √((Δy)² + (Δz)²)

Explanation

By 3D Euclidean distance formula, the true length of a segment AB in space is given by d = √((Δx)² + (Δy)² + (Δz)²).

Question 37Technical drawing: plane and spatial geometry, projections and dimensions

To find the true length of an oblique segment AB graphically using the right-triangle method on its horizontal projection r, what are the legs of the right triangle?

  1. Leg 1: length of horizontal projection r; Leg 2: absolute difference in elevation |Δz| = |zB - zA| Correct answer.
  2. Leg 1: length of frontal projection r'; Leg 2: absolute difference in departure |Δy|
  3. Leg 1: length of horizontal projection r; Leg 2: length of frontal projection r'
  4. Leg 1: x-coordinate difference Δx; Leg 2: y-coordinate difference Δy

Explanation

In the graphical right-triangle method, the horizontal projection r (which has length √((Δx)²+(Δy)²)) forms one leg. Constructing a perpendicular leg at one endpoint with length equal to the elevation difference |Δz| yields a hypotenuse equal to the true length d = √(r² + (Δz)²).

Question 38Technical drawing: plane and spatial geometry, projections and dimensions

In the right-triangle construction for true length built on the horizontal projection r of segment AB (with altitude leg |Δz|), what angle does the hypotenuse make with leg r?

  1. True inclination angle α of line AB with the horizontal projection plane HP Correct answer.
  2. True inclination angle β of line AB with the frontal projection plane VP
  3. Angle between line AB and the 1st bisector plane
  4. Angle between line AB and the profile plane

Explanation

The right triangle built on horizontal projection r with altitude |Δz| lies in a vertical plane containing segment AB. The angle between the true length hypotenuse AB and its horizontal projection r is precisely the true angle of inclination α that the line makes with the horizontal projection plane HP.

Question 39Technical drawing: plane and spatial geometry, projections and dimensions

A segment AB has horizontal projection length r = 12 cm and frontal projection length r' = 13 cm. If the difference in x-coordinates between A and B is Δx = 5 cm, what is the true length of segment AB?

  1. √119 cm (~10.91 cm)
  2. 13 cm
  3. √288 cm (~16.97 cm) Correct answer.
  4. √313 cm (~17.69 cm)

Explanation

In the dihedral (Monge) system the true spatial length is d = √(Δx² + Δy² + Δz²). The horizontal projection gives r² = Δx² + Δy², so Δy² = r² − Δx² = 12² − 5² = 144 − 25 = 119. The frontal projection gives r'² = Δx² + Δz², so Δz² = r'² − Δx² = 13² − 5² = 169 − 25 = 144. Therefore d = √(Δx² + Δy² + Δz²) = √(25 + 119 + 144) = √288 ≈ 16.97 cm. Equivalently, d = √(r² + r'² − Δx²) = √(144 + 169 − 25) = √288, since Δx² is shared by both projections and must be subtracted once.

Question 40Technical drawing: plane and spatial geometry, projections and dimensions

In descriptive geometry, what is plane abatment (abatiment / abatimiento)?

  1. Rotating a plane about one of its traces until it coincides with a projection plane to reveal true shapes Correct answer.
  2. Translating a plane along the ground line by a fixed distance
  3. Scaling a plane by a factor of √2 to compensate for isometric distortion
  4. Projecting a plane orthogonally onto the profile view

Explanation

Abatment is an auxiliary operation where an inclined plane is revolved (rotated) around one of its traces (hinge / axis of abatment) until it lies flat on a projection plane (HP or VP), allowing true lengths and true shapes of planar figures to be drawn and measured.

Question 41Technical drawing: plane and spatial geometry, projections and dimensions

What geometric relationship links the horizontal projection of a plane figure and its abatted figure (figura abatuda) when abatted around horizontal trace α1?

  1. Central Affinity (Homología Afín), where the axis of affinity is trace α1 Correct answer.
  2. Pure Translation along the ground line
  3. Inversion relative to the origin point on LT
  4. Rotational symmetry of 180° around the center of the plane

Explanation

Abatment around horizontal trace α1 establishes a plane central affinity (homologia afí) between the horizontal projection of points/lines and their abatted positions (A0, B0, etc.). The axis of affinity is the trace α1, and the direction of affinity is perpendicular to α1.

Question 42Technical drawing: plane and spatial geometry, projections and dimensions

To find the abatted position P0 of a point P lying in an oblique plane α (abatted onto HP around trace α1), what distance from the point's horizontal projection p along the affinity direction line is used?

  1. The hypotenuse of a right triangle with legs = distance from p to trace α1 and elevation zP Correct answer.
  2. The elevation zP directly measured from LT
  3. The departure yP directly measured from LT
  4. The true distance from origin (0,0,0) to P

Explanation

The true spatial distance from point P to the axis of rotation α1 forms the radius of rotation for abatment. In 3D, this radius R is the hypotenuse of a right triangle whose legs are the distance d(p, α1) in plan view and the elevation zP of point P. The abatted point P0 lies at distance R from α1 along the perpendicular to α1 through p.

Question 43Technical drawing: plane and spatial geometry, projections and dimensions

What is the reverse operation of plane abatment (desabatiment / desabatimiento)?

  1. Reconstructing the 3D projections (plan and elevation) of a geometric entity from its true-shape abatted representation Correct answer.
  2. Rotating the frontal plane onto the profile plane
  3. Projecting an abatted circle as a sphere
  4. Translating the abatted figure back to the origin

Explanation

Desabatiment (de-abatment) is the inverse graphical procedure where a figure constructed in true shape on the abatted plane (such as a true regular polygon or circle) is mapped back to determine its corresponding horizontal and frontal projections in the dihedral system.

Question 44Technical drawing: plane and spatial geometry, projections and dimensions

When an oblique plane α is abatted onto the horizontal plane around trace α1, what is the abatted position of its frontal trace α2 (denoted α2_0)?

  1. A line passing through the intersection point of traces (α1 ∩ LT) making an angle with α1 equal to the true angle between α1 and α2 in space Correct answer.
  2. A line parallel to trace α1 at a distance equal to the height of the plane
  3. A line perpendicular to the ground line passing through origin
  4. The ground line LT itself

Explanation

The vertex of the plane (where α1 and α2 intersect on LT) is a fixed point during rotation. As trace α2 rotates into HP, it passes through this vertex and its distance from LT is preserved, making α2_0 form the true spatial angle between traces α1 and α2.

Question 45Technical drawing: plane and spatial geometry, projections and dimensions

In a change of projection plane (canvi de pla) where the vertical plane of projection is replaced (changing LT to LT1), which coordinate of a point remains unchanged in the new elevation view?

  1. The elevation (z-coordinate / height) Correct answer.
  2. The departure (y-coordinate)
  3. The x-coordinate
  4. Both departure and x-coordinate

Explanation

When changing the vertical projection plane (creating a new frontal view with new ground line LT1), the horizontal plane of projection HP remains fixed. Therefore, the elevation (height z) of every point above HP remains strictly unchanged.

Question 46Technical drawing: plane and spatial geometry, projections and dimensions

To transform an oblique line r into a line of position (such as a frontal-projecting or vertical-projecting line) using changes of projection plane, how many consecutive plane changes are required?

  1. 1 plane change
  2. 2 consecutive plane changes Correct answer.
  3. 3 consecutive plane changes
  4. 4 consecutive plane changes

Explanation

Two steps are required: (1) First plane change: set a new LT parallel to r (or r') to transform the oblique line into a principal line (horizontal or frontal line showing true length). (2) Second plane change: set a new LT perpendicular to the true length projection to transform the line into a point-projecting line (vertical or end view).

Question 47Technical drawing: plane and spatial geometry, projections and dimensions

When rotating a point P in space around a vertical axis of rotation e (axis perpendicular to HP):

  1. Its horizontal projection p moves along a circle centered at e, while its elevation p' moves horizontally parallel to LT. Correct answer.
  2. Its elevation p' moves along a circle centered at e', while its horizontal projection p moves vertically.
  3. Both projections move in straight lines parallel to LT.
  4. Both projections describe concentric ellipses.

Explanation

A vertical axis e is perpendicular to HP. Any point P rotating around e describes a horizontal circle parallel to HP. Thus, its plan view p traces a circular arc centered at e, while its frontal view p' stays at constant elevation z, moving along a horizontal straight line parallel to LT.

Question 48Technical drawing: plane and spatial geometry, projections and dimensions

To transform an oblique plane α into a horizontal (level) plane using rotations (girs), what is the correct strategy?

  1. Rotate plane α around a vertical axis to make it frontal-projecting, then rotate it around a frontal axis to make it horizontal. Correct answer.
  2. Rotate plane α by 90° around the ground line LT.
  3. Rotate trace α1 by 45° and trace α2 by 45° simultaneously around origin.
  4. Rotate plane α around an oblique axis passing through the plane's centroid.

Explanation

Transforming an oblique plane to horizontal requires two principal rotations: (1) rotate around a vertical axis e1 so that horizontal trace α1 becomes perpendicular to LT (making α a frontal-projecting / edge-view plane), then (2) rotate around a frontal axis e2 so that frontal trace α2 becomes parallel to LT (making α a horizontal level plane).

Question 49Technical drawing: plane and spatial geometry, projections and dimensions

What is the true distance from a point P to an oblique plane α?

  1. The length of the segment perpendicular to plane α dropped from point P to its piercing point I on α. Correct answer.
  2. The distance between horizontal projection p and trace α1.
  3. The difference in elevation between P and the vertex of plane α.
  4. The distance between frontal projection p' and trace α2 measured along LT.

Explanation

The distance from a point P to a plane α is defined as the shortest distance, which is measured along the line r passing through P perpendicular to plane α (r ⊥ α1, r' ⊥ α2) up to its intersection point I with plane α. Finding the true length of segment PI gives the true distance.

Question 50Technical drawing: plane and spatial geometry, projections and dimensions

How can the true distance between two parallel oblique planes α and β be found graphically in dihedral projection?

  1. Pass an auxiliary line perpendicular to both planes, find its piercing points Iα and Iβ on each plane, and determine the true length of segment IαIβ. Correct answer.
  2. Measure the distance between horizontal traces α1 and β1 along a line perpendicular to LT.
  3. Measure the vertical distance between frontal traces α2 and β2.
  4. Abat plane α onto HP and measure its distance to the vertex of β.

Explanation

The true distance between two parallel planes is the length of any segment perpendicular to both planes. One constructs a perpendicular line n (n ⊥ α1, β1 and n' ⊥ α2, β2), finds its intersection points Iα = n ∩ α and Iβ = n ∩ β, and calculates/constructs the true length of segment IαIβ.