Projections of Solids Questions and Answers

Answer: A

Correct answer: all visible

Looking down on a pyramid standing on its base, the lateral faces all slope towards the observer and nothing lies above the lateral edges, so all of them are seen and all are drawn as continuous lines, running from the corners of the base to the centre. Hidden edges would appear only if the solid were tilted so that some face came to lie over another.

Answer: B

Correct answer: a circle of the same diameter

A sphere looks the same from every direction, so each orthographic view is a circle of the sphere’s own diameter, whatever the direction of viewing. It is the only common solid for which this is true, and it is why a sphere on a drawing must be located by centre lines rather than identified by its outline.

Answer: C

Correct answer: no distinguishable orientation

Because a sphere is identical about every axis through its centre, turning it changes nothing that a drawing can record: every view is the same circle. There is no axis, no base and no face to fix its attitude, so orientation is not a property it possesses. Its position, however, is definite and is given by the centre.

Answer: D

Correct answer: 50 mm

The slant height is the hypotenuse of the triangle formed by the radius and the vertical height: \[ l = \sqrt{r^2 + h^2} = \sqrt{30^2 + 40^2} \] \[ = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \mathrm{mm}. \] Adding the two, which gives 70, is the usual error; the slant height must exceed both the radius and the height but be less than their sum.

Answer: A

Correct answer: 20 mm

Rearranging the same relation for the vertical height: \[ h = \sqrt{l^2 - r^2} = \sqrt{29^2 - 21^2} \] \[ = \sqrt{841 - 441} = \sqrt{400} = 20\ \mathrm{mm}. \] The slant height is the hypotenuse, so it is the quantity squared and subtracted from; treating the radius as the hypotenuse instead would leave a negative under the root.

Answer: B

The lateral edge runs from the apex to a base corner, so the horizontal distance it spans is half the base diagonal, not half the base side. For a square base of side \( a = 40 \) mm, \[ \tfrac{1}{2}d \;=\; \frac{a\sqrt{2}}{2} \;=\; \frac{40\sqrt{2}}{2} \;=\; 20\sqrt{2} \;\approx\; 28.28\ \text{mm}. \] The edge is then the hypotenuse on that and the axis height \( h = 50 \) mm: \[ l \;=\; \sqrt{h^{2} + \left(\tfrac{1}{2}d\right)^{2}} \;=\; \sqrt{50^{2} + \left(20\sqrt{2}\right)^{2}} \;=\; \sqrt{2500 + 800} \;=\; \sqrt{3300} \;\approx\; 57.4\ \text{mm}. \] Using half the base side instead gives the slant height of a face, \[ \sqrt{50^{2} + 20^{2}} \;=\; \sqrt{2900} \;\approx\; 53.9\ \text{mm}, \] which is the distractor offered at option A.

Answer: C

The face slant height runs from the apex to the midpoint of a base edge, so its horizontal reach is half the base side, \( 20 \) mm, and not half the diagonal. It is the hypotenuse on that and the axis height \( h = 50 \) mm: \[ h_{s} \;=\; \sqrt{h^{2} + \left(\tfrac{a}{2}\right)^{2}} \;=\; \sqrt{50^{2} + 20^{2}} \;=\; \sqrt{2500 + 400} \;=\; \sqrt{2900} \;\approx\; 53.9\ \text{mm}. \] This is shorter than the lateral edge \( \sqrt{3300} \approx 57.4 \) mm because the corner of the base lies further out in plan than an edge midpoint, by a factor \[ \frac{d/2}{a/2} \;=\; \sqrt{2}. \] Using the whole base side, \( \sqrt{50^{2} + 40^{2}} = \sqrt{4100} \approx 64.0 \) mm, is the usual slip.

Answer: D

Correct answer: 50 mm by 70 mm

With the cylinder standing on its base the elevation is a rectangle whose width is the base diameter and whose height is the axis length - 50 mm by 70 mm. The plan is a circle of 50 mm diameter. Each view shows the dimensions that lie parallel to its own plane of projection, which is why the diameter appears in both and the axis in only one.

Answer: A

Correct answer: 50 mm

In a regular hexagon the distance across corners is twice the side, because the hexagon divides into six equilateral triangles with the side as every edge: \[ 2 \times 25 = 50\ \mathrm{mm}. \] Since the prism stands on its base, the plan shows the hexagon at true size, so this is also the measurement on the drawing.

Answer: B

Correct answer: 43.3 mm

The distance across flats is \( \sqrt{3} \) times the side: \[ \sqrt{3} \times 25 = 1.732 \times 25 = 43.3\ \mathrm{mm}. \] It is the shorter of the two hexagon dimensions, standing to the across-corners distance in the ratio \( \sqrt{3} : 2 \), and it is the one that matters when the prism must pass through an opening or be gripped by a spanner.