Sections of Solids

Given/Concept: Sectioning is a drawing technique used in engineering graphics.

Answer: Sectioning is the process of cutting a solid (or an object) with an imaginary cutting plane (called the section plane) in order to reveal its internal details, shape, and construction. The exposed cut surface is shown with hatching lines in the resulting view, which is called a sectional view.

Answer: Two practical examples where sectioning is used:

1. Machine parts / Engineering components – Internal features such as holes, slots, keyways, and recesses of a machine part (e.g., a pulley or a bearing housing) cannot be seen in a normal orthographic view. A sectional view reveals these hidden details clearly.

2. Building/Architectural drawings – A sectional view of a building shows the internal layout, wall thickness, floor levels, staircase details, etc., which are not visible in a plain elevation or plan view.

Answer: The cutting plane is shown as a cutting plane line (a chain line with thick ends and arrows indicating the direction of viewing).

Answer: In sectional view, the portion of the object between the section plane and the observer is assumed to be removed, exposing the internal cut surface.

Answer: The hatching lines are generally drawn inclined to the horizontal at an angle of 45°.

*(They are drawn as thin, equally spaced lines at 45° to the reference line or to the principal edges of the section surface.)*

Note: The actual figures 5.35(a) and (b) are not reproduced in the OCR text. The general procedure to draw sectional views from given orthographic projections is as follows:

Given: Orthographic projections (Front View and Top View) of two solids with section planes A-A and B-B respectively.

Concept/Procedure:

Step 1 – Identify the section plane: Locate the cutting plane line (A-A or B-B) on the given view. Note whether it is horizontal, vertical, or inclined.

Step 2 – Find Points of Intersection (POIs): Mark the points where the section plane cuts the edges/generators of the solid in the view where the section plane appears as a line (edge view).

Step 3 – Project the POIs: Project these points to the adjacent view using standard orthographic projection rules (vertical projectors for Top View, horizontal projectors for Front View).

Step 4 – Join the POIs: Connect the projected points in the correct sequence to obtain the outline of the section.

Step 5 – Apply hatching: Draw hatching lines (thin lines at 45°, equally spaced ~2–3 mm apart) within the section outline to indicate the cut surface.

Step 6 – Show visible edges: Draw all visible edges of the remaining solid (the portion between the section plane and the observer) as continuous thick lines. Hidden lines are generally omitted in sectional views.

Result: The completed sectional view shows the cut surface (hatched) along with the visible outline of the remaining solid.

Given:
- Base diameter of cylinder $= 60$ mm, so radius $= 30$ mm
- Height (axis length) $= 60$ mm
- Cylinder rests on its base on HP with axis vertical
- Section plane: parallel to HP (i.e., parallel to the base), at height $= 40$ mm from HP

Concept: A section plane parallel to HP cuts a vertical cylinder in a circle equal to the base circle. The sectional Top View shows the true shape of the section (a circle), and the Front View shows the cut.

Construction Steps:

Step 1 – Draw the Front View:
- Draw the X-Y line.
- Draw the Front View of the cylinder: a rectangle $60$ mm wide and $60$ mm tall, sitting on X-Y.
- Mark the section plane line (A-A) as a horizontal chain line at height $40$ mm from the base (i.e., $40$ mm above X-Y).
- The section plane cuts the two vertical edges (generators) of the cylinder at points $a'$ and $b'$ at height $40$ mm.
- Hatch the Front View above the section plane line (the cut face visible in FV) — actually, in the Front View, the cut portion above the section plane is removed; the remaining lower portion is shown. The top edge of the remaining solid in FV is the section line A-A.
- Draw hatching on the rectangular cut face visible in FV (the two small rectangles at the top corners are not applicable here since the section is a full horizontal cut).

Step 2 – Draw the sectional Top View:
- Since the section plane is parallel to HP, the sectional Top View shows the true shape of the section.
- The section is a circle of diameter $60$ mm (same as the base).
- Draw a circle of diameter $60$ mm as the sectional Top View.
- Apply hatching (45° lines, equally spaced) throughout the circle to indicate the cut surface.
- Also show the base circle (outline of the cylinder base) as a continuous thin line if required.

Result:
- Front View: Rectangle $60 \times 40$ mm (the lower portion of the cylinder after cutting), with the top edge representing the section plane.
- Sectional Top View: A fully hatched circle of diameter $60$ mm representing the true circular section.

Given:
- Pentagonal pyramid: base side $= 30$ mm, height $= 60$ mm
- Resting on HP with axis vertical
- One edge of base is normal (perpendicular) to VP
- Section plane: horizontal (parallel to HP), at height $= 20$ mm from base

Concept: A horizontal section plane cuts a vertical pyramid parallel to its base. The section is a smaller, similar pentagon. The sectional Top View shows the true shape.

Construction Steps:

Step 1 – Draw the Top View (base):
- Draw a regular pentagon with side $30$ mm such that one side is perpendicular to the XY line (i.e., parallel to VP means the edge is normal to VP, so that edge appears as a point in the side view and as a line perpendicular to XY in the top view).
- Label the base corners $1, 2, 3, 4, 5$ and the apex projection $o$ (centroid of pentagon).

Step 2 – Draw the Front View:
- Project the pentagon corners to get the Front View.
- The Front View of the pyramid is a triangle of base $=$ width of pentagon in FV direction and height $= 60$ mm.
- Draw the section plane line (A-A) as a horizontal chain line at $20$ mm above X-Y.
- The section plane cuts the slant edges at points $1', 2', 3', 4', 5'$ (in FV, only the visible ones are seen).

Step 3 – Find the section in Front View:
- At height $20$ mm, the section plane cuts all five slant edges.
- By similar triangles, the distance of the section from apex $= 60 - 20 = 40$ mm from apex.
- Scale factor $= \dfrac{40}{60} = \dfrac{2}{3}$
- Side of section pentagon $= 30 \times \dfrac{2}{3} = 20$ mm
- In the Front View, mark the cutting points on the visible slant edges at height $20$ mm.

Step 4 – Draw the sectional Top View:
- Project the cutting points from the Front View down to the Top View.
- The section is a regular pentagon of side $20$ mm, similar to the base and centred on the axis.
- Draw this smaller pentagon in the Top View.
- Apply hatching (45° lines) inside the smaller pentagon.
- Show the base pentagon outline as a thin line.
- Show the visible slant edges from the section pentagon to the base corners.

Result:
- Front View: Triangle showing the pyramid with a horizontal section line at $20$ mm from base. The lower portion (frustum) is retained; the top face is the section.
- Sectional Top View: A hatched regular pentagon of side $20$ mm (the true shape of the section) centred on the axis, with the base pentagon and visible edges shown.

Given:
- Cube: side $= 55$ mm
- One edge of the cube rests on HP
- Axis (body diagonal or the axis of the cube) is inclined at $60°$ to HP
- Section plane: vertical, parallel to VP, perpendicular to HP, cuts the axis at its midpoint

Concept: When a cube has its axis inclined to HP, we use the change-of-position method (two-stage projection). The section plane parallel to VP gives a sectional Front View showing the true shape of the section.

Construction Steps:

Stage 1 – Cube with axis vertical (initial position):
- Draw the Top View as a square of side $55$ mm.
- Draw the Front View as a square of side $55$ mm with base on X-Y.
- Label corners: base $1,2,3,4$ and top $5,6,7,8$.

Stage 2 – Tilt the axis to 60° to HP:
- Redraw the Front View so that the axis (vertical line through centre) makes $60°$ with X-Y.
- One edge of the cube rests on HP (on X-Y line).
- Obtain the new Top View by projecting from the tilted Front View.

Step 3 – Draw the section plane:
- The section plane is parallel to VP (appears as a vertical line in the Top View) and cuts the axis at its midpoint.
- In the Top View, draw a vertical line (parallel to Y-axis / perpendicular to X-Y) through the midpoint of the axis projection. This is the section plane line.
- Mark the Points of Intersection (POIs) of this line with the edges of the cube in the Top View: label them $a, b, c, \ldots$

Step 4 – Draw the sectional Front View:
- Project the POIs from the Top View up to the Front View.
- The section plane is parallel to VP, so the sectional Front View shows the true shape of the section.
- Join the projected points in sequence.
- Apply hatching (45° lines) to the section area.
- Show the visible edges of the remaining half of the cube.

Step 5 – Complete the drawing:
- Show the Top View with the section plane line and the cut portion indicated.
- The sectional Front View shows the cross-section (which will be a rectangle or irregular polygon depending on the orientation) with hatching.

Result: The projections show the cube inclined at $60°$ to HP with one edge on HP. The sectional Front View shows the true shape of the vertical section through the midpoint of the axis, with the cut surface hatched at $45°$.

Answer: True shape can be obtained on a plane parallel to the section plane.

Reason: When an auxiliary reference plane is drawn parallel to the section plane, and the section points are projected onto it (with distances transferred from the adjacent view), the resulting shape is the true shape of the section without any distortion.

Answer: The true shape of a section of a sphere cut by any plane (at any angle) is always a circle.

Reason: Any plane cutting a sphere produces a circular cross-section. The size of the circle depends on the distance of the cutting plane from the centre of the sphere. If the plane passes through the centre, the section is a great circle (maximum diameter); otherwise it is a smaller circle.

Answer: When a vertical pentagonal pyramid is cut by a horizontal section plane, the true shape will be a pentagon (a smaller regular pentagon similar to the base).

Reason: A horizontal section plane is parallel to the base of the vertical pyramid. Since the base is a regular pentagon, every horizontal cross-section is also a regular pentagon (smaller in size, depending on the height of the cut), similar to the base.

Correct Answer: (i) Top View

Justification: The projection of any object (or its cut portion) onto the Horizontal Plane (HP) gives the Top View. When the solid is sectioned and the cut portion is projected onto HP, the resulting view is called the Sectional Top View.

Correct Answer: (iii) frustum

Justification: When a horizontal plane cuts a vertical cone parallel to its base (but not through the apex), the portion between the cutting plane and the base is a frustum of a cone — a solid with two parallel circular faces (top and bottom) and a tapering lateral surface.

Correct Answer: (i) When the cutting plane is parallel to HP and perpendicular to VP

Justification: The Top View is the projection onto HP. When the section plane is parallel to HP, the section surface is also parallel to HP. Therefore, its projection onto HP (the sectional Top View) gives the section without any foreshortening — i.e., the true shape. A plane parallel to HP is automatically perpendicular to VP.

Correct Answer: (iv) Rectangle

Justification: A plane passing through the axis of a cylinder (axial section) cuts the cylinder along a rectangle. The width of the rectangle equals the diameter of the base ($= 2r$) and the height equals the height of the cylinder ($= r$, since height $=$ base radius). So the section is a rectangle of dimensions $2r \times r$. Since $2r \neq r$, it is a rectangle (not a square).

Correct Answer: (ii) Sphere

Justification: A sphere is the only solid among the given options where every possible plane section is a circle (since all points on a sphere are equidistant from the centre). For a cylinder or cone, the shape of the section depends on the angle of the cutting plane (it can be a circle, ellipse, parabola, or hyperbola). A circular lamina gives a circle only when cut parallel to its face.

Correct Answer: (iv) Triangle

Justification: A plane passing through the apex of a cone and the centre of the base passes through the axis of the cone. This axial section cuts the cone along two slant edges (generators) and the diameter of the base, producing an isosceles triangle (with base equal to the diameter of the cone's base and height equal to the cone's height).

Correct Answer: (iv) Rectangle

Justification: A plane parallel to the axis of a hexagonal prism cuts through the lateral faces (not the bases). Since the lateral faces of a prism are rectangles, the section through one or more lateral faces parallel to the axis produces a rectangle (the height of the rectangle equals the height of the prism and the width depends on how many faces are cut). The answer given in the textbook answer key is (iv) Rectangle.

Given:
- Hexagonal pyramid: base side $= 30$ mm, axis $= 50$ mm
- Resting on HP with axis vertical
- One base edge parallel to VP
- Section plane: parallel to VP (and perpendicular to HP), at $10$ mm in front of the axis

Concept: A section plane parallel to VP appears as a straight line in the Top View. The sectional Front View shows the true shape of the section (since the section plane is parallel to VP).

Construction Steps:

Step 1 – Draw the Top View:
- Draw a regular hexagon of side $30$ mm with one side parallel to X-Y (since one base edge is parallel to VP).
- Label the corners $1, 2, 3, 4, 5, 6$ (with edge $1$-$2$ parallel to XY) and the apex projection $o$ at the centroid.
- The axis projects as point $o$ in the Top View.

Step 2 – Draw the Front View:
- Project the hexagon corners to get the Front View.
- Draw the Front View: a triangle with base equal to the width of the hexagon in the FV direction and height $= 50$ mm.
- Label the apex $o'$ at the top.

Step 3 – Mark the section plane in Top View:
- The section plane is parallel to VP, so it appears as a vertical line in the Top View (perpendicular to X-Y, i.e., parallel to the Y-direction).
- Draw this line at $10$ mm in front of (below) the axis point $o$ in the Top View.
- Label it A-A.
- Mark the Points of Intersection (POIs) of line A-A with the edges of the hexagon in the Top View: these are points $a, b, c, \ldots$ on the edges that the section plane crosses.

Step 4 – Identify cut edges:
- The section plane at $10$ mm in front of the axis will cut through the front portion of the hexagonal pyramid.
- It will intersect the slant edges and/or base edges that lie in front of the axis.
- Mark all POIs carefully.

Step 5 – Draw the sectional Front View:
- Project all POIs from the Top View up to the Front View using vertical projectors.
- On each slant edge in the Front View, mark the projected POI.
- Join these points in sequence to get the section outline.
- Apply hatching (45° lines) to the section area.
- Show the visible edges of the remaining solid (the portion between the section plane and the observer/VP).

Step 6 – Complete the Top View:
- Show the section plane line A-A in the Top View.
- The Top View shows the base hexagon and the projection of the cut solid.

Result: The sectional Front View shows the true shape of the section (a trapezoid or triangle depending on the exact position) with hatching. The Top View shows the hexagonal base with the section plane line.

Given:
- Square pyramid: base side $= 55$ mm, height $= 70$ mm
- Lying on one of its triangular (lateral) faces on HP
- Axis is parallel to VP
- Section plane: parallel to HP, cuts the axis at its midpoint

Concept: The pyramid is in an inclined position (axis parallel to VP but inclined to HP). We use the change-of-position method. The section plane parallel to HP gives a sectional Top View.

Construction Steps:

Stage 1 – Pyramid with base on HP (initial position):
- Draw the Top View: a square of side $55$ mm with diagonals, apex $o$ at centre.
- Draw the Front View: a triangle, base $55$ mm, height $70$ mm.
- Label base corners $1,2,3,4$ and apex $o$.

Stage 2 – Tilt to lie on a triangular face:
- When the pyramid lies on a triangular face, the slant face is on HP.
- The slant height of the triangular face $= \sqrt{h^2 + (a/2)^2} = \sqrt{70^2 + 27.5^2} = \sqrt{4900 + 756.25} = \sqrt{5656.25} \approx 75.2$ mm
- Tilt the Front View so that one triangular face lies on X-Y (on HP), with the axis parallel to VP (axis appears true length in FV).
- Redraw the Front View in the tilted position.
- Obtain the new Top View by projecting from the tilted Front View.

Step 3 – Mark the section plane:
- The midpoint of the axis is at $35$ mm from the base (or apex).
- In the Front View, mark the midpoint $M$ of the axis.
- Draw the section plane line (A-A) as a horizontal line (parallel to X-Y) through $M$ in the Front View.
- Mark the POIs of A-A with all visible edges in the Front View.

Step 4 – Draw the sectional Top View:
- Project the POIs from the Front View down to the Top View.
- Join the projected points to get the section outline in the Top View.
- Apply hatching (45° lines) to the section area.
- Show the visible edges of the remaining solid.

Result: The Front View shows the pyramid lying on its triangular face with the horizontal section line at the midpoint of the axis. The sectional Top View shows the section outline with hatching.

Given:
- Pentagonal pyramid: base side $= 30$ mm, height $= 50$ mm
- Axis vertical, resting on HP
- One base edge perpendicular to VP (i.e., that edge is parallel to the X-Y line in Top View — actually perpendicular to VP means the edge goes into the VP, so it appears as a point in FV and as a line perpendicular to XY in TV)
- Section plane: horizontal (parallel to HP), at height $= 25$ mm from base

Concept: Horizontal section of a vertical pentagonal pyramid gives a smaller similar pentagon.

Construction Steps:

Step 1 – Draw the Top View:
- Draw a regular pentagon of side $30$ mm with one side perpendicular to X-Y (i.e., one edge is perpendicular to VP, so it appears perpendicular to XY in the Top View).
- Label corners $1,2,3,4,5$ and apex projection $o$.

Step 2 – Draw the Front View:
- Project the pentagon to get the Front View.
- Draw the triangular Front View with height $50$ mm.
- Mark the section plane line A-A at height $25$ mm above X-Y (horizontal chain line).
- The section plane cuts the slant edges at $25$ mm height.

Step 3 – Find the section size:
- By similar triangles: distance from apex $= 50 - 25 = 25$ mm.
- Scale factor $= \dfrac{25}{50} = \dfrac{1}{2}$
- Side of section pentagon $= 30 \times \dfrac{1}{2} = 15$ mm
- Mark the cutting points on the slant edges in the Front View at height $25$ mm.

Step 4 – Draw the sectional Top View:
- Project the cutting points from the Front View down to the Top View.
- The section is a regular pentagon of side $15$ mm, centred on the axis, similar to and parallel to the base.
- Draw this smaller pentagon in the Top View.
- Apply hatching (45° lines) inside the smaller pentagon.
- Show the base pentagon and visible slant edges.

Result:
- Front View: Triangle (pyramid) with horizontal section line at $25$ mm from base.
- Sectional Top View: Hatched regular pentagon of side $15$ mm (true shape of section) with base pentagon and edges shown.

Given:
- Pentagonal prism: base side $= 25$ mm, height $= 65$ mm
- Resting on base on HP, axis vertical
- One side of base inclined at $30°$ to VP
- Section plane: inclined at $60°$ to HP (and perpendicular to VP), passing through midpoint of axis (at $32.5$ mm from base)

Concept: The section plane inclined to HP appears as an inclined line in the Front View. The section is an oblique cut through the prism. True shape is obtained by auxiliary projection.

Construction Steps:

Step 1 – Draw the Top View:
- Draw a regular pentagon of side $25$ mm with one side inclined at $30°$ to X-Y.
- Label corners $1,2,3,4,5$ (base) and $1',2',3',4',5'$ (top, directly above).
- The Top View shows the pentagon (base and top coincide in projection).

Step 2 – Draw the Front View:
- Project the pentagon to get the Front View.
- The Front View of the prism is a rectangle (or series of vertical lines) of height $65$ mm.
- Draw the section plane line A-A: inclined at $60°$ to X-Y (to HP), passing through the midpoint of the axis at height $32.5$ mm.
- The section plane line cuts the vertical edges of the prism in the Front View at various heights.
- Mark the POIs: $a', b', c', d', e'$ on the respective edges.

Step 3 – Find POIs in Front View:
- The section plane passes through the midpoint of the axis at $32.5$ mm and is inclined at $60°$ to HP.
- Starting from the midpoint on the axis, the section plane line goes up on one side and down on the other at $60°$ to horizontal.
- Mark where this inclined line intersects each vertical edge of the prism in the Front View.

Step 4 – Draw the sectional Top View:
- Project the POIs from the Front View down to the Top View.
- On each edge in the Top View, mark the projected POI.
- Join the POIs in sequence to get the section outline.
- Apply hatching (45° lines) to the section area.
- Show visible edges of the remaining solid.

Step 5 – Draw the true shape of the section:
- Draw an auxiliary reference line $X_1Y_1$ parallel to the section plane line A-A in the Front View.
- Draw projectors from each POI ($a', b', c', d', e'$) perpendicular to $X_1Y_1$.
- Transfer the distances of the corresponding points ($a, b, c, d, e$) from X-Y in the Top View to the respective projectors measured from $X_1Y_1$.
- Join the transferred points $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}$ in sequence.
- Apply hatching to the enclosed area.

Result: The true shape of the section is an irregular pentagon (oblique section of a pentagonal prism). The Front View shows the prism with the inclined section line. The sectional Top View shows the apparent section with hatching.

Given:
- Hexagonal prism: base side $= 24$ mm, axis $= 55$ mm
- Resting on an edge of base on HP
- Axis inclined at $60°$ to HP and parallel to VP
- Section plane: inclined at $45°$ to VP (and perpendicular to HP), passing through a point on the axis $25$ mm from the top end

Concept: The prism is in an inclined position (axis inclined to HP). The section plane is inclined to VP, so it appears as an inclined line in the Top View. True shape requires auxiliary projection.

Construction Steps:

Stage 1 – Prism with axis horizontal and parallel to VP:

Step 1 – Initial position (axis vertical):
- Draw Top View: regular hexagon of side $24$ mm.
- Draw Front View: rectangle $\approx 48$ mm wide, $55$ mm tall.

Step 2 – Tilt axis to 60° to HP (axis parallel to VP):
- In the Front View, redraw the prism with its axis inclined at $60°$ to X-Y, with one base edge on X-Y (on HP).
- The axis appears true length in the Front View (since it is parallel to VP).
- Obtain the new Top View by projecting from the tilted Front View.

Step 3 – Mark the section plane:
- The section plane is inclined at $45°$ to VP, so it appears as an inclined line in the Top View.
- Locate the point on the axis $25$ mm from the top end in the Front View. Project this to the Top View to get the corresponding point.
- Draw the section plane line (A-A) through this point at $45°$ to X-Y in the Top View.
- Mark the POIs of A-A with the edges of the prism in the Top View.

Step 4 – Draw the sectional Top View:
- Mark all POIs on the prism edges in the Top View.
- Join the POIs to get the section outline.
- Apply hatching (45° lines) to the section area.
- Show visible edges.

Step 5 – Draw the Front View:
- Project the POIs from the Top View up to the Front View.
- Mark the POIs on the respective edges in the Front View.
- Join them and show the sectional Front View.

Step 6 – True shape of the section:
- Draw auxiliary reference line $X_1Y_1$ parallel to the section plane line A-A in the Top View.
- Draw projectors from each POI perpendicular to $X_1Y_1$.
- Transfer distances from X-Y in the Front View to the projectors measured from $X_1Y_1$.
- Join the transferred points and hatch the area.

Result: The true shape of the section is an irregular hexagon (oblique section through the hexagonal prism). The Front View and sectional Top View show the cut solid with hatching.

Given:
- Cone: base diameter $= 50$ mm (radius $= 25$ mm), axis $= 60$ mm
- Axis parallel to VP and inclined at $45°$ to HP
- Section plane: horizontal (parallel to HP), passing through midpoint of axis

Concept: The cone is inclined with axis at $45°$ to HP. The horizontal section plane cuts the cone obliquely (not parallel to the base), producing an elliptical section.

Construction Steps:

Stage 1 – Cone with axis vertical:
- Draw Top View: circle of diameter $50$ mm with centre $o$.
- Draw Front View: isosceles triangle, base $50$ mm, height $60$ mm.
- Divide the base circle into $12$ equal parts and label generators.

Stage 2 – Tilt axis to 45° to HP (parallel to VP):
- In the Front View, redraw the cone with axis inclined at $45°$ to X-Y.
- The base is now inclined; one point of the base circle touches HP (or the base is positioned as given).
- Obtain the new Top View by projecting from the tilted Front View.
- Draw all generators in both views.

Step 3 – Mark the section plane:
- The midpoint of the axis is at $30$ mm from each end.
- In the Front View, mark the midpoint $M$ of the axis.
- Draw the section plane line A-A as a horizontal line (parallel to X-Y) through $M$.
- Mark the POIs of A-A with all generators in the Front View: $a', b', c', \ldots$

Step 4 – Draw the sectional Top View:
- Project the POIs from the Front View down to the Top View.
- On each generator in the Top View, mark the projected POI.
- Join the POIs with a smooth curve (ellipse) to get the section outline.
- Apply hatching to the section area.

Step 5 – True shape of the section:
- Since the section plane is horizontal (parallel to HP), the sectional Top View IS the true shape.
- The true shape is an ellipse.
- Alternatively, draw auxiliary projection parallel to A-A to confirm.

Result:
- Front View: Inclined cone with horizontal section line at midpoint of axis.
- Sectional Top View: Elliptical section with hatching (true shape since section plane is parallel to HP).

Given:
- Cylinder: diameter $= 40$ mm (radius $= 20$ mm), height $= 65$ mm
- Resting on base on HP, axis vertical
- Section plane: inclined at $60°$ to HP (perpendicular to VP), cutting the axis at $20$ mm from the top (i.e., at height $= 65 - 20 = 45$ mm from base)

Concept: An oblique section plane cuts a vertical cylinder to produce an ellipse. The section plane appears as an inclined line in the Front View.

Construction Steps:

Step 1 – Draw the Top View:
- Draw a circle of diameter $40$ mm.
- Divide it into $12$ equal parts and label the points $1, 2, 3, \ldots, 12$.

Step 2 – Draw the Front View:
- Draw a rectangle $40$ mm wide and $65$ mm tall (the cylinder).
- Project the $12$ division points to the Front View as vertical lines (generators).
- Mark the section plane line A-A: inclined at $60°$ to X-Y, passing through the axis at height $45$ mm from base.
- The section plane line cuts each generator at a different height.
- Mark the POIs: $1', 2', 3', \ldots, 12'$ on the respective generators.

Step 3 – Find heights of POIs:
- The section plane passes through the axis at height $45$ mm and is inclined at $60°$ to HP.
- For each generator at horizontal distance $x$ from the axis, the height of the POI $= 45 + x \cdot \tan 60° = 45 + x\sqrt{3}$.
- Calculate for each of the $12$ generators (using their $x$-coordinates from the Top View).
- Mark these heights in the Front View.

Step 4 – Draw the sectional Top View (sectional plan):
- Project the POIs from the Front View down to the Top View.
- On each generator point in the Top View, mark the projected POI.
- Join the POIs with a smooth elliptical curve.
- Apply hatching (45° lines) to the section area.

Step 5 – True shape of the section:
- Draw auxiliary reference line $X_1Y_1$ parallel to the section plane line A-A in the Front View.
- Draw projectors from each POI ($1', 2', \ldots, 12'$) perpendicular to $X_1Y_1$.
- Transfer the distances of the corresponding points ($1, 2, \ldots, 12$) from X-Y in the Top View to the respective projectors measured from $X_1Y_1$.
- Join the transferred points with a smooth curve.
- Apply hatching.

Result: The true shape of the section is an ellipse. The major axis of the ellipse $= \dfrac{\text{diameter}}{\sin 60°} = \dfrac{40}{\sin 60°} = \dfrac{40}{0.866} \approx 46.2$ mm. The minor axis $=$ diameter $= 40$ mm.

Given:
- Sphere: diameter $= 50$ mm, radius $R = 25$ mm
- Resting on HP (centre is at height $25$ mm above HP)
- Section plane: perpendicular to HP, inclined at $45°$ to VP, at distance $10$ mm from the centre of the sphere

Concept: Any plane cutting a sphere gives a circular section. The radius of the section circle $r = \sqrt{R^2 - d^2}$ where $d$ is the perpendicular distance from the centre to the cutting plane.

Radius of section:
$$r = \sqrt{25^2 - 10^2} = \sqrt{625 - 100} = \sqrt{525} = 5\sqrt{21} \approx 22.9 \text{ mm}$$

Construction Steps:

Step 1 – Draw the Front View:
- Draw the Front View of the sphere: a circle of diameter $50$ mm with centre $o'$ at height $25$ mm above X-Y.

Step 2 – Draw the Top View:
- Draw the Top View of the sphere: a circle of diameter $50$ mm with centre $o$.
- The section plane is perpendicular to HP and inclined at $45°$ to VP, so it appears as an inclined straight line in the Top View.
- Draw the section plane line A-A at $45°$ to X-Y in the Top View, at a perpendicular distance of $10$ mm from the centre $o$.
- The section plane line intersects the base circle (Top View of sphere) at two points. Mark these as $a$ and $b$.

Step 3 – Draw the sectional Front View:
- The section plane is perpendicular to HP, so it appears as a vertical line in the Front View (if it were parallel to VP) or as an inclined line. Since it is inclined at $45°$ to VP and perpendicular to HP, it appears as a vertical line in the Front View (its edge view in FV is vertical).
- Project points $a$ and $b$ from the Top View up to the Front View.
- The section in the Front View is bounded by the chord $a'b'$ (vertical line) and the arc of the sphere.
- The sectional Front View shows the portion of the sphere on the observer's side of the section plane, with the cut face shown as a chord.
- Apply hatching to the cut face (the chord area in FV).

Step 4 – True shape of the section:
- Draw auxiliary reference line $X_1Y_1$ parallel to the section plane line A-A in the Top View.
- The section is a circle of radius $r \approx 22.9$ mm.
- Draw projectors from $a$ and $b$ perpendicular to $X_1Y_1$.
- Transfer the heights of $a'$ and $b'$ from X-Y in the Front View to the projectors.
- The true shape is a circle of radius $\approx 22.9$ mm (diameter $\approx 45.8$ mm).
- Draw this circle and apply hatching.

Result: The true shape of the section is a circle of diameter $\approx 45.8$ mm. The sectional Front View shows the sphere with the cut face (chord) hatched.

Given:
- Triangular pyramid (tetrahedron): base side $= 45$ mm, slant height $= 70$ mm
- Base on HP, axis vertical
- One side of base perpendicular to VP
- Section plane: inclined at $60°$ to VP (perpendicular to HP), intersecting the axis at $35$ mm from base

Note: Slant height $= 70$ mm refers to the slant edge (lateral edge) of the triangular pyramid.

Concept: The section plane is perpendicular to HP and inclined to VP, so it appears as an inclined line in the Top View. The sectional Top View shows the apparent section. True shape requires auxiliary projection.

Construction Steps:

Step 1 – Draw the Top View:
- Draw an equilateral triangle of side $45$ mm with one side perpendicular to X-Y (one edge perpendicular to VP).
- Label base corners $1, 2, 3$ and apex projection $o$ (centroid).

Step 2 – Draw the Front View:
- Project the triangle to get the Front View.
- Calculate the height of the pyramid:
- For equilateral triangle base, centroid to vertex $= \dfrac{45}{\sqrt{3}} = 25.98$ mm
- Height $h = \sqrt{70^2 - 25.98^2} = \sqrt{4900 - 675} = \sqrt{4225} = 65$ mm (approx., using slant edge $= 70$ mm)
- Draw the triangular Front View with height $\approx 65$ mm.

Step 3 – Mark the section plane in Top View:
- The section plane is perpendicular to HP, so it appears as a straight line in the Top View.
- It is inclined at $60°$ to VP, so the section plane line makes $60°$ with X-Y in the Top View.
- The section plane intersects the axis at $35$ mm from the base. Project this point from the Front View to the Top View to locate it on the axis projection $o$.
- Draw the section plane line A-A through this point at $60°$ to X-Y in the Top View.
- Mark the POIs of A-A with the edges of the pyramid in the Top View.

Step 4 – Draw the sectional Top View:
- Mark all POIs on the pyramid edges in the Top View.
- Join the POIs to get the section outline.
- Apply hatching (45° lines) to the section area.
- Show visible edges of the remaining solid.

Step 5 – Draw the Front View:
- Project the POIs from the Top View up to the Front View.
- Mark the POIs on the respective edges in the Front View.
- Show the sectional Front View.

Step 6 – True shape of the section:
- Draw auxiliary reference line $X_1Y_1$ parallel to the section plane line A-A in the Top View.
- Draw projectors from each POI perpendicular to $X_1Y_1$.
- Transfer the heights of the corresponding points from X-Y in the Front View to the projectors measured from $X_1Y_1$.
- Join the transferred points and hatch the area.

Result: The true shape of the section is an irregular triangle or quadrilateral (depending on how many edges the section plane cuts). The Front View and sectional Top View show the cut pyramid with hatching.

Given:
- Cone: base diameter $= 42$ mm (radius $= 21$ mm), axis $= 54$ mm
- Resting on base on HP, axis vertical
- Section plane: vertical (perpendicular to HP), inclined at $60°$ to X-Y (i.e., inclined at $60°$ to VP), at $10$ mm from the axis in the Top View

Concept: The section plane is perpendicular to HP, so it appears as an inclined straight line in the Top View. The sectional Front View shows the apparent section. True shape requires auxiliary projection.

Construction Steps:

Step 1 – Draw the Top View:
- Draw a circle of diameter $42$ mm (the base of the cone) with centre $o$.
- Divide the circle into $12$ equal parts and label the points $1, 2, 3, \ldots, 12$.
- Mark the apex projection at $o$ (centre).

Step 2 – Draw the Front View:
- Draw the Front View of the cone: an isosceles triangle, base $42$ mm, height $54$ mm.
- Draw all $12$ generators in the Front View.

Step 3 – Mark the section plane in Top View:
- The section plane is perpendicular to HP (vertical), inclined at $60°$ to X-Y.
- Draw the section plane line A-A at $60°$ to X-Y in the Top View, at a perpendicular distance of $10$ mm from the centre $o$.
- The line A-A intersects the base circle at two points and also cuts some generators.
- Mark the POIs of A-A with the base circle and generators: $a, b, c, \ldots$

Step 4 – Draw the sectional Front View:
- Project the POIs from the Top View up to the Front View.
- On each generator in the Front View, mark the projected POI at the appropriate height.
- Join the POIs with a smooth curve to get the section outline in the Front View.
- Apply hatching (45° lines) to the section area.
- Show visible edges of the remaining solid.

Step 5 – True shape of the section:
- Draw auxiliary reference line $X_1Y_1$ parallel to the section plane line A-A in the Top View.
- Draw projectors from each POI ($a, b, c, \ldots$) perpendicular to $X_1Y_1$.
- Transfer the heights of the corresponding points from X-Y in the Front View to the projectors measured from $X_1Y_1$.
- Join the transferred points with a smooth curve.
- Apply hatching to the enclosed area.

Nature of the section:
- Since the section plane is $10$ mm from the axis (which is less than the radius $21$ mm) and does not pass through the apex, the section is an ellipse (a conic section where the cutting plane is not parallel to any generator and does not pass through the apex).

Result: The true shape of the section is an ellipse. The Top View shows the base circle with the inclined section plane line. The sectional Front View shows the cone with the elliptical section outline and hatching.