Chapter 7 of 8

Updated 6 August 2026

NCERT Solutions

Isometric Projection — NCERT Solutions

CBSE · Class 11 · Engineering Graphics

NCERT Solutions for Isometric Projection, CBSE Class 11 Engineering Graphics: 6 textbook questions solved step by step.

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Assignment – Isometric Projection

1Draw the isometric projection of an equilateral triangle of base side 50 mm in V.P.Show solution

Given: Equilateral triangle, base side = 50 mm, plane in V.P. (Vertical Plane).

Concept: When a plane figure lies in the V.P., its isometric projection is drawn on the isometric front face. All lines parallel to the isometric axes are foreshortened by the isometric scale factor $\approx 0.816$ (i.e., multiply true length by 0.816). In isometric drawing (full-size method), true lengths are used directly.

Step-by-step procedure:

Step 1 – Draw the reference axes.
Draw the isometric axes: one vertical axis and two axes at 30° to the horizontal (left and right). Since the triangle is in the V.P., it will appear on the vertical isometric plane (the front face).

Step 2 – Enclose the triangle in a rectangle.
Enclose the equilateral triangle of side 50 mm in a rectangle.

Height of equilateral triangle $= \dfrac{\sqrt{3}}{2} \times 50 = \dfrac{1.732}{2} \times 50 \approx 43.3$ mm.
Rectangle dimensions: width = 50 mm, height = 43.3 mm.

Step 3 – Draw the isometric rectangle on the V.P. face.
On the isometric drawing, the V.P. face uses the vertical axis and the 30° right (or left) axis.

Mark point $A$ at the bottom-left corner.
Along the 30° axis, mark $B$ at 50 mm from $A$ (base of rectangle).
From $A$ and $B$ , draw vertical lines of height 43.3 mm to get points $D$ and $C$ respectively.
$ABCD$ is the isometric rectangle enclosing the triangle.

Step 4 – Locate key points of the triangle inside the rectangle.

The base of the triangle coincides with $AB$ .
The apex $P$ is at the midpoint of $DC$ , i.e., midpoint of the top side of the rectangle.
Mark $M$ = midpoint of $AB$ ; the apex $P$ is directly above $M$ at height 43.3 mm.

Step 5 – Draw the triangle.
Join:

$A$ to $B$ (base, along 30° axis).
$A$ to $P$ (left slant side — draw as a straight line between the two points).
$B$ to $P$ (right slant side — draw as a straight line between the two points).

Step 6 – Isometric scale (if isometric projection, not drawing).
Multiply all dimensions by 0.816:

Base $= 50 \times 0.816 = 40.8$ mm.
Height $= 43.3 \times 0.816 = 35.3$ mm.

Redraw with these scaled dimensions following the same procedure.

Result: The isometric projection of the equilateral triangle (base 50 mm) in V.P. is obtained — it appears as a triangle on the isometric front face with the base along the 30° axis and the apex located vertically above the midpoint of the base.

2Draw the isometric projection of a square lamina having side 55 mm in H.P.Show solution

Given: Square lamina, side = 55 mm, plane in H.P. (Horizontal Plane).

Concept: When a plane figure lies in the H.P., its isometric projection is drawn on the top (horizontal) isometric face. The two isometric axes used are the two axes at 30° to the horizontal (left and right). Vertical measurements do not apply here.

Step-by-step procedure:

Step 1 – Draw the isometric axes.
Draw the three isometric axes from a common origin $O$ : one vertical, one at 30° to the left, and one at 30° to the right.

Step 2 – Draw the isometric parallelogram (rhombus) for the square.
The square in H.P. appears as a rhombus on the isometric top face.

From origin $O$ , mark point $A$ along the right 30° axis at 55 mm.
From origin $O$ , mark point $B$ along the left 30° axis at 55 mm.
From $A$ , draw a line parallel to $OB$ for 55 mm to get point $C$ .
From $B$ , draw a line parallel to $OA$ for 55 mm to get point $C$ (same point).
$OACB$ (or $OABC$ ) forms the isometric rhombus representing the square.

Step 3 – Check angles.
In the isometric rhombus, the angles are 60° and 120° (since the two 30° axes are at 60° to each other, the parallelogram has included angles of 60° and 120°).

Step 4 – Isometric scale (for true isometric projection).
Multiply side by 0.816:
$\text{Isometric side} = 55 \times 0.816 = 44.9 \text{ mm} \approx 45 \text{ mm}$
Redraw the rhombus with side 44.9 mm along both 30° axes.

Step 5 – Complete the figure.
Darken the four sides of the rhombus $OACB$ . Label the corners.

Result: The isometric projection of the square lamina (side 55 mm) in H.P. is a rhombus with sides $\approx 44.9$ mm drawn on the horizontal isometric face, with sides along the two 30° isometric axes.

3Draw the isometric projection of a regular pentagon of base side 35 mm in V.P.Show solution

Given: Regular pentagon, base side = 35 mm, plane in V.P.

Concept: For irregular plane figures (pentagon), enclose the figure in a rectangle, locate key points by their coordinates (offsets), transfer those coordinates onto the isometric face using the isometric axes, and join the points.

Step-by-step procedure:

Step 1 – Draw the true shape and find coordinates.
Draw the regular pentagon $ABCDE$ with base $AB = 35$ mm in true shape (orthographic view).

Interior angle of regular pentagon $= 108°$ .
Draw $AB$ as the base (horizontal, 35 mm).
Locate all five vertices. Using geometry:
Let $A = (0, 0)$ , $B = (35, 0)$ .
$C = (35 + 35\cos 72°,\; 35\sin 72°) \approx (45.8,\; 33.3)$ mm.
$E = (-35\cos 72°,\; 35\sin 72°) \approx (-10.8,\; 33.3)$ mm.
$D$ = apex at midpoint of $CE$ horizontally $= (17.5,\; 35 + 35\sin 72° \times \tan 54°)$ .
More practically: height of pentagon $= \dfrac{35}{2}(1 + \sqrt{5})\sin 54° \approx 35 \times 1.539 \approx 53.9$ mm (total height from base to apex).
Enclose in a rectangle of width = 35 + 2(35 cos 72°) $\approx$ 46.6 mm and height $\approx$ 53.9 mm.
Mark all five vertices with their $x$ (horizontal) and $y$ (vertical) offsets from the bottom-left corner of the enclosing rectangle.

Step 2 – Draw the isometric enclosing rectangle on the V.P. face.

On the isometric drawing, the V.P. face uses the 30° axis (horizontal direction) and the vertical axis.
Draw the isometric rectangle: width $\approx 46.6$ mm along the 30° axis, height $\approx 53.9$ mm along the vertical axis.
(For isometric projection, multiply all dimensions by 0.816 before drawing.)

Step 3 – Transfer all five vertices.
For each vertex, measure its $x$ -offset along the 30° axis and its $y$ -offset along the vertical axis from the reference corner, and mark the point on the isometric rectangle.

Step 4 – Join the vertices.
Join $A \to B \to C \to D \to E \to A$ with straight lines in sequence.

Step 5 – Isometric scale.
All offsets used in Step 3 should be multiplied by $0.816$ for true isometric projection (or used as-is for isometric drawing).

Result: The isometric projection of the regular pentagon (side 35 mm) in V.P. is obtained as an irregular five-sided figure on the isometric front face.

4Draw the isometric projection of a regular hexagon of base side 30 mm in H.P.

Given: Regular hexagon, base side = 30 mm, plane in H.P.

Concept: Enclose the hexagon in a rectangle, find key point coordinates, and transfer them onto the isometric top face using the two 30° axes.

Step-by-step procedure:

Step 1 – Draw the true shape and find the enclosing rectangle.
Draw the regular hexagon with side 30 mm in true shape (flat, in H.P.).

Place the hexagon with one pair of sides horizontal (flat orientation).
Width of hexagon $= 2 \times 30 = 60$ mm (distance between two parallel vertical sides).
Height of hexagon $= \sqrt{3} \times 30 = 51.96 \approx 52$ mm (distance between two parallel horizontal sides).
Enclosing rectangle: $60$ mm wide $\times$ $52$ mm tall (in true shape).
Label the six vertices $A, B, C, D, E, F$ and note their offsets from the bottom-left corner of the rectangle:
$A = (15, 0)$ , $B = (45, 0)$ , $C = (60, 26)$ , $D = (45, 52)$ , $E = (15, 52)$ , $F = (0, 26)$ .

Step 2 – Draw the isometric enclosing rectangle on the H.P. face.

On the isometric drawing, the H.P. face uses the right 30° axis (for width = 60 mm) and the left 30° axis (for depth = 52 mm).
From origin $O$ :
Mark $P$ at 60 mm along the right 30° axis.
Mark $Q$ at 52 mm along the left 30° axis.
Complete the parallelogram $OPRQ$ (where $R = P + Q$ vectorially).

Step 3 – Transfer all six vertices onto the isometric parallelogram.
For each vertex, its $x$ -offset is measured along the right 30° axis and its $y$ -offset along the left 30° axis:

$A$ : 15 mm along right axis, 0 along left axis.
$B$ : 45 mm along right axis, 0 along left axis.
$C$ : 60 mm along right axis, 26 mm along left axis.
$D$ : 45 mm along right axis, 52 mm along left axis.
$E$ : 15 mm along right axis, 52 mm along left axis.
$F$ : 0 along right axis, 26 mm along left axis.

Step 4 – Join the vertices.
Join $A \to B \to C \to D \to E \to F \to A$ with straight lines.

Step 5 – Isometric scale.
For true isometric projection, multiply all dimensions by $0.816$ :

Side $= 30 \times 0.816 = 24.5$ mm; all offsets scaled similarly.

Result: The isometric projection of the regular hexagon (side 30 mm) in H.P. is obtained as a six-sided figure on the isometric top face, appearing as a flattened hexagon due to the 30° projection.

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5Draw the isometric projection of a circle of dia 50 mm in H.P.

Given: Circle, diameter = 50 mm (radius = 25 mm), plane in H.P.

Concept: A circle in isometric projection appears as an ellipse. It is drawn using the four-centre method (approximate method), which constructs the ellipse using four circular arcs.

Step-by-step procedure:

Step 1 – Determine the isometric square.
The circle is enclosed in a square of side = diameter = 50 mm.
In isometric projection (using isometric scale): side $= 50 \times 0.816 = 40.8$ mm $\approx 41$ mm.
(For isometric drawing, use 50 mm directly.)

Step 2 – Draw the isometric square (rhombus) on the H.P. face.

From origin $O$ , draw two lines at 30° (right and left) each of length 50 mm (or 41 mm for projection).
Complete the rhombus $ABCD$ where $AB$ and $CD$ are along the right 30° axis and $AD$ , $BC$ are along the left 30° axis.
The rhombus has angles of 60° and 120°.

Step 3 – Find the four centres for the four-centre ellipse method.

Mark the midpoints of all four sides of the rhombus: $M_1$ (mid of $AB$ ), $M_2$ (mid of $BC$ ), $M_3$ (mid of $CD$ ), $M_4$ (mid of $AD$ ).
The two obtuse-angle corners of the rhombus (120° corners) are centres $O_1$ and $O_2$ .
Draw perpendiculars from $O_1$ to sides $AB$ and $AD$ ; they meet at $M_1$ and $M_4$ .
Draw perpendiculars from $O_2$ to sides $BC$ and $CD$ ; they meet at $M_2$ and $M_3$ .
The intersections of lines $O_1 M_1$ extended and $O_2 M_2$ extended give centres $O_3$ and $O_4$ (the other two centres, located near the acute-angle corners).

Step 4 – Draw the four arcs.

Arc 1: Centre $O_1$ , radius $= O_1 M_1 = O_1 M_4$ , draw arc from $M_4$ to $M_1$ (large arc, near obtuse corner).
Arc 2: Centre $O_2$ , radius $= O_2 M_2 = O_2 M_3$ , draw arc from $M_2$ to $M_3$ (large arc, near other obtuse corner).
Arc 3: Centre $O_3$ , radius $= O_3 M_1 = O_3 M_2$ , draw small arc from $M_1$ to $M_2$ .
Arc 4: Centre $O_4$ , radius $= O_4 M_3 = O_4 M_4$ , draw small arc from $M_3$ to $M_4$ .

Step 5 – Complete the ellipse.
The four arcs join smoothly at $M_1, M_2, M_3, M_4$ to form the approximate ellipse representing the circle in isometric projection.

Result: The isometric projection of the circle (dia 50 mm) in H.P. is an ellipse drawn by the four-centre method, inscribed in the isometric rhombus on the horizontal face.

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6Draw the isometric projection of a semi-circle of radius 30 mm in H.P.

Given: Semi-circle, radius = 30 mm (diameter = 60 mm), plane in H.P.

Concept: The semi-circle is half of a full circle. First, construct the full isometric ellipse of the circle (diameter 60 mm) using the four-centre method on the H.P. face, then retain only the half that represents the semi-circle.

Step-by-step procedure:

Step 1 – Determine dimensions.

Diameter of full circle $= 2 \times 30 = 60$ mm.
Isometric scale factor $= 0.816$ .
Isometric diameter $= 60 \times 0.816 = 48.96 \approx 49$ mm (for isometric projection).

(Use 60 mm for isometric drawing.)

Step 2 – Draw the enclosing isometric rectangle for the semi-circle.

The semi-circle fits in a rectangle of width = 60 mm and height = 30 mm (radius).
On the H.P. isometric face, draw the isometric rectangle:
From origin $O$ , mark 60 mm along the right 30° axis to point $B$ .
From $O$ , mark 30 mm along the left 30° axis to point $D$ .
From $B$ , mark 30 mm parallel to $OD$ to point $C$ .
$OBCD$ is the isometric rectangle (parallelogram) enclosing the semi-circle.

Step 3 – Construct the full isometric rhombus (for reference).
Extend the rectangle to a full rhombus of side 60 mm to use the four-centre method:

From $O$ , mark 60 mm along the left 30° axis to point $A$ .
Complete the full rhombus $OABC'$ (where $C'$ is the fourth corner of the full rhombus).

Step 4 – Apply the four-centre method to the full ellipse.
Follow the same four-centre method as in Question 5 (with diameter 60 mm) to locate all four centres $O_1, O_2, O_3, O_4$ and midpoints $M_1, M_2, M_3, M_4$ .

Step 5 – Draw only the half ellipse (semi-circle portion).

The diameter line $OB$ (60 mm along the right 30° axis) divides the full ellipse into two halves.
Retain only the upper half (or the half corresponding to the semi-circle in the original orientation) of the four-centre ellipse.
This consists of:
The large arc (centre $O_1$ or $O_2$ ) on the far side of the diameter.
The two small arcs ( $O_3$ and $O_4$ ) at the ends of the diameter.
Draw the straight diameter line $OB$ (60 mm along 30° axis, scaled to 49 mm for projection).

Step 6 – Complete the figure.

The semi-ellipse (half ellipse) plus the straight diameter line $OB$ together form the isometric projection of the semi-circle.
Darken the semi-ellipse and the diameter line.

Result: The isometric projection of the semi-circle (radius 30 mm) in H.P. is a half-ellipse on the horizontal isometric face, with the straight diameter edge along the 30° isometric axis and the curved portion drawn using the relevant arcs of the four-centre method.

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