Construction

Unit 14 · Compulsory Mathematics · 14 min read

Construction

What the syllabus asks for 2
  • Construction: concepts, formulas and worked problems
  • Exercises of the SEE type

A construction is a drawing made with only two tools: an unmarked ruler (straightedge) for straight lines and a compass for circles and arcs. A protractor may be used only when an angle is given in degrees. Constructions turn the theorems of the previous chapter into practice: since a triangle between the same parallels keeps its area, we can replace a quadrilateral by a triangle of exactly the same area, or change a triangle into a parallelogram or a rectangle with the same area. Surveyors do this to simplify the shape of a plot without changing its size. In the SEE, construction questions are worth 3 or 4 marks and are scored on accuracy, clear arcs, labelled letters, and a short statement of the steps.

What the SEE asks from this chapter

Expect one of: (i) construct a triangle equal in area to a given quadrilateral; (ii) construct a parallelogram equal in area to a triangle with a given angle; (iii) construct a rectangle or a square equal in area to a given figure; (iv) construct a triangle or quadrilateral from the given sides and angles first, then do (i)–(iii). Always write Steps of construction in order, leave all compass arcs visible, and add a line of justification ("Proof") explaining why the areas are equal.

photograph
Figure 1. Technical drawing pens. Every construction in this chapter is done with a compass and a straight edge, and the result is only as accurate as the instrument and the hand that holds it. Source: Lucasbosch / Wikimedia Commons (CC BY-SA 4.0)

14.1 Construction: concepts, formulas and worked problems

Equipment and conventions

Tool Used for
Sharp pencil (HB / 2H) thin, accurate lines and arcs
Ruler straight lines, measuring given lengths
Compass arcs of a given radius, copying a length
Protractor drawing a given angle (e.g. 60^\circ)
Set squares parallel lines and right angles (optional)

Conventions: label every point; draw given figures with thin full lines; draw construction lines (parallels, perpendiculars) in light lines; keep arcs visible; the required figure should be drawn darker. Write the given data at the top of the page.

Basic constructions you need

Four basic constructions
  1. Perpendicular bisector of AB: arcs of the same radius (more than \frac{1}{2}AB) from A and from B meet at two points; join them.
  2. Bisector of an angle: arc cutting both arms; two equal arcs from those points meet; join the vertex to the intersection.
  3. Perpendicular from a point to a line: arc from the point cutting the line twice; then bisect the segment between the cut points.
  4. Line through a point parallel to a given line — shown below, using a rhombus.

Parallel line through a point

figure
Figure 2. To draw the line through P parallel to l: arc from Q through P meets l at R; arcs of the same radius from P and R meet at S. Then PQRS is a rhombus, so PS ∥ QR.

Construct a line parallel to l through P

Steps.

  1. Draw a line l and mark a point P not on it.
  2. Mark any point Q on l. Join PQ.
  3. With centre Q and radius QP draw an arc cutting l at R.
  4. With centres P and R and the same radius QP draw two arcs that cross at S.
  5. Join P and S and extend both ways. PS\parallel l.

Proof. PQ=QR=RS=SP (all equal to the radius), so PQRS is a rhombus; opposite sides of a rhombus are parallel, hence PS\parallel QR, that is PS\parallel l.

A triangle equal in area to a quadrilateral

The idea: the diagonal AC splits quadrilateral ABCD into \Delta ABC and \Delta ACD. We replace \Delta ACD by a triangle \Delta ACE with the same base AC and the apex E on a line through D parallel to AC. By the theorem of Chapter 13, \Delta ACE=\Delta ACD.

figure
Figure 3. Quadrilateral ABCD is replaced by triangle ABE of the same area: DE ∥ AC meets BC produced at E, so ΔACD = ΔACE.

Triangle equal to a quadrilateral

Problem. Construct a quadrilateral ABCD with AB=5 cm, BC=4 cm, CD=4 cm, DA=3.5 cm and \angle B=70^\circ, and construct a triangle equal in area to it.

Steps of construction.

  1. Draw AB=5 cm. At B construct \angle ABX=70^\circ and cut BC=4 cm along BX.
  2. With centre A radius 3.5 cm and centre C radius 4 cm draw arcs meeting at D. Join AD and CD. ABCD is the quadrilateral.
  3. Join the diagonal AC.
  4. Through D draw a line parallel to AC (use the rhombus method above).
  5. Produce BC to meet that line at E.
  6. Join AE. \Delta ABE is the required triangle.

Proof. \Delta ACD and \Delta ACE are on the same base AC and between the same parallels AC and DE, so \text{area}(\Delta ACD)=\text{area}(\Delta ACE). Adding \text{area}(\Delta ABC) to both:

\text{area}(ABCD)=\text{area}(\Delta ABC)+\text{area}(\Delta ACD)

=\text{area}(\Delta ABC)+\text{area}(\Delta ACE)=\text{area}(\Delta ABE)

Numerical check. For the quadrilateral A(0,0), B(5,0), C(6,3.5), D(1.5,3) of the figure, the shoelace formula gives area 15.125\text{ cm}^2, and the triangle ABE with E=(6.73,6.05) has area \frac{1}{2}\times5\times6.05=15.125\text{ cm}^2. ✓

Common mistakes in constructions
  • Rubbing out arcs. The examiner needs them as evidence of the method.
  • A line through D that is only nearly parallel to AC — use the rhombus or set-square method carefully.
  • Forgetting to produce (extend) BC far enough to meet the parallel line.
  • Giving no steps or no proof: construction marks are usually split among figure, steps and justification.
  • Measuring the given angle with the wrong scale on the protractor (70^\circ vs 110^\circ).

A parallelogram equal in area to a triangle

The idea: \text{area}(\Delta)=\frac{1}{2}\times\text{base}\times\text{height} and \text{area}(\text{parallelogram})=\text{base}\times\text{height}. So a parallelogram with half the base and the same height has the same area as the triangle.

figure
Figure 4. Parallelogram MCZY equal to ΔABC: M is the midpoint of BC, MY makes the given angle 60° with MC, AY ∥ BC, and CZ ∥ MY.

Parallelogram equal to a triangle, with a given angle

Problem. Construct \Delta ABC with BC=6 cm, AB=3.8 cm and \angle B=57^\circ. Then construct a parallelogram equal in area to it having one angle of 60^\circ.

Steps of construction.

  1. Construct \Delta ABC from the given data.
  2. Bisect BC (perpendicular bisector): M is the midpoint of BC.
  3. At M construct \angle CMX=60^\circ (the given angle).
  4. Through A draw a line parallel to BC meeting MX at Y.
  5. Through C draw a line parallel to MY meeting AY produced at Z.
  6. MCZY is the required parallelogram.

Proof. MC\parallel YZ and MY\parallel CZ, so MCZY is a parallelogram with \angle YMC=60^\circ. It stands on base MC=\frac{1}{2}BC and has the same height as \Delta ABC (the distance between the parallels BC and AY). So

\text{area}(MCZY)=MC\times h=\frac{1}{2}BC\times h=\text{area}(\Delta ABC)

Numerical check (figure: BC=6, height h=3.2): triangle \frac{1}{2}\times6\times3.2=9.6; parallelogram 3\times3.2=9.6. ✓

Replacing a figure by one with the same area
Start with Replace by Key theorem
Triangle ACD Triangle ACE (apex slid along a parallel to the base) same base, same parallels
Quadrilateral Triangle (one diagonal used) as above, then add \Delta ABC
Triangle Parallelogram on half the base (any angle) triangle =\frac{1}{2} parallelogram
Triangle Rectangle on half the base (angle 90^\circ) special case of above
Pentagon Quadrilateral, then triangle apply the "slide the apex" step twice

A rectangle equal to a triangle

Rectangle equal to a triangle

Problem. Construct a rectangle equal in area to \Delta ABC (BC=6 cm, height 3.2 cm).

Steps. (1) Draw the triangle. (2) Bisect BC at M. (3) At M draw MX\perp BC. (4) Through A draw AY\parallel BC meeting MX at Y. (5) Through C draw CZ\perp BC meeting AY produced at Z. MCZY is a rectangle.

Proof. Its sides are MC=\frac{1}{2}BC=3 cm and MY=3.2 cm, so area =3\times3.2=9.6\text{ cm}^2=\frac{1}{2}\times6\times3.2=\text{area}(\Delta ABC).

A square equal to a rectangle

A square of side x equal to a rectangle a\times b needs x^2=ab, i.e. x=\sqrt{ab}: the geometric mean of the sides. The semicircle gives this length.

figure
Figure 5. Square equal to a rectangle 4 cm × 2.25 cm: on PQ = 4 + 2.25 draw a semicircle; the perpendicular at the junction O meets it at T; OT = √(4 × 2.25) = 3 cm.

Square equal to a rectangle

Problem. Construct a square equal in area to a rectangle of sides 4 cm and 2.25 cm.

Steps. (1) Draw PO=4 cm and produce it to Q so that OQ=2.25 cm. (2) Bisect PQ and draw a semicircle on PQ as diameter. (3) At O draw OT\perp PQ to meet the semicircle at T. (4) Construct a square on OT as a side.

Proof. The angle in a semicircle is a right angle: \angle PTQ=90^\circ. In the right triangle PTQ with altitude TO, the triangles POT and TOQ are similar, so \frac{PO}{OT}=\frac{OT}{OQ} and OT^2=PO\times OQ=4\times2.25=9. So OT=3 cm, and the square on OT has area 9\text{ cm}^2=4\times2.25. ✓

Constructing the triangle first

Many questions start with a triangle or quadrilateral described by lengths and angles. The standard cases are:

Given Method
SSS: three sides base; two arcs from its ends
SAS: two sides and the included angle base, angle at one end, cut off the side
ASA: two angles and the included side base; two angles at its ends
Quadrilateral: 4 sides and 1 diagonal two triangles on the diagonal
Quadrilateral: 4 sides and 1 angle triangle from SAS, then arcs
Triangle from SSS, then half its area

Problem. Construct \Delta PQR with PQ=7 cm, QR=5 cm and PR=6 cm. Find its area by Heron's formula and then construct a rectangle of the same area.

Solution. Area: s=\frac{7+5+6}{2}=9; \text{area}=\sqrt{9\times2\times4\times3}=\sqrt{216}=14.70\text{ cm}^2. Construct the triangle by drawing PQ=7 cm and arcs of radius 6 cm from P and 5 cm from Q. The rectangle has base \frac{1}{2}PQ=3.5 cm and height equal to the height of R above PQ: h=\frac{2\times14.70}{7}=4.20 cm. Check: 3.5\times4.20=14.7. ✓

14.2 Exercises of the SEE type

Method for any question

  1. Read the data and draw a rough sketch with all measurements.
  2. Construct the given figure accurately (SSS/SAS/ASA, etc.).
  3. Draw the diagonal or the median that the method needs, in a light line.
  4. Draw the parallel line through the vertex you are replacing; produce the side to meet it.
  5. Darken the required figure; label every vertex; keep the arcs.
  6. Write the steps and a two-line proof referring to "same base, same parallels".
SEE type: quadrilateral to triangle

Problem. Construct a quadrilateral ABCD with AB=4.5 cm, BC=3.5 cm, CD=4 cm, DA=3 cm and diagonal AC=5 cm. Construct a triangle equal in area to it.

Solution (outline). Triangle ABC from three sides (4.5, 3.5, 5). Triangle ACD on the other side of AC from sides AD=3 and CD=4 (arcs from A and C). Through D draw DE\parallel AC meeting BC produced at E. Join AE: \Delta ABE is the answer, because \Delta ACD=\Delta ACE.

Area check. \Delta ACD has sides 3,4,5, a right-angled triangle of area 6\text{ cm}^2. \Delta ABC: s=6.5, area =\sqrt{6.5\times2\times3\times1.5}=\sqrt{58.5}=7.65\text{ cm}^2. Quadrilateral area =13.65\text{ cm}^2, which equals \text{area}(\Delta ABE).

SEE type: parallelogram from a triangle

Problem. Construct \Delta ABC with AB=5 cm, BC=6 cm, CA=4.5 cm and a parallelogram equal in area to it with one angle of 45^\circ.

Solution (outline). \Delta ABC from SSS. Mark the midpoint M of BC. At M construct 45^\circ to MC. Through A draw AY\parallel BC meeting the new ray at Y. Through C draw CZ\parallel MY meeting AY produced at Z. MCZY is the parallelogram, with area MC\times h=\frac{1}{2}BC\times h=\text{area}(\Delta ABC).

Area check. s=7.75; area =\sqrt{7.75\times2.75\times1.75\times3.25}=\sqrt{121.2}=11.01\text{ cm}^2, height on BC is \frac{2\times11.01}{6}=3.67 cm; parallelogram 3\times3.67=11.01. ✓

Chapter summary

  • Constructions use only a ruler and compass (a protractor only for a given angle); keep all arcs.
  • Parallel through a point: rhombus method with equal radii.
  • Triangle = quadrilateral: join diagonal AC; through D draw DE\parallel AC meeting BC produced at E; \Delta ABE has the same area.
  • Parallelogram = triangle: bisect the base; draw the given angle at the midpoint; parallels through the apex and through the end of the base.
  • Rectangle = triangle: same, with 90^\circ. Square = rectangle: semicircle on a+b; the perpendicular at the junction is \sqrt{ab}.
  • The proofs rest on: triangles on the same base between the same parallels are equal in area; a triangle is half a parallelogram.
  • Steps + figure + proof earns the full marks.

Practice questions

Group A — Multiple choice (1 mark each)

  1. Which tool may be used to draw an arc in a construction? [1]arulerbcompassccalculatordset square only
  2. To change a quadrilateral into a triangle of equal area we draw a line through a vertex parallel to [1]aa sidebthe diagonal that does not pass through that vertexcthe perpendicular bisectordthe angle bisector
  3. A parallelogram equal in area to a triangle is constructed on [1]athe full basebhalf the basecdouble the basedthe hypotenuse
  4. The side of a square equal in area to a rectangle 9\times4 is [1]a5b6c36d13
  5. The angle in a semicircle is [1]a60^\circb45^\circc90^\circd180^\circ
Answers to Group A

1. (b).

2. (b) — parallel to the diagonal AC through D.

3. (b) — base halves, height unchanged.

4. (b) — \sqrt{9\times4}=6.

5. (c).

Group B — Short answer (2 marks each)

  1. Describe how to draw a line through a point P parallel to a line l using only compass and ruler. [2]
  2. A triangle has base 8 cm and height 5 cm. What base and height does a rectangle of equal area have if its base is half the triangle's base? [2]
  3. Why are triangles ACD and ACE equal in area in the quadrilateral construction? [2]
  4. Find the side of a square equal in area to a rectangle 12\text{ cm}\times3\text{ cm}. [2]
Answers to Group B

1. Mark Q on l; arc centre Q radius QP cuts l at R; arcs of the same radius from P and R meet at S; line PS is parallel to l (rhombus PQRS).

2. Base 4 cm, height 5 cm: 4\times5=20=\frac{1}{2}\times8\times5.

3. Same base AC and same parallels AC, DE, hence the same height.

4. \sqrt{36}=6 cm.

Group C — Short answer (3 marks each)

  1. Construct \Delta ABC with AB=5 cm, BC=6 cm, \angle B=60^\circ. Bisect BC at M. [3]
  2. Construct a parallelogram equal in area to the triangle of question 1, having an angle of 60^\circ at M. [3]
  3. Construct a rectangle of sides 5 cm and 3.2 cm; state the steps to construct a square of equal area. [3]
  4. Show by calculation that a triangle with base 10 cm and height 7.2 cm equals a parallelogram of base 5 cm and height 7.2 cm. [3]
Answers to Group C

1. Draw BC=6, angle 60^\circ at B, cut BA=5; the perpendicular bisector of BC gives M.

2. As in the worked example: MY at 60^\circ to MC, AY\parallel BC, CZ\parallel MY.

3. Extend 5 to 5+3.2=8.2; semicircle; perpendicular at the junction gives \sqrt{5\times3.2}=\sqrt{16}=4 cm; square of side 4 cm (area 16 = 5\times3.2).

4. Triangle: \frac{1}{2}\times10\times7.2=36; parallelogram: 5\times7.2=36.

Group D — Long answer (4 marks each)

  1. Construct quadrilateral ABCD with AB=5 cm, BC=4 cm, CD=4 cm, DA=3.5 cm and \angle B=70^\circ, and construct a triangle equal in area to it. Give steps and proof. [4]
  2. Construct \Delta ABC with BC=6 cm, \angle B=57^\circ, AB=3.8 cm. Construct a parallelogram equal in area with one angle of 60^\circ. [4]
  3. Construct a square equal in area to a rectangle 6\text{ cm}\times 2\text{ cm} and prove that your square has the right area. [4]
  4. Construct \Delta PQR with sides 7, 5, 6 cm and a rectangle equal to it, giving the rectangle's dimensions. [4]
Answers to Group D

1. Follow the worked example; area check 15.125\text{ cm}^2 for the coordinates used there.

2. Follow the worked example; area of the triangle is \frac{1}{2}\times6\times3.8\times\sin57^\circ\approx9.56\text{ cm}^2.

3. \sqrt{6\times2}=\sqrt{12}\approx3.46 cm: semicircle on PQ=8 with the perpendicular at the junction of 6 and 2; OT^2=12.

4. Area \sqrt{216}\approx14.70\text{ cm}^2; rectangle 3.5\text{ cm}\times4.2\text{ cm}.

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