Construction

Every construction in this unit rests on three facts from Unit 10, for figures standing on the same base and lying between the same parallels:

• two parallelograms are equal in area;
• a triangle is half of a parallelogram;
• two triangles are equal in area.

To see them again, draw two parallel grid lines 4 squares apart, a base of 6 squares on the lower one, and count the squares inside a parallelogram, a triangle and another triangle on that base.

The tools and the steps

You need a sharp pencil, a ruler marked in centimeters and millimeters, a compass, and a protractor only for checking (the angles 60∘°, 120∘°, 30∘°, 90∘°, 45∘° and 75∘° are all constructed with the compass, as in Grade 9).

Every construction goes in three steps:

1. Draw a rough sketch freehand with a ruler and pencil, as the finished figure should look.
2. Write the given measures on the sketch, so you can see which side to draw first and where the new figure goes.
3. Construct with compass and ruler, naming each point as it is made, and leave every arc showing: the arcs are your working, as in an algebra answer.

The one move used again and again is a line through a point parallel to a given line. To draw ST through A parallel to BC: copy the angle ∠ABC at A on the other side of AB, so that ∠BAS = ∠ABC. These are alternate angles, and equal alternate angles make the lines parallel.

Each construction below starts by drawing the given figure, then draws one line parallel to a base. Every point on that line is the same distance from the base, so any figure on that base with its top on the line has the same height. The second figure’s side or angle then picks out one point on the line.

A. Parallelograms equal in area

Why the areas are equal: ABCD and ABQP stand on the same base AB, and their tops DC and PQ are on the same line $ST \parallel AB$. By Theorem 1, parallelograms on the same base and between the same parallels are equal in area.

Measure the height (the distance between AB and ST): it is about 4.8 cm. Both parallelograms have the area 4 × 4.8 ≈ 19 cm², though one leans one way and the other the other way.

B. Triangles equal in area

Why the areas are equal: △ABC and △DBC stand on the same base BC, and A and D are on the same line $ST \parallel BC$. By Theorem 3 their areas are equal.

C. Triangles and parallelograms equal in area

A triangle is half the parallelogram on the same base. So to get a parallelogram equal to a triangle, give the parallelogram half the base; to get a triangle equal to a parallelogram, give the triangle twice the base.

Why the areas are equal: D is the midpoint of BC, so △ABD is half of △ABC (AD is a median). △ABD is also half of parallelogram PBDQ: the same base BD, between $BD \parallel ST$. So

$$\text{area of } PBDQ = 2 \times \text{area of } \triangle ABD = \text{area of } \triangle ABC.$$

Why the areas are equal: BE = 2BC. △PBC is half of ABCD (the same base BC, between $BC \parallel AD$), and △PCE has the same area as △PBC (equal bases BC = CE, the same corner P). So △PBE = 2 × △PBC = ABCD.

D. A triangle equal to a quadrilateral

The steps are in the player at the top of the unit. Here is the idea.

Why the areas are equal: △QRS and △QGS stand on the same base SQ, and R and G are on the line $RH \parallel SQ$, so they have the same area. Then

$$\begin{aligned} \text{area of } PQRS &= \triangle PQS + \triangle QRS \\ &= \triangle PQS + \triangle QGS \\ &= \triangle PSG. \end{aligned}$$

The quadrilateral’s corner R has slid along the parallel to G, onto the line PQ, and one corner has gone. This is Exercise 10.1, Question 4b, drawn.

Back to the plot

The four-sided plot becomes a triangle of the same area with one parallel line, as in Activity 5: draw a diagonal, then through the far corner a line parallel to it, to meet a side produced. Measure the triangle’s base and height on the drawing, and ½ × base × height, times the square of the scale, is the plot’s area.

For a plot with five sides, do it twice: each time one corner slides onto the line of its neighbor and the figure loses a side, until a triangle is left.

In the SEE’s Compulsory Mathematics paper, Geometry carries 13 marks in three questions (CDC’s specification grid). The construction question is worth 4 marks:

Part	Level	Marks	What it asks
a.	application	3	construct a figure from its measures, then a figure of equal area with a given side or angle
b.	higher ability	1	state the relation between the two areas, with a reason

The marks in a. go to the given figure correctly constructed, the parallel line, and the new figure; keep the arcs showing.

Each answer shows the finished construction (blue lines are compass arcs, the new figure shaded) and its steps. Draw a rough sketch with the measures first; it is left out of the steps.

Question 1: parallelograms

Question 2: triangles

Question 3: a parallelogram equal to a triangle

Question 4: a triangle equal to a parallelogram

Question 5: a triangle equal to a quadrilateral