Geometric Constructions Worked ExamplesIGCSE Maths 0580

(a) Measure $PQ$ on the drawing. The ruler starts at $0$ at $P$, and $Q$ is level with the fourth small mark after $7$.

Each centimetre on the drawing represents $2.5\text{ km}$, so multiply the measured length by $2.5$. Without a calculator, multiply by $10$ and divide by $4$.

(b) Now go the other way: a real distance becomes a drawn length by dividing by the scale.

Sense check: $3.6\text{ cm}$ is just under half of $7.4\text{ cm}$, and $9\text{ km}$ is just under half of $18.5\text{ km}$. ✓

(a) Convert both real distances to lengths on the drawing by dividing by $5$.

Open the compasses to $6\text{ cm}$ against a ruler. With the point on $R$, draw an arc above $RS$ where $T$ could be.

Reset the compasses to $5\text{ cm}$. With the point on $S$, draw an arc that crosses the first one. Full circles would cross twice, once on each side of $RS$; $T$ is north of $RS$, so it is the crossing above the line. Label it $T$ and leave both arcs showing.

(b) Draw the line $RT$. Place the protractor at $R$ and measure clockwise from the north line to $RT$.

Check by calculation: the cosine rule in triangle $RST$ (sides $8$, $6$ and $5\text{ cm}$) gives angle $SRT = 38.6^{\circ}$. $RS$ points due east, a bearing of $090^{\circ}$, so the bearing of $T$ is $90^{\circ} - 38.6^{\circ} = 51.4^{\circ}$. ✓

Convert every real length to a drawn length by dividing by $2$. $PQ$ is already drawn, and it should measure $8\text{ cm}$.

Match each length to the correct end of $PQ$. $PR$ starts at $P$, so the $6.5\text{ cm}$ arc is centred at $P$. $QR$ starts at $Q$, so the $5.5\text{ cm}$ arc is centred at $Q$.

Set the compasses to $6.5\text{ cm}$ and draw an arc from $P$ above the line. Set them to $5.5\text{ cm}$ and draw an arc from $Q$ that crosses it. The crossing point is $R$.

Join $PR$ and $QR$ with a ruler, and leave both arcs visible. Check with the ruler: $PR$ should measure $6.5\text{ cm}$ and $QR$ $5.5\text{ cm}$.

(a) A rhombus is two congruent isosceles triangles, $ABC$ and $ADC$, sharing the diagonal $AC$. Start by drawing $AC = 8\text{ cm}$ with a ruler.

Set the compasses to $5\text{ cm}$. From $A$, draw an arc above $AC$ and an arc below it. Keeping the same radius, do the same from $C$.

The arcs cross below $AC$ at $B$ and above $AC$ at $D$. Join $AB$, $BC$, $CD$ and $DA$, leaving all four arcs visible.

(b) Measure $BD$ with a ruler. Check: the diagonals of a rhombus bisect each other at right angles, so half of $BD$ is found with Pythagoras, using half of $AC$.

(c) The area of a rhombus is half the product of its diagonals (it is the two triangles $ABC$ and $ADC$, each with base $8$ and height $3$).

(a) Convert the real distance to a drawn length by dividing by the scale.

(b) Mark $P$ and draw a north line vertically upwards from it.

Measure $62^{\circ}$ clockwise from that north line using a protractor, and draw a faint ray.

Measure $9\text{ cm}$ along the ray from $P$ and mark $Q$.

(a) The prism has three rectangular faces, each $4\text{ cm}$ by $6\text{ cm}$ (one for each side of the triangle), and two triangular ends. Draw the three rectangles side by side with a ruler: a strip $12\text{ cm}$ wide and $6\text{ cm}$ tall, divided at $4\text{ cm}$ and $8\text{ cm}$.

An angle of $60^{\circ}$ is not measured with a protractor here; the triangles are constructed. Set the compasses to $4\text{ cm}$ and draw arcs from both ends of the top edge of the middle rectangle. They cross at the apex, so join it to both ends.

Repeat on the bottom edge of the same rectangle so the second triangle points downwards. Both triangles must be on the same rectangle, one on each side, or the net will not fold into a closed prism.

(b) Find the height of one equilateral triangle. It splits into two right-angled triangles with hypotenuse $4\text{ cm}$ and base $2\text{ cm}$.

Area of one triangle, keeping the unrounded height:

The surface area is the total area of the net: two triangles and three rectangles.

(a) The net has two identical triangles and three rectangles. The triangles are the ends and the rectangles join them, so the solid is a triangular prism.

(b) Each triangle has a $3\text{ cm}$ edge and a $4\text{ cm}$ edge along grid lines, meeting at a right angle. The sloping edge crosses the grid diagonally, so use Pythagoras.

When the net is folded, the sloping edge of each triangle joins the free edge of the right-hand rectangle, so the two lengths must be equal. That rectangle is $5$ squares wide. ✓

(c) The cross-section is the right-angled triangle, and the length of the prism is the $6\text{ cm}$ height of the rectangles.

(d) The surface area is the total area of the net: two triangles and three rectangles, each $6\text{ cm}$ long.

(a) Divide each real length by $20$ to get the drawn length.

$AB$ is drawn first with a ruler. With the compasses set to $7\text{ cm}$ she drew an arc centred at $A$, and with them set to $5\text{ cm}$ an arc centred at $B$. $C$ is where the arcs cross — the point $7\text{ cm}$ from $A$ and $5\text{ cm}$ from $B$. The arcs are left visible as evidence of the construction.

(b) A scale drawing keeps the angles of the real field, so use the cosine rule with the drawn lengths. Angle $B$ is opposite $AC$.

The angle whose cosine is $\tfrac{1}{2}$ is $60^{\circ}$.

(c) $B$ is due east of $A$, so from $B$ the direction of $A$ is due west, a bearing of $270^{\circ}$. $C$ is on the north side of $AB$, so turning from $BA$ to $BC$ is a clockwise turn of $60^{\circ}$.

(d) Use the real lengths $AB$ and $BC$ with the angle between them in $\tfrac{1}{2}ab\sin C$.

Check using the drawing: its area is $\tfrac{1}{2} \times 8 \times 5 \times \sin 60^{\circ} = 17.32\ldots\text{ cm}^2$. Each $1\text{ cm}^2$ on the drawing represents $20 \times 20 = 400\text{ m}^2$.

(a) Read from the grid: the base is $6\text{ cm}$ by $6\text{ cm}$, and each triangle has base $6\text{ cm}$ and height $5\text{ cm}$ (from the edge of the square to the tip).

(b) When the net is folded, the $5\text{ cm}$ height of each triangle becomes the slant height of the pyramid: it runs up the face from the midpoint of a base edge to the apex. The apex sits directly above the centre of the base, which is $3\text{ cm}$ from that midpoint.

The vertical height, the $3\text{ cm}$ and the slant height form a right-angled triangle with the slant height as hypotenuse.

(c) A pyramid is one third of base area times vertical height.

(d) Work backwards. The four triangles now have total area $156 - 36 = 120\text{ cm}^2$, so each is $30\text{ cm}^2$. Find the new slant height $l$.

Use the same right-angled triangle as in (b).