The scale drawing shows two boundaries, AB and BC, of a field ABCD.
The scale of the drawing is 1cm represents 8m.
(a) The boundaries CD and AD of the field are each 72m long.
(i) Work out the length of CD and AD on the scale drawing.
(ii) Using a ruler and compasses only, complete accurately the scale drawing of the field.
(b) A tree in the field is
• equidistant from A and B
and
• equidistant from AB and BC.
On the scale drawing, construct two lines to find the position of the tree.
Use a straight edge and compasses only and leave in your construction arcs
▶️ Answer/Explanation
(a)(i) Ans: 9 cm
Given scale: 1cm = 8m. For 72m, length on drawing = $\frac{72}{8} = 9$ cm.
(a)(ii) Using compasses, mark points D completing the rectangle ABCD with sides AB, BC, CD = 9cm, AD = 9cm.
(b) Construct perpendicular bisector of AB for equidistant condition. Then construct angle bisector of ∠ABC for the second condition. The tree’s position is where these two lines intersect.
In this question use a ruler and compasses only.
Show all your construction arcs.
The diagram shows a triangular field ABC.
The scale is 1 centimetre represents 50 metres.
(a) Construct the locus of points that are equidistant from A and B.
(b) Construct the locus of points that are equidistant from the lines AB and AC.
(c) The two loci intersect at the point E.
Construct the locus of points that are 250m from E.
(d) Shade any region inside the field ABC that is
• more than 250m from E
and
• closer to AC than to AB.
▶️ Answer/Explanation
(a) The locus of points equidistant from A and B is the perpendicular bisector of AB. Construct this by drawing arcs from A and B with the same radius, intersecting above and below AB, then joining these points.
(b) The locus of points equidistant from AB and AC is the angle bisector of ∠BAC. Construct this by drawing arcs from A intersecting AB and AC, then from these points draw intersecting arcs to find the bisector.
(c) The locus of points 250m from E is a circle with radius 5cm (since 250m ÷ 50 = 5cm). Draw this circle centered at E.
(d) The shaded region is the area inside triangle ABC, outside the circle from (c), and on the side of the angle bisector closer to AC. This satisfies both conditions.
