Hard GCSE Constructions and loci Questions

Deal with the first corner

Inside the field, the points within 10 m of corner $A$ form a quarter circle.

Deal with the second corner

The same happens at the other corner, giving a second quarter circle of equal size.

Check the two regions do not overlap

The two corners are 24 m apart, and $10 + 10 = 20$ is less than $24$, so the two quarter circles never meet and nothing is double-counted.

Add them

The total cut area is $50\pi$ m squared.

Check the decomposition a different way

Rebuilding the region from its pieces gives the same total, so nothing has been counted twice and nothing has been left out.

Keep the answer exact

Leaving $\pi$ in the answer keeps it exact. Rounding at this stage would throw away accuracy the question has not asked us to lose.

Check the size is sensible

Replacing $\pi$ by about $3.14$ turns $50\pi$ into a sensible everyday number, which is a quick way to spot a decomposition error.

Note the common error

Mixing the area formula with the circumference formula, or forgetting that only part of the circle is available, are the usual mistakes here.

Note the units

An area is measured in square metres and a length in metres; the units are part of the answer.

Restate the locus in words

The locus is the boundary curve; the region is everything the boundary encloses. The question asked about the region.

Relate the region back to a construction

On paper each curved edge would be drawn as a single compass arc, with the compass point at the post or corner.

Sanity check against a simpler region

Comparing with the full circle of the same radius shows the answer is the right sort of size.

Say why no measuring was needed

Every length used came from the question, so the answer does not depend on measuring any drawing.

Check the pieces one last time

Listing the pieces again and re-adding them reproduces the same exact total.

Conclude

The exact answer is $50\pi$.

Deal with the first condition

The points within $5$ m of the segment $PQ$ form a racetrack shape: a rectangle along the fence, capped by a half-circle of radius $5$ m at each end.

Note it is a segment, not a full line

Because the fence stops at $P$ and at $Q$, the region curves round each end. If the fence were an endless straight line, the region would just be a strip.

Deal with the second condition

The points the same distance from $P$ as from $Q$ lie on the perpendicular bisector of $PQ$.

Identify the correct side

Being nearer to $P$ means lying on the $P$ side of that bisector.

Combine the two conditions

The coin lies in the overlap: the half of the racetrack on the $P$ side of the perpendicular bisector.

Check the bisector cuts the racetrack in half

The racetrack is symmetrical about the perpendicular bisector of $PQ$, so the bisector splits it into two equal halves.

Reject the whole racetrack

The whole racetrack includes points nearer to $Q$, which the detector has ruled out.

Reject the Q side

That is the opposite half: those points are nearer to $Q$ than to $P$.

Reject the circle at P

The circle of radius $5$ m centred on $P$ describes points within $5$ m of the point $P$, not within $5$ m of the whole fence. The two regions are quite different.

Reject the fence-side option

Splitting the racetrack along the fence itself separates one side of the field from the other; it says nothing about being nearer to $P$ than to $Q$.

Work out the area of the answer as a check

The region is exactly half the racetrack, so its area is half of the rectangle plus half of the circle formed by the two end caps — a real, computable area.

Note which loci were used

Two standard loci are combined here, which is exactly what a combined-loci exam question asks for.

Note the boundary

The curved and straight edges $5$ m from the fence are included; the bisector itself is not, because points on it are the same distance from $P$ and $Q$.

Shade rather than outline

The answer is an area to be shaded, not a curve to be drawn.

Conclude

The coin lies in the half of the racetrack region on the $P$ side of the perpendicular bisector of $PQ$.

Set up the two triangles

Joining $R$ to $M$ splits triangle $RAB$ into two smaller triangles, $RMA$ and $RMB$.

Use the given equal sides

This is given: $R$ is equidistant from $A$ and $B$.

Use the midpoint

This is what "midpoint" means.

Use the common side

The side $RM$ is shared by both triangles.

Apply SSS congruence

Three pairs of equal sides give congruent triangles.

Read off the equal angles at M

Corresponding angles of congruent triangles are equal.

Use the straight line at M

The points $A$, $M$ and $B$ are in a straight line, so the two angles at $M$ are angles on a straight line.

Solve for the angle

Two equal angles adding to $180^{\circ}$ must each be $90^{\circ}$.

Conclude the perpendicularity

So $RM$ meets $AB$ at a right angle at the midpoint: $R$ is on the perpendicular bisector of $AB$.

Reject the equilateral argument

Equal distances to $A$ and $B$ make the triangle isosceles. It is equilateral only in the one special case $RA = RB = AB$.

Reject the half-of-AB argument

$RM$ can be any length at all — $R$ may be a millimetre above $AB$ or a mile above it. It has no fixed relationship to $AB$.

Reject the centre-of-a-circle argument

It is true that $R$ is the centre of a circle through $A$ and $B$, but such centres lie on the perpendicular bisector, not on the segment $AB$. The claim is false.

Reject the right-angle-at-R argument

Angle $ARB$ changes as $R$ moves up and down the bisector; it approaches $180^{\circ}$ near $AB$ and shrinks towards $0^{\circ}$ far away. It is not always $90^{\circ}$.

State the converse too

The argument also runs backwards: any point on the perpendicular bisector is equidistant from $A$ and $B$, again by congruent triangles. So the locus is exactly the bisector.

Conclude

The SSS congruence argument is the correct one.

Start with the longest side

The base is drawn first with a ruler, because it is the only side whose two endpoints are both known at the start.

Check the triangle exists

The two shorter sides add to more than the base, so the arcs will cross.

Set the compasses to AC

The vertex $C$ must be $5$ cm from $A$, so the arc from $A$ has radius $5$ cm.

Set the compasses to BC

The vertex $C$ must also be $6$ cm from $B$, so the arc from $B$ has radius $6$ cm.

Mark the crossing point

Where the two arcs cross is the only point that is $5$ cm from $A$ and $6$ cm from $B$ (on that side of $AB$). That is $C$.

Join the sides

Finally join $C$ to $A$ and to $B$ with a ruler.

Reject the swapped-radii order

Drawing the $6$ cm arc from $A$ and the $5$ cm arc from $B$ produces a triangle in which $AC = 6$ cm and $BC = 5$ cm. The side lengths are swapped, so it is not the triangle asked for.

Reject the equal-arcs order

Using $5$ cm from both ends gives an isosceles triangle with $BC = 5$ cm, not $6$ cm.

Reject the straight-up method

Measuring $5$ cm vertically from $A$ and $6$ cm vertically from $B$ gives two different points, and neither is the vertex; joining them does not build the triangle at all.

Reject the trial-and-error method

Drawing the $5$ cm and $6$ cm sides from $C$ at a guessed angle leaves the third side to chance; it will only come out at $8$ cm by luck, so it is not a construction.

Note why arcs and not measurements

An arc marks every point at the right distance from a vertex at once, so the crossing point satisfies both conditions exactly.

Note there are two crossing points

The arcs cross above and below the base, giving two mirror-image triangles. Either is acceptable; they are congruent.

Check the construction against the data

Reading the finished drawing back, all three side lengths are as required.

Note the leftover arcs

Exam papers award marks for the visible arcs, so they must not be rubbed out.

Conclude

Base first, then the $5$ cm arc from $A$, then the $6$ cm arc from $B$, then join.

Deal with the first condition

Points less than $4$ m from $A$ lie inside the circle of radius $4$ m centred on $A$.

Cut that circle down to the garden

The corner $A$ is a right angle of the rectangle, so only a quarter of that circle lies inside the garden.

Deal with the second condition

The points equidistant from the two sides $AB$ and $AD$ form the bisector of the angle between them — the angle at $A$.

Identify the correct side of the bisector

Being nearer to $AB$ than to $AD$ means being on the $AB$ side of that bisector.

Combine the two conditions

The tree must be inside the quarter circle and on the $AB$ side of the bisector, so the region is the part of the quarter circle between $AB$ and the bisector.

Work out what fraction of a circle that is

The bisector splits the right angle into two equal $45^{\circ}$ pieces, so the region is half of the quarter circle, that is one eighth of a full circle.

Find its area as a check

The region has exact area $2\pi$ square metres, which confirms the description is a genuine, measurable region.

Check the boundary

The circle itself is excluded because the distance must be less than $4$ m.

Reject the whole quarter circle

The whole quarter circle contains points nearer to $AD$ than to $AB$, so it is too big.

Reject the AD side

The points between $AD$ and the bisector are nearer to $AD$, which is the opposite of what was asked.

Reject the perpendicular bisector option

The perpendicular bisector of $AB$ is the locus of points equidistant from the two points $A$ and $B$, not from the two sides $AB$ and $AD$. It is a different locus entirely, and it is $5$ m away from $A$, outside the circle.

Reject the two-circles option

Points within $4$ m of $A$ and within $4$ m of $B$ would need the two circles to overlap, but $A$ and $B$ are $10$ m apart while the radii only total $8$ m. That region is empty.

Note which two constructions were used

This is a typical combined-loci question: one circular locus and one angle-bisector locus, intersected.

Note the region is not the locus

The answer is a region (an area), bounded by two straight edges and an arc.

Conclude

The possible positions form one eighth of the circle of radius $4$ m centred on $A$, between $AB$ and the angle bisector at $A$.