GCSE Constructions and loci Practice Questions

Recall what a locus is

A locus is the set of every point that follows a given rule — here, every point at a fixed distance of $4\,\mathrm{cm}$ from $P$.

Picture the compass point at $P$

Putting the compass point on $P$ and opening it to $4\,\mathrm{cm}$, one full sweep marks every point that is $4\,\mathrm{cm}$ away. That sweep is a circle of radius $4\,\mathrm{cm}$.

Rule out the other descriptions

A radius of $8\,\mathrm{cm}$ would be points $8\,\mathrm{cm}$ away; the region inside the circle contains points closer than $4\,\mathrm{cm}$; and no straight line or single point keeps the distance fixed in every direction.

Name the locus

Again the locus of points equidistant from $A$ and $B$ is the perpendicular bisector of $AB$.

Find the midpoint

The midpoint of $AB$ is $(3, 6)$.

Use the fact that $AB$ is vertical

A line at right angles to a vertical segment is horizontal, so the locus is $y = 6$.

Write the distance from the post

By Pythagoras, the distance from the origin to the point $(x, y)$ is $\sqrt{x^2 + y^2}$.

Write the rope condition

The goat can reach a point if it is no further than the rope length, so the distance is at most $5\,\mathrm{m}$.

Square both sides

Squaring is safe because both sides are positive, and it removes the square root.

Check the boundary is included

A taut rope lets the goat stand exactly $5\,\mathrm{m}$ away, so the circle itself is part of the region and the inequality is not strict.

Reject the other options

An equation would give only the boundary circle; the reversed inequality gives everything the goat cannot reach; $x + y \le 5$ is a straight-line region, not a circular one; and $x^2 + y^2 \le 5$ has radius $\sqrt{5}$, not $5$.

State the answer

The reachable region is the disc $x^2 + y^2 \le 25$.

Say what point is being asked for

A point the same distance from all three must lie on the perpendicular bisector of every pair. Two bisectors are enough to fix it; the third then passes through it automatically.

Construct the perpendicular bisector of $AB$

$AB$ is horizontal, so its perpendicular bisector is the vertical line through the midpoint of $AB$: $x = 5$.

Construct the perpendicular bisector of $BC$

$BC$ is vertical, so its perpendicular bisector is the horizontal line through the midpoint of $BC$: $y = 4$.

Solve the two equations together

The two bisectors cross at $(5, 4)$, which is the circumcentre of the triangle.

Check the distance to A

So $OA = 5$.

Check the distance to $B$

So $OB = 5$, the same as $OA$.

Check the distance to $C$

So $OC = 5$ too: the point really is equidistant from all three.

State the radius

All three vertices are $5$ units from the centre.

Note why only two bisectors are needed

Once a point is equidistant from $A$ and $B$, and also equidistant from $B$ and $C$, it is automatically equidistant from $A$ and $C$.

Conclude

The point equidistant from all three is $(5, 4)$.

Deal with the first corner

Inside the field, the points within $10\,\mathrm{m}$ 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 $24\,\mathrm{m}$ 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 $50\pi$.