THE GEOMETRY OF POLYGONS

QUESTION 1 4 marks Criterion A

Medium

Answer each of the following, using known geometry facts:

a. Two angles on a straight line are $x$ and $115°$. Find $x$.

[1]

b. Two vertically opposite angles are $x$ and $73°$. Find $x$.

[1]

c. Angles around a point are $90°$, $x$, $x$, and $110°$. Find $x$.

[2]

Show complete worked solution

(a)

$$x=180-115=65°$$

(b)

$$x=73° \text{ (vertically opposite angles are equal)}$$

(c)

$$90+x+x+110=360 \Rightarrow 2x=160 \Rightarrow x=80°$$

QUESTION 2 3 marks Criterion A

Medium

Two parallel lines are cut by a transversal. One angle formed is $58°$.

a. Find the co-interior (allied) angle, which is supplementary to it.

[1]

b. Find the alternate angle, which is equal to it.

[1]

c. Find the corresponding angle, which is also equal to it.

[1]

Show complete worked solution

(a)

$$180-58=122°$$

(b)

$$58°$$

(c)

$$58°$$

QUESTION 3 4 marks Criterion B

Medium

Investigate the relationship between co-interior angles and alternate angles, when parallel lines are cut by a transversal.

a. If one angle is $70°$, find its co-interior angle and its alternate angle.

[2]

b. Explain, using the fact that co-interior angles are supplementary (sum to 180°) and alternate angles are equal, why the co-interior angle and the alternate angle must ALSO be supplementary to each other.

[2]

Show complete worked solution

(a)

Co-interior: $180-70=110°$. Alternate: $70°$ (equal).

(b)

Since the alternate angle equals the original angle (70°), and the co-interior angle is $180°$ minus the original angle (110°), the co-interior and alternate angles must sum to $70+110=180°$ — they are supplementary, following logically from the two facts combined.

QUESTION 4 5 marks Criterion C

Medium

A student says vertically opposite angles and corresponding angles are 'the same thing, just with different names'.

a. Explain the difference between these two angle relationships, referring to WHERE each type of angle pair is formed (at a single intersection vs. across two parallel lines).

[3]

b. Give an example showing vertically opposite angles can exist even with NO parallel lines involved.

[2]

Show complete worked solution

(a)

Vertically opposite angles occur at a SINGLE intersection of two lines (opposite each other across the crossing point) — they don't require any parallel lines at all. Corresponding angles occur specifically when a transversal crosses TWO PARALLEL lines, comparing an angle at one intersection to the angle in the 'matching' position at the other intersection.

(b)

Any two straight lines crossing at a single point create two pairs of vertically opposite angles, regardless of whether any other lines are parallel — e.g. simply drawing an X shape creates vertically opposite angles with no parallel lines needed at all.

QUESTION 5 4 marks Criterion D

Medium

A road crosses two parallel railway tracks. The angle between the road and the first track is measured as $63°$.

a. Assuming the road acts as a transversal cutting the two parallel tracks, find the corresponding angle at the second track.

[2]

b. A surveyor measures the angle at the second track as $65°$ instead of the expected $63°$. Explain what this discrepancy might suggest about the two railway tracks.

[2]

Show complete worked solution

(a)

$$63° \text{ (corresponding angles are equal)}$$

(b)

If the corresponding angle isn't exactly equal, this suggests the two tracks are NOT perfectly parallel — a genuine surveying result like this could indicate the tracks have a very slight, real-world deviation from true parallel alignment.

QUESTION 6 5 marks Criterion A

Medium

Two parallel lines are cut by a transversal, creating an angle of $(3x+15)°$ and its co-interior (allied) angle of $(2x+25)°$.

a. Write an equation using the fact that co-interior angles are supplementary (sum to 180°), and solve for $x$.

[3]

b. Find the size of both angles.

[2]

Show complete worked solution

(a)

$(3x+15)+(2x+25)=180 \Rightarrow 5x+40=180 \Rightarrow 5x=140 \Rightarrow x=28$.

(b)

First angle: $3(28)+15=99°$. Second angle: $2(28)+25=81°$. (Check: $99+81=180$.)

QUESTION 7 5 marks Criterion A

Medium

At a point, four angles meet: $x°$, $2x°$, $75°$, and $(x+15)°$.

a. Write an equation using the fact that angles at a point sum to 360°, and solve for $x$.

[3]

b. Find all four angle values, and verify they sum to 360°.

[2]

Show complete worked solution

(a)

$x+2x+75+(x+15)=360 \Rightarrow 4x+90=360 \Rightarrow 4x=270 \Rightarrow x=67.5$.

(b)

$67.5°$, $135°$, $75°$, $82.5°$. Sum: $67.5+135+75+82.5=360°$.

QUESTION 8 5 marks Criterion B

Medium

Investigate the relationship between alternate angles and co-interior angles at TWO different transversal crossings of the same pair of parallel lines.

a. A transversal crosses two parallel lines, creating an angle of 62° at the first crossing. State the alternate angle (at the second crossing) and the co-interior angle (also at the second crossing).

[2]

b. A SECOND, different transversal crosses the SAME two parallel lines at a different angle, creating a 40° angle at ITS first crossing. Without needing more information, explain whether the 62° and 40° transversal-angle relationships are INDEPENDENT of each other (i.e. does knowing about one transversal tell you anything about the other transversal's angles)?

[3]

Show complete worked solution

(a)

Alternate angle: $62°$ (equal). Co-interior angle: $180-62=118°$ (supplementary).

(b)

The two transversals are entirely INDEPENDENT — each transversal creates its OWN set of related angles (alternate, corresponding, co-interior) based on ITS OWN angle of crossing, using the SAME parallel-line angle rules. Knowing the 62° transversal's angles gives no direct information about the 40° transversal's angles, since they are separate, unrelated intersecting lines, even though both interact with the same pair of parallel lines and follow the same general angle rules.

QUESTION 9 5 marks Criterion B

Medium

Investigate whether the SUM of an angle and its vertically opposite angle is always a FIXED value, or varies.

a. If one angle is 50°, find its vertically opposite angle, and their sum.

[2]

b. Test with a DIFFERENT angle, say 73°, finding its vertically opposite angle and sum. Does the SUM stay the same value (100°) as before, or does it change? Explain why, referencing the definition of vertically opposite angles.

[3]

Show complete worked solution

(a)

Vertically opposite: $50°$ (equal). Sum: $50+50=100°$.

(b)

Vertically opposite: $73°$. Sum: $73+73=146°$ — DIFFERENT from the earlier 100°. Since vertically opposite angles are always EQUAL to each other (not a fixed value like 100°), their sum is always DOUBLE the original angle — which changes depending on what the original angle actually is; there's no single fixed 'sum' value for all vertically-opposite-angle pairs.

QUESTION 10 3 marks Criterion C

Medium

A student says: 'alternate angles and co-interior angles are basically opposites of each other, since one is EQUAL and the other adds to 180°.'

a. Explain why describing them as 'opposites' is a potentially confusing way to think about the relationship, and instead explain the ACTUAL connection between them (both apply to the SAME general position relative to the transversal, just measured slightly differently).

[3]

Show complete worked solution

(a)

Calling them 'opposites' is misleading because they're not fundamentally contrasting concepts — alternate angles and co-interior angles are actually closely CONNECTED: for any pair of parallel lines cut by a transversal, the alternate angle and co-interior angle on the SAME side are always SUPPLEMENTARY to each other (summing to 180°), which follows directly from the fact that alternate angles are equal to the original angle, while co-interior angles are supplementary to it — they're two related consequences of the same parallel-line geometry, not opposing ideas.

QUESTION 11 3 marks Criterion C

Medium

A surveyor measures the angle between a road and a railway line crossing it as 58°, and needs to communicate to a colleague the angle on the OTHER side of the railway line (co-interior, at a different point) without drawing a diagram, using only words and angle facts.

a. Write a clear, precise written explanation (a few sentences) that would allow a colleague to correctly determine the co-interior angle WITHOUT seeing a diagram, referencing the specific angle rule being used.

[3]

Show complete worked solution

(a)

'The road acts as a transversal crossing two parallel railway tracks. Since co-interior (allied) angles between parallel lines cut by a transversal always sum to 180°, the co-interior angle on the other side is $180°-58°=122°$. This follows from the property that interior angles on the same side of a transversal are supplementary when the two lines being crossed are parallel.'

QUESTION 12 6 marks Criterion D

Hard

A roof truss design requires two support beams to meet the horizontal base at equal angles on either side (like an isosceles arrangement), with the APEX angle where the two beams meet measuring 100°.

a. Using the angle sum of a triangle, find the size of EACH base angle where a beam meets the horizontal base.

[2]

b. Building regulations require the base angle to be AT LEAST 35° for structural stability. Find the MAXIMUM apex angle allowed while still satisfying this regulation, and explain the trade-off between apex angle and base angle in this design.

[4]

Show complete worked solution

(a)

$$\frac{180-100}{2}=40° \text{ each}$$

(b)

If base angle $=35°$ (the minimum allowed), apex angle $=180-2(35)=110°$. So the MAXIMUM apex angle while still meeting the 35° base-angle minimum is $110°$ — note this means a LARGER apex angle actually corresponds to the smallest allowed base angle, since apex and base angles trade off inversely: increasing the apex angle decreases each base angle (and vice versa), so the architect must balance a taller-looking roof (larger apex) against maintaining sufficient base-angle support.

QUESTION 13 3 marks Criterion D

Medium

A ladder leans against a wall, making a 72° angle with the ground. A parallel second ladder (same length, leaning the same way) is placed elsewhere, also making some angle with the ground.

a. If the two ladders are truly PARALLEL to each other, and a third rope acts as a 'transversal' connecting corresponding points on both ladders, explain what you can conclude about the angle the SECOND ladder makes with the ground, using angle facts for parallel lines.

[3]

Show complete worked solution

(a)

If the two ladders are parallel, and BOTH meet the (also parallel, since both ladders meet the same flat ground) horizontal ground, then by corresponding angle rules, the second ladder MUST also make exactly a 72° angle with the ground — parallel lines crossing another line (or a common direction like 'the ground') always create equal corresponding angles.

QUESTION 1 2 marks Criterion A

Medium

The diagram shows a triangle with two known angles.

a. Find the value of $x$, using the angle sum of a triangle.

[2]

Show complete worked solution

(a)

$$x=180-55-70=55°$$

QUESTION 2 3 marks Criterion A

Medium

The diagram shows a right-angled triangle.

a. Find the value of $x$.

[2]

b. State the name given to the third angle in ANY triangle that contains a right angle (i.e. what type of angle must the OTHER two angles be, individually)?

[1]

Show complete worked solution

(a)

$$x=180-90-35=55°$$

(b)

The other two angles must each be acute (less than 90°), since they must sum to exactly 90° together, and neither can be 0° or more than 90°.

QUESTION 3 4 marks Criterion B

Medium

Investigate the angle sum of a triangle by tearing (conceptually) the three corners off a triangle and placing them together.

a. If you tear off the three corners of ANY triangle and place them together so their vertices meet at a point, they form a straight line (180°). Using two different specific triangles — one with angles 60°,60°,60° and one with angles 40°,50°,90° — verify that both sets of angles sum to 180°.

[2]

b. Explain why this 'tear and rearrange' demonstration provides good evidence (though not a full formal proof) that the angle sum of ANY triangle is 180°.

[2]

Show complete worked solution

(a)

$60+60+60=180°$. $40+50+90=180°$. Both confirm the angle sum rule.

(b)

Since the torn corners always seem to form a straight line (180°) regardless of the triangle's shape, this suggests the property holds universally, not just for special triangles — though a fully rigorous mathematical proof (e.g. using parallel line angle facts) is needed to confirm it holds for absolutely every possible triangle.

QUESTION 4 4 marks Criterion C

Medium

A student says: 'A triangle can have two obtuse angles, like 100° and 95°, as long as the third angle makes it add up to 180°.'

a. Test the student's claim: if two angles are 100° and 95°, what would the third angle need to be?

[2]

b. Explain why a triangle can never have two obtuse angles (angles greater than 90°), using the angle sum rule.

[2]

Show complete worked solution

(a)

Third angle $=180-100-95=-15°$ — a NEGATIVE angle, which is impossible for a real triangle.

(b)

If two angles were both obtuse (each greater than 90°), their sum alone would already exceed 180° — leaving no room (or even a negative amount) for the third angle. Since all three angles must be positive and sum to exactly 180°, at most ONE angle in any triangle can be obtuse.

QUESTION 5 5 marks Criterion D

Medium

A surveyor measures two angles of a triangular plot of land: one angle is 48°, and by using an instrument at a second corner, determines that angle equals exactly twice the THIRD (unmeasured) angle.

a. Let the third angle be $y$. Write an equation using the angle sum of a triangle, and solve for $y$.

[3]

b. State the size of all three angles, and classify the triangle (acute, right, or obtuse) based on its largest angle.

[2]

Show complete worked solution

(a)

$48 + 2y + y = 180 \Rightarrow 3y=132 \Rightarrow y=44°$. So the second angle is $2(44)=88°$.

(b)

Angles: $48°, 88°, 44°$. Since the largest angle (88°) is less than 90°, this is an ACUTE triangle.

QUESTION 6 2 marks Criterion A

Easy

The triangle shown has two known angles.

a. Find $x$.

[2]

Show complete worked solution

(a)

$$x=180-65-48=67°$$

QUESTION 7 3 marks Criterion A

Medium

A right-angled triangle has one other known angle.

a. Find $x$, and state whether the triangle is acute, right, or obtuse (already knowing it contains a right angle, classify by the LARGEST angle).

[3]

Show complete worked solution

(a)

$x=180-90-27=63°$. Since the triangle contains a 90° angle (and no angle exceeds 90°), it is classified as a RIGHT triangle.

QUESTION 8 5 marks Criterion B

Medium

Investigate the relationship between the EXTERIOR angle of a triangle and the two INTERIOR angles NOT adjacent to it (the 'exterior angle theorem').

a. A triangle has interior angles 55°, 70°, and 55°. Find the exterior angle formed by extending the side adjacent to the two 55° angles (i.e. the exterior angle at the 70° vertex... more precisely: the exterior angle SUPPLEMENTARY to the 70° angle).

[2]

b. Compare this exterior angle (110°) to the SUM of the two interior angles NOT adjacent to it (the two 55° angles). What do you notice, and does this match the general 'exterior angle theorem' (exterior angle = sum of the two non-adjacent interior angles)?

[3]

Show complete worked solution

(a)

Exterior angle $=180-70=110°$.

(b)

$55+55=110°$ — EXACTLY matches the exterior angle found. This confirms the exterior angle theorem: the exterior angle of a triangle always equals the sum of the two interior angles that are NOT adjacent to it.

QUESTION 9 5 marks Criterion B

Hard

Investigate WHY the exterior angle theorem (exterior angle = sum of the two non-adjacent interior angles) must always be true, using the angle sum of a triangle (180°) and the fact that an exterior angle and its adjacent interior angle are supplementary (sum to 180°).

a. Let a triangle have interior angles $A$, $B$, $C$. Write an equation using the fact that $A+B+C=180°$.

[1]

b. The exterior angle at vertex $C$ (call it $E$) is supplementary to the interior angle $C$, i.e. $E+C=180°$. Using BOTH equations, prove algebraically that $E=A+B$ (the exterior angle theorem).

[4]

Show complete worked solution

(a)

$$A+B+C=180°$$

(b)

From $A+B+C=180°$: $C=180-A-B$. Substituting into $E+C=180°$: $E+(180-A-B)=180 \Rightarrow E=A+B$. This proves the exterior angle theorem holds for ANY triangle, not just specific tested examples — it follows directly and necessarily from the two fundamental facts (angle sum of a triangle, and supplementary angles on a straight line).

QUESTION 10 3 marks Criterion C

Medium

A student says a triangle's exterior angle 'must always be bigger than both of the two non-adjacent interior angles individually', after checking one example where this happened to be true.

a. Using the exterior angle theorem ($E=A+B$, where $A$ and $B$ are both positive angles), explain WHY this claim is actually a GUARANTEED mathematical fact (not just something that happened to be true in one example), for ANY triangle.

[3]

Show complete worked solution

(a)

Since $E=A+B$, and BOTH $A$ and $B$ must be positive angles (every angle in a triangle is greater than 0°), adding a positive amount to either one INDIVIDUALLY must always make $E$ LARGER than that angle alone — i.e. $E=A+B>A$ (since $B>0$) and similarly $E=A+B>B$ (since $A>0$). This isn't a coincidence limited to specific examples; it's a guaranteed consequence of the exterior angle theorem itself, true for every possible triangle.

QUESTION 11 4 marks Criterion C

Medium

A student calculates the third angle of a triangle with known angles 47° and 68° by writing '47+68=115, so the third angle is also 115°' (forgetting to subtract from 180°).

a. Explain the error, and give the correct third angle.

[2]

b. Explain why the student's answer (115°) being LARGER than 90° should have raised suspicion, given the triangle already has two angles that sum to 115° themselves — what would be wrong about a triangle having angles 47°, 68°, AND 115°?

[2]

Show complete worked solution

(a)

The student found the SUM of the two given angles but incorrectly used this sum AS the third angle, instead of subtracting it from 180° (the full angle sum of a triangle). Correct: $180-47-68=65°$.

(b)

If the third angle were genuinely 115°, the total sum would be $47+68+115=230°$, far exceeding the required 180° angle sum for ANY triangle — recognizing that all three angles must sum to EXACTLY 180° (not more) is a quick way to catch this kind of error before even doing the correct calculation.

QUESTION 12 6 marks Criterion D

Hard

A surveyor measuring a triangular plot of land finds one angle is 38°, and determines that a second angle is exactly 15° more than THREE TIMES the third (unmeasured) angle.

a. Let the third angle be $y$. Write an equation using the triangle's angle sum, and solve for $y$.

[3]

b. Find all three angles, and classify the triangle (acute/right/obtuse) based on its largest angle.

[3]

Show complete worked solution

(a)

$38+(3y+15)+y=180 \Rightarrow 4y+53=180 \Rightarrow 4y=127 \Rightarrow y=31.75°$.

(b)

Angles: $38°$, $3(31.75)+15=110.25°$, $31.75°$. Since the largest angle (110.25°) exceeds 90°, this is an OBTUSE triangle.

QUESTION 13 6 marks Criterion D

Hard

A ramp for wheelchair access forms a right-angled triangle with the ground, where the angle of incline must be between 4.5° and 8.5° by accessibility regulations.

a. If the ramp's incline angle is 6°, find the OTHER non-right angle in the triangle formed by the ramp, the ground, and the vertical support.

[2]

b. The ramp needs to rise 0.9m over its length. Using the incline angle of 6°, estimate the REQUIRED HORIZONTAL LENGTH of the ramp using the tangent ratio ($\tan(6°)=\frac{\text{rise}}{\text{run}}$), correct to 1 decimal place, and comment on whether a longer or shorter ramp results from choosing the STEEPEST allowed angle (8.5°) instead of 6°.

[4]

Show complete worked solution

(a)

$$180-90-6=84°$$

(b)

$\tan(6°)=\frac{0.9}{\text{run}} \Rightarrow \text{run}=\frac{0.9}{\tan(6°)}\approx8.6$m. A STEEPER angle (8.5° instead of 6°) would require a SHORTER horizontal run to achieve the same 0.9m rise, since a steeper incline gains height more quickly over a shorter horizontal distance — so choosing 8.5° would result in a shorter (but steeper, less gentle) ramp compared to using 6°.

QUESTION 1 3 marks Criterion A

Medium

The isosceles triangle shown has two equal sides (marked) and an apex angle of 48°.

a. Explain why the two base angles must be EQUAL to each other.

[1]

b. Find the size of each base angle.

[2]

Show complete worked solution

(a)

The base angles are opposite the two equal (marked) sides, and angles opposite equal sides in a triangle are always equal.

(b)

$$\frac{180-48}{2}=66° \text{ each}$$

QUESTION 2 2 marks Criterion A

Medium

The isosceles triangle shown has two equal base angles of 72° each.

a. Find the apex angle $x$.

[2]

Show complete worked solution

(a)

$$x=180-72-72=36°$$

QUESTION 3 6 marks Criterion B

Medium

Investigate the relationship between the apex angle and base angles of an isosceles triangle as the apex angle changes.

a. If the apex angle is 20°, 60°, and 100°, find the base angles in each case.

[3]

b. Describe the pattern: as the apex angle increases, what happens to the base angles?

[1]

c. In the apex-60° case, all three angles turned out equal (60°, 60°, 60°). What special type of triangle is this, and is this a coincidence?

[2]

Show complete worked solution

(a)

Apex 20°: base angles $=\frac{180-20}{2}=80°$ each. Apex 60°: base angles $=\frac{180-60}{2}=60°$ each. Apex 100°: base angles $=\frac{180-100}{2}=40°$ each.

(b)

As the apex angle increases, the base angles decrease (and vice versa) — they change in opposite directions.

(c)

This is an EQUILATERAL triangle. It's not a coincidence — an equilateral triangle is actually a special case of an isosceles triangle where ALL sides (not just two) are equal, which happens precisely when the apex angle is 60°, making all angles equal too.

QUESTION 4 4 marks Criterion C

Medium

A student calculates the base angles of an isosceles triangle with apex angle 50° by writing '$180-50=130°$ for EACH base angle' (forgetting to divide by 2).

a. Explain the error, and give the correct base angle.

[2]

b. Verify the correct answer by checking all three angles sum to 180°.

[2]

Show complete worked solution

(a)

The student found the combined total of BOTH base angles together (130°) but forgot to divide by 2 to find EACH individual base angle (since they are equal, sharing the 130° between them). Correct: $130\div2=65°$ each.

(b)

$50+65+65=180°$ (while the student's version, $50+130+130=310°$, is clearly wrong since it exceeds 180°).

QUESTION 5 5 marks Criterion D

Medium

A tent is shaped like an isosceles triangle when viewed from the front, with the two equal sides being the sloped fabric panels, and a base angle of 65° where each panel meets the ground.

a. Find the apex angle at the top of the tent.

[2]

b. The tent manufacturer wants to REDUCE the apex angle to make the tent taller and narrower, while keeping it isosceles. If the new apex angle is 34°, find the new base angles, and state whether the tent becomes 'pointier' or 'flatter' at the top compared to before.

[3]

Show complete worked solution

(a)

$$180-65-65=50°$$

(b)

New base angles $=\frac{180-34}{2}=73°$ each. Since the apex angle decreased (50° to 34°), the tent becomes POINTIER (narrower and taller) at the top.

QUESTION 6 2 marks Criterion A

Easy

An isosceles triangle has apex angle 52°.

a. Find each base angle.

[2]

Show complete worked solution

(a)

$$\frac{180-52}{2}=64° \text{ each}$$

QUESTION 7 2 marks Criterion A

Easy

An isosceles triangle has base angles of 63° each.

a. Find the apex angle.

[2]

Show complete worked solution

(a)

$$180-2(63)=54°$$

QUESTION 8 5 marks Criterion B

Hard

Investigate whether an isosceles triangle can EVER also be a right triangle, and if so, what its angles must be.

a. If an isosceles triangle has a right angle (90°) as its APEX angle, find the two equal base angles.

[2]

b. Now investigate whether the right angle could instead be one of the BASE angles (which must be EQUAL to each other). If one base angle is 90°, what would the OTHER base angle and apex angle need to be, and is this actually possible for a valid triangle?

[3]

Show complete worked solution

(a)

$$\frac{180-90}{2}=45° \text{ each}$$

(b)

If one base angle is 90°, the OTHER base angle must ALSO be 90° (since base angles are equal) — but two 90° angles already sum to 180°, leaving 0° for the apex angle, which is impossible (an angle can't be 0° in a real triangle). So a right angle CANNOT be a base angle of an isosceles triangle; it can only be the apex angle, giving the unique 45-45-90 triangle found in part (a).

QUESTION 9 6 marks Criterion B

Hard

Investigate the relationship between the apex angle and the RATIO of apex-to-base angle, as the isosceles triangle becomes more 'stretched' (larger apex) or more 'pointed' (smaller apex).

a. Find the base angles for apex angles of 20°, 80°, and 140°.

[3]

b. Notice that the apex=20°/base=80° case and the apex=140°/base=20° case involve the SAME three numbers (20 and 80) but swapped between apex and base roles. Explain why this 'swap' pattern makes sense, connecting it to the general formula for base angle in terms of apex angle.

[3]

Show complete worked solution

(a)

Apex 20°: base $=\frac{160}{2}=80°$ each. Apex 80°: base $=\frac{100}{2}=50°$ each. Apex 140°: base $=\frac{40}{2}=20°$ each.

(b)

Base angle $=\frac{180-\text{apex}}{2}$. When apex=20°, base=80°. When apex=140°, base=20° — the SPECIFIC swap occurs here because $140=180-2(20)$, a special numerical coincidence in this case (not a general rule for all isosceles triangles) — but it usefully illustrates that apex and base angles are linked through the fixed 180° total, so extreme values of one naturally correspond to extreme (but different) values of the other.

QUESTION 10 4 marks Criterion C

Medium

A student calculates the base angles of an isosceles triangle with apex 38° by writing '$180-38=142°$, so each base angle is 142°' (forgetting to divide by 2, and not noticing the impossibility of the result).

a. Explain the error, and give the correct base angle.

[2]

b. Explain why the student's answer of 142° for EACH base angle should have immediately been recognized as impossible, without needing to check the arithmetic further.

[2]

Show complete worked solution

(a)

The student found the COMBINED total of both base angles (142°) but didn't divide by 2 to find each INDIVIDUAL angle. Correct: $142\div2=71°$ each.

(b)

If each base angle were 142°, the two base angles ALONE would sum to $142\times2=284°$ — already exceeding the maximum possible total of 180° for an entire triangle, before even including the apex angle. Any single angle in a valid triangle must be less than 180°, and certainly two of them can't sum to more than 180° on their own — this alone signals an error.

QUESTION 11 3 marks Criterion C

Medium

A classmate says: 'in an isosceles triangle, the two EQUAL sides are always the two LONGEST sides.'

a. Explain why this claim is not always true, by considering an isosceles triangle with a very SMALL apex angle (making it tall and narrow) versus one with a very LARGE apex angle (making it short and wide), and how the relative lengths of the equal sides versus the base can change.

[3]

Show complete worked solution

(a)

This claim is false in general — in a 'wide, flat' isosceles triangle (large apex angle, small base angles), the BASE (the non-equal side) can actually be LONGER than the two equal sides. For example, with a very large apex angle close to 180°, the triangle becomes almost flat, with the base stretching out to be much longer than the two short equal sides. The relative length of the equal sides versus the base depends entirely on the SPECIFIC apex angle, not a fixed rule.

QUESTION 12 5 marks Criterion D

Medium

A decorative garden trellis is built in the shape of an isosceles triangle, with the apex angle measuring 36° (a 'golden triangle' shape, associated with pentagons).

a. Find each base angle.

[2]

b. The trellis manufacturer wants to CHANGE the design so the apex angle EQUALS one of the base angles (making the triangle EQUILATERAL, all angles equal). Determine what apex angle would be needed for this, and find the SINGLE value all three angles would share.

[3]

Show complete worked solution

(a)

$$\frac{180-36}{2}=72°\text{ each}$$

(b)

For equilateral (all angles equal), each angle must be $180\div3=60°$. So the apex angle would need to change from 36° to 60°, and ALL three angles (apex and both base angles) would then equal 60° each.

QUESTION 13 5 marks Criterion D

Medium

A tent's cross-section is an isosceles triangle. The base angles must each be AT LEAST 55° for the tent to shed rain effectively (steep enough sides), but the manufacturer wants the LARGEST possible apex angle (for maximum interior headroom) while still meeting this rain-shedding requirement.

a. Find the apex angle when the base angles are EXACTLY at the minimum allowed (55° each).

[2]

b. Explain why this 70° apex angle represents the MAXIMUM allowed (not minimum), given the requirement is a MINIMUM base angle — i.e. explain the inverse relationship between base angle and apex angle that makes 'largest apex' correspond to 'smallest allowed base angle'.

[3]

Show complete worked solution

(a)

$$180-2(55)=70°$$

(b)

Since base angle $=\frac{180-\text{apex}}{2}$, a LARGER apex angle directly causes a SMALLER base angle (they're inversely related through this fixed formula) — so achieving the LARGEST possible apex angle while still meeting the 'base angle $\ge55^\circ$' requirement means using the SMALLEST allowed base angle exactly at its minimum (55°), which is precisely what gives the maximum apex angle of 70°; any larger apex angle would push the base angle below the required 55° minimum.

QUESTION 1 3 marks Criterion A

Medium

The quadrilateral shown has three known angles.

a. State the angle sum of any quadrilateral.

[1]

b. Find the value of $x$.

[2]

Show complete worked solution

(a)

360°

(b)

$$x=360-85-95-110=70°$$

QUESTION 2 3 marks Criterion A

Medium

A quadrilateral has two right angles and two other EQUAL unknown angles, $x$.

a. Write and solve an equation for $x$.

[2]

b. Given all four angles are 90°, what special type of quadrilateral must this be?

[1]

Show complete worked solution

(a)

$$90+90+x+x=360 \Rightarrow 2x=180 \Rightarrow x=90°$$

(b)

A rectangle (or square, if the sides are also all equal) — any quadrilateral with all four angles equal to 90° qualifies.

QUESTION 3 5 marks Criterion B

Medium

Investigate why the angle sum of a quadrilateral (360°) is exactly DOUBLE the angle sum of a triangle (180°).

a. Draw a diagonal across any quadrilateral, splitting it into two triangles. How many triangles are formed?

[1]

b. Since each triangle has an angle sum of 180°, and the quadrilateral's angles are made up of exactly these two triangles' angles combined (with no overlap or gap), what must the quadrilateral's total angle sum be?

[2]

c. Using this same 'split into triangles' idea, predict the angle sum of a PENTAGON (5 sides), which can be split into 3 triangles from one vertex.

[2]

Show complete worked solution

(a)

2 triangles.

(b)

$2 \times 180° = 360°$ — this matches the known angle sum of a quadrilateral exactly.

(c)

$$3 \times 180° = 540°$$

QUESTION 4 4 marks Criterion C

Medium

A student calculates a quadrilateral's fourth angle, given three angles of 100°, 85°, and 90°, by writing '$100+85+90=275°$, so the fourth angle is also $275°$' (repeating the sum instead of subtracting from 360°).

a. Explain the student's error, and find the correct fourth angle.

[2]

b. Verify the correct answer by checking all four angles sum to 360°.

[2]

Show complete worked solution

(a)

The student found the sum of the three KNOWN angles but then incorrectly used that same number as the fourth angle, instead of subtracting it from the total 360° to find what remains. Correct: $360-275=85°$.

(b)

$100+85+90+85=360°$

QUESTION 5 4 marks Criterion D

Medium

A four-sided garden plot has three measured corner angles: 88°, 92°, and 95°.

a. Find the size of the fourth angle.

[2]

b. A landscaper claims the plot 'must be a perfect rectangle' since three of the angles are 'close to 90°'. Evaluate this claim using your answer.

[2]

Show complete worked solution

(a)

$$x=360-88-92-95=85°$$

(b)

The claim is incorrect — a true rectangle requires ALL FOUR angles to be exactly 90°. Here, the angles (88°, 92°, 95°, 85°) are close to but not exactly 90°, so the plot is an irregular quadrilateral, not a true rectangle.

QUESTION 6 2 marks Criterion A

Easy

The quadrilateral shown has three known angles.

a. Find $x$.

[2]

Show complete worked solution

(a)

$$x=360-92-88-95=85°$$

QUESTION 7 3 marks Criterion A

Medium

A quadrilateral has three known angles: 70°, 110°, 70°.

a. Find $x$, and identify a special property of this quadrilateral if two pairs of angles turn out to be equal (70°,70° and 110°,$x$° if $x=110$).

[3]

Show complete worked solution

(a)

$x=360-70-110-70=110°$. Since the angles form two EQUAL pairs (70°,70° and 110°,110°), this could represent a parallelogram (opposite angles equal) — though confirming this fully would require checking the SIDE lengths too, not just angles.

QUESTION 8 5 marks Criterion B

Medium

Investigate how many DIAGONALS a quadrilateral has, and connect this to why splitting it into 2 triangles gives the 360° angle sum.

a. Draw (describe) the diagonals of a quadrilateral $ABCD$ from vertex $A$. How many diagonals can be drawn from a SINGLE vertex of a quadrilateral?

[2]

b. Explain how this single diagonal splits the quadrilateral into exactly 2 triangles, and use this to explain WHY the angle sum of a quadrilateral must be $2\times180°=360°$.

[3]

Show complete worked solution

(a)

From vertex $A$, only ONE diagonal can be drawn (to the opposite vertex $C$) — diagonals to the two ADJACENT vertices ($B$ and $D$) would just be the existing SIDES of the quadrilateral, not diagonals.

(b)

The diagonal from $A$ to $C$ divides quadrilateral $ABCD$ into triangle $ABC$ and triangle $ACD$ — together, these two triangles' angles EXACTLY make up all four angles of the original quadrilateral (with no gaps or overlaps). Since each triangle has an angle sum of 180°, the two triangles combined give a total angle sum of $2\times180°=360°$, which is exactly the quadrilateral's angle sum.

QUESTION 9 4 marks Criterion B

Hard

Investigate whether a quadrilateral can have angles $200°, 60°, 50°, 50°$ (summing correctly to 360°), and what this reveals about angle size limits within a valid quadrilateral shape.

a. Verify that $200+60+50+50=360°$, confirming the angle SUM is technically correct.

[1]

b. Explain why these angles cannot form a convex quadrilateral. State what the $200^\circ$ interior angle tells you about the shape of a simple quadrilateral with these angles.

[3]

Show complete worked solution

(a)

$200+60+50+50=360°$ — the sum is correct.

(b)

A convex quadrilateral has every interior angle less than $180^\circ$, so an angle of $200^\circ$ is impossible for a convex quadrilateral. An interior angle of $200^\circ$ is reflex, which makes the quadrilateral concave at that vertex. It can still be a simple, non-self-intersecting quadrilateral.

QUESTION 10 4 marks Criterion C

Medium

A student calculates a quadrilateral's fourth angle, given three angles 88°, 95°, 102°, by writing '$88+95+102=285°$, so the fourth angle equals 285° too' (repeating the sum as the answer, instead of subtracting from 360°).

a. Explain the error, and find the correct fourth angle.

[2]

b. Verify the correct answer by checking all four angles sum to 360°.

[2]

Show complete worked solution

(a)

The student found the SUM of the three known angles but then incorrectly used that number AS the fourth angle, rather than subtracting it from 360° to find the REMAINING angle. Correct: $360-285=75°$.

(b)

$88+95+102+75=360°$.

QUESTION 11 3 marks Criterion C

Medium

A classmate says: 'a quadrilateral with all four angles equal MUST be a square.'

a. Explain why this claim is incorrect, giving a specific example of a quadrilateral with all four angles equal (each 90°) that is NOT a square.

[3]

Show complete worked solution

(a)

A RECTANGLE (that isn't also a square) has all four angles equal to 90° each, but its SIDES are not all equal length (it has two pairs of different-length sides) — a square specifically requires BOTH all angles equal to 90° AND all sides equal in length; equal angles alone (as in any rectangle) is not sufficient to guarantee a square.

QUESTION 12 5 marks Criterion D

Medium

A four-sided plot of land for a community garden has measured angles 85°, 95°, and 90°, with one corner angle unmeasured.

a. Find the fourth (unmeasured) angle.

[2]

b. The garden planning committee wants to install a rectangular raised bed in the corner with the 90° angle. Given the OTHER angles are 85° and 95° (not exactly 90°), explain what practical challenge this creates for fitting a perfectly rectangular raised bed elsewhere in the garden, and suggest how the design might need to adapt.

[3]

Show complete worked solution

(a)

$$x=360-85-95-90=90°$$

(b)

Since only ONE corner (90°) matches a true right angle, a rectangular raised bed would fit PERFECTLY only in that specific corner — attempting to place a rectangular bed elsewhere (near the 85° or 95° corners) would leave AWKWARD GAPS or require the bed to be angled/cut to match the plot's actual irregular shape. The design might need custom-shaped beds (trapezoidal or otherwise irregular) for the non-90° corners, or the raised beds could be positioned centrally, away from the irregular edges entirely, to avoid the fitting problem.

QUESTION 13 5 marks Criterion D

Hard

A regular quadrilateral (i.e. a square) has all sides length 8m. A LANDSCAPE architect wants to redesign it into an IRREGULAR quadrilateral with the SAME perimeter (32m), but with one angle increased to 130° while keeping the area as large as possible.

a. If three of the new quadrilateral's angles are 130°, 70°, 70° (chosen to balance the shape), find the fourth angle.

[2]

b. Explain, in general terms (without detailed area calculations), why a SQUARE typically encloses the MAXIMUM possible area for a GIVEN fixed perimeter among all quadrilaterals, meaning this redesign to an irregular shape (even with the same 32m perimeter) will likely result in a SMALLER enclosed area than the original square.

[3]

Show complete worked solution

(a)

$$360-130-70-70=90°$$

(b)

Among all quadrilaterals with a fixed perimeter, the SQUARE (with all sides and angles equal) is known to maximize the enclosed area — any deviation from this perfectly regular, symmetric shape (like stretching one angle to 130° while compressing others) typically REDUCES the enclosed area for the same total perimeter, since irregular or elongated shapes tend to 'waste' boundary length relative to the area they enclose, compared to a more compact, symmetric shape like a square.

QUESTION 1 3 marks Criterion A

Medium

The diagram shows a regular hexagon (6 sides).

a. Using the formula (angle sum) $=(n-2)\times180°$, find the sum of the interior angles.

[2]

b. Since the hexagon is REGULAR (all angles equal), find the size of each interior angle.

[1]

Show complete worked solution

(a)

$$(6-2)\times180=720°$$

(b)

$$720\div6=120°$$

QUESTION 2 3 marks Criterion A

Medium

The diagram shows a regular nonagon (9 sides).

a. Find the sum of the interior angles.

[2]

b. Find the size of each interior angle (since it's regular).

[1]

Show complete worked solution

(a)

$$(9-2)\times180=1260°$$

(b)

$$1260\div9=140°$$

QUESTION 3 5 marks Criterion B

Medium

Investigate the pattern in the angle sum formula $(n-2)\times180°$ by splitting polygons into triangles from one vertex.

a. A quadrilateral ($n=4$) splits into 2 triangles from one vertex. A pentagon ($n=5$) splits into 3 triangles. Complete the pattern: how many triangles does a hexagon ($n=6$) and a heptagon ($n=7$) split into?

[2]

b. Explain why the number of triangles formed is always exactly $n-2$, by describing how the triangles are formed from one vertex.

[2]

c. Hence explain why the angle sum formula is $(n-2)\times180°$.

[1]

Show complete worked solution

(a)

Hexagon: 4 triangles. Heptagon: 5 triangles — the number of triangles is always $n-2$.

(b)

From one vertex, you can draw a diagonal to every OTHER vertex except the two adjacent to it (since those would just be the polygon's existing sides, not diagonals) and itself — this creates $n-3$ diagonals, which divide the polygon into $n-2$ triangles.

(c)

Since each triangle contributes $180°$ to the total angle sum, and there are always $(n-2)$ triangles, the total angle sum must be $(n-2)\times180°$.

QUESTION 4 5 marks Criterion C

Medium

A student uses the formula for a regular polygon's interior angle as simply '$180-n$' (subtracting the number of sides from 180), getting $174°$ for a hexagon ($n=6$).

a. Explain why this formula is incorrect, and calculate the CORRECT interior angle of a regular hexagon.

[3]

b. Test whether the student's WRONG formula happens to give a sensible-looking answer for a triangle ($n=3$), and explain why checking a known simple case is a useful way to test whether a formula might be wrong.

[2]

Show complete worked solution

(a)

The student's formula has no real mathematical basis — the correct formula is $\frac{(n-2)\times180}{n}$ (angle sum divided by number of sides). Correct hexagon interior angle: $\frac{(6-2)\times180}{6}=\frac{720}{6}=120°$, not 174°.

(b)

Student's formula for $n=3$: $180-3=177°$ — but a real EQUILATERAL triangle should have interior angles of exactly $60°$ each, so $177°$ is clearly wrong. Testing a formula against a simple, well-known case (like an equilateral triangle) is a quick, reliable way to catch an incorrect formula before using it further.

QUESTION 5 5 marks Criterion D

Medium

A company logo is designed as a regular polygon with 12 sides (a dodecagon), shown in the diagram.

a. Find the size of each interior angle of the logo.

[2]

b. A designer wants to modify the logo to use a regular polygon where each interior angle is exactly $160°$ instead. Set up and solve an equation to find the required number of sides $n$.

[3]

Show complete worked solution

(a)

$$\frac{(12-2)\times180}{12}=\frac{1800}{12}=150°$$

(b)

$$\frac{(n-2)\times180}{n}=160 \Rightarrow 180n-360=160n \Rightarrow 20n=360 \Rightarrow n=18 \text{ sides}$$

QUESTION 6 3 marks Criterion A

Medium

The diagram shows a regular 14-sided polygon (a tetradecagon).

a. Find the sum of the interior angles.

[2]

b. Find the size of each interior angle, correct to 1 decimal place.

[1]

Show complete worked solution

(a)

$$(14-2)\times180=2160°$$

(b)

$$2160\div14\approx154.3°$$

QUESTION 7 3 marks Criterion A

Medium

The diagram shows a regular 11-sided polygon (a hendecagon).

a. Find the sum of the interior angles, and the size of each interior angle, correct to 1 decimal place.

[3]

Show complete worked solution

(a)

Sum: $(11-2)\times180=1620°$. Each: $1620\div11\approx147.3°$.

QUESTION 8 6 marks Criterion B

Hard

Investigate the relationship between the EXTERIOR angles of a regular polygon and the number of sides, $n$.

a. For a regular hexagon ($n=6$), find the interior angle, then find the exterior angle (supplementary to the interior angle, since they lie on a straight line).

[2]

b. Repeat for a regular nonagon ($n=9$) and a regular dodecagon ($n=12$). State the pattern connecting exterior angle to $n$ directly (without needing to find the interior angle first).

[4]

Show complete worked solution

(a)

Interior: $\frac{(6-2)\times180}{6}=120°$. Exterior: $180-120=60°$.

(b)

Nonagon: interior $=140°$, exterior $=40°$. Dodecagon: interior $=150°$, exterior $=30°$. Pattern: exterior angle $=\dfrac{360°}{n}$ directly — verify: hexagon $360\div6=60°$, nonagon $360\div9=40°$, dodecagon $360\div12=30°$.

QUESTION 9 5 marks Criterion B

Hard

Investigate WHY the sum of the EXTERIOR angles of ANY convex polygon (regardless of the number of sides) always equals exactly 360°, using the relationship between interior and exterior angles.

a. Using the interior angle sum formula $(n-2)\times180°$ and the fact that each exterior angle is $(180°-\text{interior angle})$, write an expression for the TOTAL sum of all $n$ exterior angles, in terms of $n$.

[3]

b. Explain why this result (360°, independent of $n$) makes intuitive sense by imagining 'walking around' the polygon's perimeter, turning by each exterior angle at every corner.

[2]

Show complete worked solution

(a)

Sum of exterior angles $= n\times180° - \text{(sum of interior angles)} = 180n - (n-2)\times180 = 180n - 180n + 360 = 360°$.

(b)

If you walk all the way around ANY closed polygon's perimeter, returning to your starting point facing the SAME direction you began, the TOTAL amount you've turned (the sum of all the exterior 'turning' angles at each corner) must equal exactly one full rotation, $360°$ — this is true regardless of how many sides the polygon has, which is why the exterior angle sum is always 360° universally.

QUESTION 10 4 marks Criterion C

Medium

A student calculates the interior angle sum of an octagon ($n=8$) by writing '$8\times180=1440°$' (multiplying $n$ by 180 directly, forgetting to subtract 2 from $n$ first).

a. Explain the error, and give the correct interior angle sum.

[2]

b. Verify the correct answer by checking it against a SQUARE ($n=4$, which is a well-known case with angle sum 360°), applying the SAME (correct) formula.

[2]

Show complete worked solution

(a)

The formula requires subtracting 2 from $n$ BEFORE multiplying by 180 (since the polygon splits into $(n-2)$ triangles, not $n$ triangles) — the student forgot this subtraction. Correct: $(8-2)\times180=1080°$.

(b)

$(4-2)\times180=360°$ — matches the well-known fact that a quadrilateral's angles sum to 360°, confirming the formula $(n-2)\times180$ is correct.

QUESTION 11 3 marks Criterion C

Medium

A classmate insists: 'a REGULAR polygon and an IRREGULAR polygon with the same number of sides must have different total interior angle sums, since irregular ones look so different.'

a. Explain why this claim is incorrect, clarifying that the interior angle sum formula $(n-2)\times180°$ depends ONLY on the number of sides $n$, not on whether the polygon is regular or irregular.

[3]

Show complete worked solution

(a)

The angle SUM formula depends only on $n$ (the number of sides), because ANY simple polygon with $n$ sides — regular or irregular — can always be divided into exactly $(n-2)$ triangles from one vertex, each contributing 180° to the total. What DOES differ between regular and irregular polygons is how that TOTAL is DISTRIBUTED among the individual angles (equal in a regular polygon, unequal in an irregular one) — but the total itself is always the same for a given $n$.

QUESTION 12 7 marks Criterion D

Hard

A company logo is designed as a regular polygon where each interior angle measures exactly 156°.

a. Set up and solve an equation to find the number of sides $n$.

[3]

b. The design team wants a SECOND version of the logo using a polygon where each interior angle is 12° LARGER (168°). Find the number of sides for this new version, and comment on how the polygon's overall SHAPE changes as the interior angle gets closer to 180° (i.e. does it look more like a circle, or more like a sharp star shape?).

[4]

Show complete worked solution

(a)

$$\frac{(n-2)\times180}{n}=156 \Rightarrow 180n-360=156n \Rightarrow 24n=360 \Rightarrow n=15$$

(b)

$\frac{(n-2)\times180}{n}=168 \Rightarrow 180n-360=168n \Rightarrow 12n=360 \Rightarrow n=30$ sides. As the interior angle approaches 180° (requiring more and more sides), a regular polygon increasingly resembles a smooth CIRCLE, rather than a sharp, star-like or angular shape — more sides with larger interior angles create a rounder, smoother overall appearance.

QUESTION 13 6 marks Criterion D

Hard

A stop-sign style regular polygon sign has 9 equal sides, forming a nonagon, for a specialized road warning sign.

a. Find each interior angle of the sign.

[2]

b. A manufacturer cuts the largest possible regular nonagon from a square sheet. Explain why some sheet metal must remain unused near the square's corners. Do not estimate a percentage without further dimensions or an exact orientation.

[4]

Show complete worked solution

(a)

$$\frac{(9-2)\times180}{9}=140°$$

(b)

A square has four right-angled corners, while the boundary of a regular nonagon meets at nine equal obtuse interior angles. Scaling the nonagon cannot make its boundary coincide with all four square corners and sides. Therefore some metal remains unused near the corners. A numerical percentage cannot be determined from the information given because it depends on the exact scale and orientation.

Review of geometrical facts 13 questions

Triangles 13 questions

Isosceles triangles 13 questions

Quadrilaterals 13 questions

Angles of an n-sided polygon 13 questions