PSLE Triangle Heuristics: 10 Question Types, Isosceles to Area Ratios

Hidden Isosceles Triangle from a Shared Side

propertyEqual sides mean equal base angles; a side shared by two triangles carries that into the second

B, C and D lie on a straight line. AB = AC and AC = CD. ∠ BAC = 40^°. Find ∠ ADC and ∠ BAD.

Ticks mark the equal sides: AB = AC, and AC = CD.

AB = AC, so triangle ABC is isosceles with apex A: ∠ ABC = ∠ ACB.

step 1 of 6

Two isosceles triangles share the side AC. The first gives the base angles at B and C; the second, on the other side of C, has its apex at C and inherits its angle from the straight line.

AB = AC, so triangle ABC is isosceles with apex A: ∠ ABC = ∠ ACB.
∠ ABC = ∠ ACB = (180^° − 40^°) ÷ 2 = 70^°.
BCD is a straight line, so ∠ ACD = 180^° − 70^° = 110^°.
AC = CD, so triangle ACD is isosceles with apex C: ∠ CAD = ∠ CDA.
∠ ADC = (180^° − 110^°) ÷ 2 = 35^°.
∠ BAD = ∠ BAC + ∠ CAD = 40^° + 35^° = 75^°.

answer∠ ADC = 35^°; ∠ BAD = 75^°

Common pitfalls

Taking ∠ ACD = 70^° because it is next to the 70^°: it is the other angle on the straight line, 110^°.
Putting the apex of the second triangle at A and making ∠ ACD a base angle: the equal sides are AC and CD, which meet at C.

Equilateral Triangle on a Side of a Square

propertyThe square's side is the triangle's side, so two more sides are equal and a 75° base angle appears

ABCD is a square. E is a point inside it such that triangle ABE is equilateral. Find ∠ ADE and ∠ DEC.

The corner at A is 90°; the triangle takes 60°, leaving ∠ DAE = 30°.

∠ DAB = 90^° and ∠ EAB = 60^°, so ∠ DAE = 90^° − 60^° = 30^°.

step 1 of 6

The corner at A is 90^° and the triangle takes 60^° of it. AD and AE are both equal to AB, so triangle ADE is isosceles; the same happens at B, and the angles round E finish it.

∠ DAB = 90^° and ∠ EAB = 60^°, so ∠ DAE = 90^° − 60^° = 30^°.
AD = AB (square) and AE = AB (equilateral triangle), so AD = AE and triangle ADE is isosceles with apex A.
∠ ADE = ∠ AED = (180^° − 30^°) ÷ 2 = 75^°.
By the same argument at B: ∠ CBE = 30^° and ∠ BEC = 75^°.
The angles round E make a full turn: ∠ AEB + ∠ AED + ∠ BEC + ∠ DEC = 360^°.
∠ DEC = 360^° − 60^° − 75^° − 75^° = 150^°.

answer∠ ADE = 75^°; ∠ DEC = 150^°

Common pitfalls

Assuming E is the centre of the square and calling ∠ ADE = 45^°: the triangle's height is less than half the side, so E sits below the centre.
Answering ∠ DEC = 60^° by symmetry with ∠ AEB: triangle DEC is isosceles but not equilateral.

Square Inside a Right-Angled Triangle

propertyThe square's sides are parallel to the legs, so the angles at the hypotenuse copy the triangle's own

Triangle ABC has a right angle at C, with ∠ CAB = 34^°. A square CDEF has D on CA, F on CB and E on AB. Find ∠ AED and ∠ BEF.

The right angle at C and 34° at A leave 56° at B.

In triangle ABC: ∠ ABC = 180^° − 90^° − 34^° = 56^°.

step 1 of 6

DE is parallel to CB and EF to CA, so the two small triangles at A and B are copies of the big one. Their angles at E are the triangle's angles at B and A.

In triangle ABC: ∠ ABC = 180^° − 90^° − 34^° = 56^°.
CDEF is a square, so DE ∥ CF, that is DE ∥ CB.
∠ AED and ∠ ABC are corresponding angles: ∠ AED = 56^°.
Check in triangle ADE: ∠ ADE = 90^°, so ∠ AED = 90^° − 34^° = 56^°.
EF ∥ DC, that is EF ∥ CA, so ∠ BEF and ∠ BAC are corresponding angles: ∠ BEF = 34^°.
Check along the straight line AEB: 56^° + 90^° + 34^° = 180^°.

answer∠ AED = 56^°; ∠ BEF = 34^°

Common pitfalls

Swapping the two: ∠ AED sits in the small triangle at A, so it copies the angle at B, not at A.
Forgetting the square's right angle at E when adding along AB, and getting 56^° + 34^° = 90^° for a straight line.

Isosceles Triangle with an Exterior Angle at the Apex

propertyThe exterior angle at the apex is the two equal base angles added

In triangle PQR, PQ = PR. QP is extended to S and ∠ SPR = 116^°. Find ∠ PQR and ∠ QPR.

QP is extended to S; the 116° sits outside the triangle at its apex.

QPS is a straight line, so ∠ SPR is an exterior angle of triangle PQR at P.

step 1 of 6

The exterior angle at P equals the sum of the two interior angles opposite it, and those two are the equal base angles. Halve it.

QPS is a straight line, so ∠ SPR is an exterior angle of triangle PQR at P.
An exterior angle equals the sum of the two interior angles opposite it: ∠ SPR = ∠ PQR + ∠ PRQ.
PQ = PR, so ∠ PQR = ∠ PRQ.
∠ PQR = 116^° ÷ 2 = 58^°.
∠ QPR = 180^° − 116^° = 64^°, on the straight line QPS.
Check: 58^° + 58^° + 64^° = 180^°.

answer∠ PQR = 58^°; ∠ QPR = 64^°

Common pitfalls

Halving 180^° − 116^° = 64^° and answering 32^°: 64^° is the apex angle, not the sum of the base angles.
Reading ∠ SPR as an angle inside the triangle.

Chain of Isosceles Triangles

propertyEach exterior angle doubles into the next triangle: 20°, 40°, 60°

A, C and E lie on one straight line, and A, B and D on another. AB = BC = CD = DE and ∠ BAC = 20^°. Find ∠ CDB and ∠ DEC.

Four equal segments in a zigzag. Triangle ABC, apex B: ∠ BCA = 20°.

AB = BC: triangle ABC is isosceles with apex B, so ∠ BCA = ∠ BAC = 20^°.

step 1 of 6

Three isosceles triangles in a row, each standing on the last. The base angle of each is an exterior angle of the one before, so the angles climb in steps of 20^°.

AB = BC: triangle ABC is isosceles with apex B, so ∠ BCA = ∠ BAC = 20^°.
∠ CBD is an exterior angle of triangle ABC at B: ∠ CBD = 20^° + 20^° = 40^°.
BC = CD: triangle BCD is isosceles with apex C, so ∠ CDB = ∠ CBD = 40^°.
∠ DCE is an exterior angle of triangle ACD at C: ∠ DCE = ∠ DAC + ∠ ADC = 20^° + 40^° = 60^°.
CD = DE: triangle CDE is isosceles with apex D, so ∠ DEC = ∠ DCE = 60^°.
Then ∠ CDE = 180^° − 60^° − 60^° = 60^° too: the third triangle is equilateral.

answer∠ CDB = 40^°; ∠ DEC = 60^°

Common pitfalls

Keeping every base angle at 20^°: each triangle stands on an exterior angle of the one before, which is a sum, not a copy.
Using ∠ BCD as the base angle of triangle BCD: C is its apex, since the equal sides BC and CD meet there.

Equilateral Triangle in a Circle

propertyRadii to the vertices cut the triangle into three equal isosceles triangles with 120° at the centre

Triangle ABC is equilateral and its vertices lie on a circle with centre O. Find ∠ AOB and ∠ OAB.

Three radii from O to the vertices split the full turn into three equal angles.

The triangle is equilateral, so its three vertices are evenly spaced round the circle and the three angles at O are equal.

step 1 of 5

The three radii OA, OB, OC split the full turn at O into three equal parts, and each radius pair makes an isosceles triangle.

The triangle is equilateral, so its three vertices are evenly spaced round the circle and the three angles at O are equal.
∠ AOB = 360^° ÷ 3 = 120^°.
OA = OB (radii), so triangle OAB is isosceles with apex O.
∠ OAB = ∠ OBA = (180^° − 120^°) ÷ 2 = 30^°.
Check: ∠ OAB + ∠ OAC = 30^° + 30^° = 60^°, the triangle's own angle at A.

answer∠ AOB = 120^°; ∠ OAB = 30^°

Common pitfalls

Answering ∠ AOB = 60^° as if O were a vertex of an equilateral triangle: the angle at the centre is twice the 60^° at the rim.
Making triangle OAB equilateral: AB is a side of the big triangle, longer than a radius.

Diagonal of a Square and a 45° Angle

propertyA diagonal cuts a square's corner into 45° and 45°

ABCD is a square with diagonal AC. E is a point on BC such that ∠ BAE = 28^°. Find ∠ EAC and ∠ AEC.

The diagonal AC makes 45° with AB.

AC is a diagonal of the square, so ∠ BAC = 45^°.

step 1 of 5

The diagonal halves the right angle at A, so the line AE sits between AB and AC with a known gap. The angle at E comes from the right-angled triangle ABE and the straight line BEC.

AC is a diagonal of the square, so ∠ BAC = 45^°.
∠ EAC = ∠ BAC − ∠ BAE = 45^° − 28^° = 17^°.
In triangle ABE, ∠ ABE = 90^°, so ∠ AEB = 180^° − 90^° − 28^° = 62^°.
BEC is a straight line, so ∠ AEC = 180^° − 62^° = 118^°.
Check in triangle AEC: 17^° + 118^° + ∠ ACE = 180^° gives ∠ ACE = 45^°, the diagonal's other half.

answer∠ EAC = 17^°; ∠ AEC = 118^°

Common pitfalls

Taking ∠ BAC = 60^° or 30^° from a badly drawn diagonal: a square's diagonal always makes 45^° with its sides.
Stopping at ∠ AEB = 62^° when the question asks for ∠ AEC, on the other side of E.

Two Identical Triangles with a Shared Vertex

propertyA rotated copy keeps its angle at the shared vertex, so the two wings are equal

Triangles PAB and PCD are identical, with PCD being PAB turned about P. The rays from P are PA, PC, PB, PD in that order. ∠ APB = 50^° and ∠ BPC = 22^°. Find ∠ APC and ∠ APD.

Turning triangle PAB about P gives PCD; its angle at P is still 50°.

The triangles are identical, so ∠ CPD = ∠ APB = 50^°.

step 1 of 5

Turning a triangle does not change its angles, so ∠ CPD = ∠ APB. The overlap ∠ BPC is inside both, and what is left on each side is the turn.

The triangles are identical, so ∠ CPD = ∠ APB = 50^°.
∠ APB is made of ∠ APC and ∠ CPB: ∠ APC = 50^° − 22^° = 28^°.
∠ CPD is made of ∠ CPB and ∠ BPD: ∠ BPD = 50^° − 22^° = 28^°, the same turn.
∠ APD is the whole fan: ∠ APC + ∠ CPB + ∠ BPD = 28^° + 22^° + 28^° = 78^°.
Check: ∠ APD = ∠ APB + ∠ BPD = 50^° + 28^° = 78^°.

answer∠ APC = 28^°; ∠ APD = 78^°

Common pitfalls

Adding the two 50^° angles and answering ∠ APD = 100^°: the 22^° overlap is counted twice that way.
Reading ∠ BPC = 22^° as the turn: the turn is what sits outside the overlap on each side.

Possible Lengths of the Third Side

propertyThe third side is longer than the difference and shorter than the sum of the other two

Two sides of a triangle are 7 cm and 12 cm long. The third side is a whole number of centimetres. What is the greatest possible length of the third side, and how many different whole-number lengths are possible?

Swing the 7 cm side about C and watch the third side change.

Angle at C: 70°

Any two sides of a triangle add to more than the third, or the two short ones could not meet.

step 1 of 5

Hinge the 7 cm side on the end of the 12 cm side and swing it: the third side grows from just over 5 cm to just under 19 cm, and never reaches either end.

Any two sides of a triangle add to more than the third, or the two short ones could not meet.
Third side < 7 + 12 = 19 cm; and 7 + third > 12, so third side > 12 − 7 = 5 cm.
So the third side is longer than 5 cm and shorter than 19 cm; 5 and 19 themselves give a flat, closed-up figure, not a triangle.
The greatest whole number below 19 is 18: greatest possible length 18 cm.
The whole numbers from 6 to 18: 18 − 6 + 1 = 13 possible lengths.

answer18 cm; 13 possible lengths

Common pitfalls

Including 19 and 5: at those lengths the three sides lie flat along one line.
Counting 18 − 6 = 12 lengths: a run from 6 to 18 has 13 numbers, both ends included.

Areas of Triangles with the Same Height

propertySame height, so the areas are in the ratio of the bases

In triangle ABC, D is a point on BC with BD : DC = 2 : 3. The area of triangle ABD is 16 cm². Find the area of triangle ADC and the area of triangle ABC.

Both triangles stand on BC and reach up to A: one height, h.

Triangles ABD and ADC both have apex A and bases along BC, so they have the same height from A.

step 1 of 5

Both small triangles have their apex at A and their base on BC, so they share one height. Their areas are then in the same ratio as their bases.

Triangles ABD and ADC both have apex A and bases along BC, so they have the same height from A.
Area of a triangle = 12 × base × height; with the same height, area is proportional to base.
Area ABD : Area ADC = BD : DC = 2 : 3.
2 units = 16 cm², so 1 unit = 8 cm² and Area ADC = 3 × 8 = 24 cm².
Area ABC = 16 + 24 = 40 cm², that is 5 units.

answer24 cm²; 40 cm²

Common pitfalls

Halving or doubling the area because the base ratio is applied to a square of something: with one shared height, area scales exactly like the base.
Giving the area of ADC as 16 × 3 = 48 cm²: 16 is two units, not one.