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The interior angle of a regular polygon is \(156^\circ\). How many sides does it have?
- A\(24\)
- B\(20\)
- C\(10\)
- D\(12\)
- E\(18\)
- F\(15\)
Worked solution
Exterior angle \(24^\circ\); \(360/24 = 15\) sides.
- Exterior angle \(= 180^\circ - 156^\circ = 24^\circ\).Interior and exterior angles at a vertex lie on a straight line.
- The exterior angles of any polygon sum to \(360^\circ\), and for a regular polygon they are all equal.Walking once around the shape turns you through a full circle.
- Number of sides \(= \dfrac{360}{24} = 15\).Divide the total turn by the turn per vertex.
- Using \(\frac{360}{156}\) or \(\frac{156}{24}\).
- Using the interior-angle-sum formula and then having to solve \(\frac{180(n-2)}{n} = 156\) - correct but slower.
A polygon has interior angles that sum to \(1980^\circ\). How many sides does it have?
- A\(10\)
- B\(15\)
- C\(14\)
- D\(11\)
- E\(13\)
- F\(12\)
Worked solution
\(180(n - 2) = 1980 \Rightarrow n - 2 = 11 \Rightarrow n = 13\).
- Recall the formula: the interior angles of an \(n\)-gon sum to \(180(n-2)^\circ\).The polygon splits into \(n - 2\) triangles, each contributing \(180^\circ\).
- Set up: \(180(n - 2) = 1980\).Substitute the given total.
- Divide: \(n - 2 = 11\), so \(n = 13\).\(1980 \div 180 = 11\); remember to add the \(2\) back.
- Forgetting to add \(2\) and answering \(11\) - an offered option.
- Using \(180n\) instead of \(180(n-2)\).
\(\mathbf{a} = \binom{2}{-1}\) and \(\mathbf{b} = \binom{-3}{4}\). Find \(2\mathbf{a} - \mathbf{b}\).
- A\(\binom{5}{-5}\)
- B\(\binom{-7}{6}\)
- C\(\binom{-1}{3}\)
- D\(\binom{1}{2}\)
- E\(\binom{7}{-2}\)
- F\(\binom{7}{-6}\)
Worked solution
\(\binom{4}{-2} - \binom{-3}{4} = \binom{7}{-6}\).
- Double \(\mathbf a\): \(2\binom{2}{-1} = \binom{4}{-2}\).A scalar multiplies both components.
- Subtract \(\mathbf b\) component by component: \(\binom{4 - (-3)}{-2 - 4}\).Subtracting a negative adds - the top component becomes \(4 + 3\).
- Simplify: \(\binom{7}{-6}\).Both components computed independently.
- Adding \(\mathbf b\) instead of subtracting, giving \(\binom{1}{2}\) - an offered option.
- Getting the sign of the bottom component wrong: \(-2 - 4 = -6\), not \(+2\).
Two similar triangles have corresponding sides of length \(6\) cm and \(9\) cm. The smaller triangle has area \(20\) cm\(^2\). Find the area of the larger triangle.
- A\(30\) cm\(^2\)
- B\(67.5\) cm\(^2\)
- C\(40\) cm\(^2\)
- D\(45\) cm\(^2\)
- E\(27\) cm\(^2\)
- F\(60\) cm\(^2\)
Worked solution
Area scale factor \(\left(\frac96\right)^2 = \frac94\): \(20 \times \frac94 = 45\).
- Length scale factor: \(\dfrac96 = \dfrac32\).From the smaller to the larger, so the factor is greater than \(1\).
- Area scale factor: \(\left(\dfrac32\right)^2 = \dfrac94\).Both dimensions scale, so the area scales by the square.
- Multiply: \(20 \times \dfrac94 = 45\) cm\(^2\).\(20 \div 4 = 5\), then \(\times 9\).
- Using the length factor directly: \(20 \times \frac32 = 30\) - an offered option.
- Cubing (that would be for volumes), giving \(67.5\).
The triangle with vertices \((1, 1)\), \((3, 1)\), \((1, 4)\) is enlarged with scale factor \(-2\), centre the origin. Which of the following is a vertex of the image?
- A\((-1, -4)\)
- B\((6, 2)\)
- C\((-2, -8)\)
- D\((-6, -8)\)
- E\((-3, -1)\)
- F\((2, 8)\)
Worked solution
Multiply each coordinate by \(-2\): \((-2, -2)\), \((-6, -2)\), \((-2, -8)\).
- With centre the origin, enlargement by \(k\) sends \((x,y)\) to \((kx, ky)\).Position vectors scale directly when the centre is the origin.
- Apply \(k = -2\) to each vertex: \((1,1) \to (-2,-2)\); \((3,1) \to (-6,-2)\); \((1,4) \to (-2,-8)\).A negative factor also rotates the shape through \(180^\circ\) - the image is on the opposite side of the origin.
- Check the options: \((-2,-8)\) appears.Match against the computed image vertices.
- Using \(+2\) and choosing \((2,8)\) - an offered option.
- Multiplying only one coordinate.
Which of the following sets of information is NOT sufficient to prove that two triangles are congruent?
- Atwo sides and a non-included angle equal (SSA)
- Btwo sides and the included angle equal (SAS)
- Ctwo angles and a corresponding side equal (ASA)
- Dright angle, hypotenuse and one other side equal (RHS)
- Ethree sides equal (SSS)
Worked solution
SSA is not a congruence condition (it is the ambiguous case of the sine rule); the other four are the standard criteria.
- Recall the valid criteria: SSS, SAS, ASA (or AAS), RHS.These four are the ones that fix a triangle completely.
- Consider SSA: two sides and a non-included angle. This is the ambiguous case of the sine rule, where two different triangles can share the same data.The side opposite the known angle can 'swing' to two positions - exactly the situation in TRI-07.
- So SSA is the odd one out.RHS is really a special case of SSA that works, because the right angle removes the ambiguity.
- Thinking RHS is invalid because it looks like SSA - the right angle makes it unambiguous.
- Confusing AAA (similar but not congruent) with ASA.
In the diagram (not shown), \(AB\) is parallel to \(CD\). A transversal makes an angle of \((3x + 10)^\circ\) with \(AB\) and the co-interior angle with \(CD\) is \((2x - 5)^\circ\). Find \(x\).
- A\(25\)
- B\(55\)
- C\(15\)
- D\(30\)
- E\(35\)
- F\(45\)
Worked solution
Co-interior angles sum to \(180^\circ\): \(5x + 5 = 180 \Rightarrow x = 35\).
- Identify the relationship: co-interior angles lie on the same side of the transversal, between the parallel lines.Their names matter: alternate angles are equal, corresponding angles are equal, co-interior angles are supplementary.
- Write the equation: \((3x + 10) + (2x - 5) = 180\).Supplementary means summing to \(180^\circ\).
- Simplify: \(5x + 5 = 180\), so \(5x = 175\) and \(x = 35\).Combine like terms, then divide.
- Setting the angles equal (treating them as alternate), which gives \(x = 15\) - an offered option.
- Arithmetic: \(10 - 5 = 5\), not \(15\).
Which of the following statements is/are true?
- IEvery rhombus is a parallelogram.
- IIEvery rectangle is a rhombus.
- IIIThe diagonals of a kite are perpendicular.
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
A rhombus has two pairs of parallel sides (I). A rectangle need not have equal sides (II false). The diagonals of a kite cross at right angles (III).
- I: a rhombus has two pairs of parallel sides, which is the definition of a parallelogram. True.Rhombus \(=\) parallelogram with all sides equal, so it is a special parallelogram.
- II: a rectangle has equal angles but not necessarily equal sides - a \(2 \times 5\) rectangle is not a rhombus. False.Only the square lies in both families.
- III: a kite's diagonals meet at right angles (the axis of symmetry bisects the other diagonal perpendicularly). True.This is a standard kite property.
- I and III only.Match to the option list.
- Reversing II: every rhombus is not a rectangle either, but the claim as stated is about rectangles.
- Thinking a kite's diagonals bisect each other - only one bisects the other.
A shape is reflected in the line \(y = x\) and then rotated \(90^\circ\) anticlockwise about the origin. The point \((3, 1)\) ends up at
- A\((1, 3)\)
- B\((-3, 1)\)
- C\((-1, 3)\)
- D\((1, -3)\)
- E\((3, -1)\)
- F\((-1, -3)\)
Worked solution
Reflection in \(y = x\): \((3, 1) \to (1, 3)\). Rotation \(90^\circ\) anticlockwise: \((x, y) \to (-y, x)\), so \((1, 3) \to (-3, 1)\).
- Reflection in \(y = x\) swaps the coordinates: \((3,1) \to (1,3)\).The line \(y = x\) exchanges the roles of the two axes.
- Rotation \(90^\circ\) anticlockwise about the origin sends \((x,y) \to (-y,x)\).Worth memorising; check it on \((1,0) \to (0,1)\) ✓.
- Apply it: \((1,3) \to (-3,1)\).Here \(x = 1\), \(y = 3\), so the image is \((-3, 1)\).
- Applying the transformations in the wrong order.
- Using the clockwise rule \((x,y) \to (y,-x)\), giving \((3,-1)\) - an offered option.
In triangle \(ABC\), angle \(A\) is twice angle \(B\), and angle \(C\) is \(20^\circ\) more than angle \(A\). Find angle \(B\).
- A\(64^\circ\)
- B\(30^\circ\)
- C\(32^\circ\)
- D\(36^\circ\)
- E\(40^\circ\)
- F\(28^\circ\)
Worked solution
\(B + 2B + (2B + 20) = 180 \Rightarrow 5B = 160 \Rightarrow B = 32^\circ\).
- Let angle \(B = b\). Then \(A = 2b\) and \(C = A + 20 = 2b + 20\).Express everything in terms of the angle being asked for.
- Angle sum: \(b + 2b + (2b + 20) = 180\).The three angles of a triangle sum to \(180^\circ\).
- Simplify: \(5b + 20 = 180\), so \(5b = 160\) and \(b = 32^\circ\).Check: \(32 + 64 + 84 = 180\) ✓.
- Setting \(C = 2b + 20\) but forgetting that \(C\) is \(20\) more than \(A\), not than \(B\).
- Arithmetic: \(180 - 20 = 160\), not \(200\).
In triangle \(ABC\), \(D\) lies on \(AB\) and \(E\) lies on \(AC\) with \(DE\) parallel to \(BC\). \(AD = 4\), \(DB = 6\) and \(DE = 5\). Find \(BC\).
- A\(12.5\)
- B\(8\)
- C\(15\)
- D\(10\)
- E\(7.5\)
- F\(12\)
Worked solution
Triangles \(ADE\) and \(ABC\) are similar with ratio \(AD : AB = 4 : 10\). \(BC = 5 \times \frac{10}{4} = 12.5\).
- \(DE \parallel BC\) makes triangles \(ADE\) and \(ABC\) similar (equal corresponding angles).Parallel lines give equal corresponding angles, which is the AAA similarity condition.
- Compute the ratio: \(AB = AD + DB = 4 + 6 = 10\), so \(AD : AB = 4 : 10 = 2 : 5\).The ratio must compare corresponding whole sides - \(AD\) to \(AB\), not \(AD\) to \(DB\).
- Scale up \(DE\): \(BC = 5 \times \dfrac{10}{4} = 12.5\).Multiply by the reciprocal of the ratio to go from small to large.
- Using the ratio \(4 : 6\) and answering \(7.5\) - an offered option.
- Dividing instead of multiplying, giving \(2\).
\(ABCD\) is a quadrilateral. \(P\): 'the diagonals of \(ABCD\) bisect each other'. \(Q\): '\(ABCD\) is a rectangle'. Which one of the following is true?
- AP is necessary and sufficient for Q.
- BP is necessary but not sufficient for Q.
- CP is sufficient but not necessary for Q.
- DP is neither necessary nor sufficient for Q.
Worked solution
Every rectangle is a parallelogram, and a parallelogram's diagonals bisect each other: \(P\) is necessary. A non-rectangular parallelogram has bisecting diagonals without being a rectangle: not sufficient.
- Every rectangle is a parallelogram, and a parallelogram's diagonals bisect each other. So \(Q \Rightarrow P\): \(P\) is necessary.Opposite sides equal and parallel forces the diagonals to share a midpoint.
- Is \(P\) sufficient? Take a non-rectangular parallelogram, e.g. a rhombus with angles \(60^\circ/120^\circ\): diagonals bisect each other, but it is not a rectangle.Bisecting diagonals say nothing about the angles.
- So \(P\) is necessary but not sufficient.For a rectangle you also need the diagonals to be equal in length.
- Answering 'necessary and sufficient' by confusing 'bisect each other' with 'bisect each other and are equal'.
- Getting the direction backwards: the broader property (parallelogram) is necessary for the narrower one (rectangle).
In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\) and \(P\) is the point on \(OM\) with \(OP : PM = 2 : 1\). Express \(\overrightarrow{AP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
- A\(\dfrac{2}{3}(\mathbf{a} + \mathbf{b})\)
- B\(\dfrac{2}{3}\mathbf{b} - \dfrac{1}{3}\mathbf{a}\)
- C\(\dfrac{1}{3}(\mathbf{b} - \mathbf{a})\)
- D\(\dfrac{1}{3}\mathbf{b} - \dfrac{2}{3}\mathbf{a}\)
- E\(\dfrac{1}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)
- F\(\dfrac{1}{2}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\)
Worked solution
\(\overrightarrow{OM} = \frac12(\mathbf a + \mathbf b)\), \(\overrightarrow{OP} = \frac23\overrightarrow{OM} = \frac13(\mathbf a + \mathbf b)\). \(\overrightarrow{AP} = \overrightarrow{OP} - \mathbf a = \frac13\mathbf b - \frac23\mathbf a\).
- Find \(\overrightarrow{OM}\): \(M\) is the midpoint of \(AB\), so \(\overrightarrow{OM} = \frac12(\mathbf a + \mathbf b)\).The midpoint of \(AB\) is the average of the two position vectors.
- Use the ratio \(OP : PM = 2 : 1\): \(P\) is \(\frac23\) of the way from \(O\) to \(M\), so \(\overrightarrow{OP} = \frac23 \times \frac12(\mathbf a + \mathbf b) = \frac13(\mathbf a + \mathbf b)\).\(2 : 1\) means \(2\) parts out of \(3\) - convert the ratio into a fraction of the whole.
- Travel from \(A\) to \(P\) via \(O\): \(\overrightarrow{AP} = \overrightarrow{AO} + \overrightarrow{OP} = -\mathbf a + \frac13(\mathbf a + \mathbf b)\).\(\overrightarrow{AO} = -\overrightarrow{OA} = -\mathbf a\); reversing a vector negates it.
- Simplify: \(-\mathbf a + \frac13\mathbf a + \frac13\mathbf b = \frac13\mathbf b - \frac23\mathbf a\).Collect the \(\mathbf a\) terms: \(-1 + \frac13 = -\frac23\).
- Using \(\frac12\) instead of \(\frac23\) for the ratio point.
- Forgetting the \(-\mathbf a\) and giving \(\overrightarrow{OP}\) instead of \(\overrightarrow{AP}\) - an offered option.
A ship sails from \(P\) on a bearing of \(060^\circ\) for \(10\) km to \(Q\), then on a bearing of \(150^\circ\) for \(10\) km to \(R\). What is the bearing of \(P\) from \(R\)?
- A\(255^\circ\)
- B\(195^\circ\)
- C\(315^\circ\)
- D\(285^\circ\)
- E\(075^\circ\)
- F\(105^\circ\)
Worked solution
The turn at \(Q\) is \(90^\circ\) and \(PQ = QR\), so triangle \(PQR\) is right-angled isosceles and \(PR\) makes \(45^\circ\) with \(QR\). Bearing of \(R\) from \(P\) is \(060 + 45 = 105^\circ\); the back bearing is \(105 + 180 = 285^\circ\).
- The bearing changes from \(060^\circ\) to \(150^\circ\), a turn of \(90^\circ\), so the path turns through a right angle at \(Q\).The interior angle \(PQR\) is therefore \(90^\circ\).
- Both legs are \(10\) km, so triangle \(PQR\) is right-angled isosceles, with the two base angles \(45^\circ\) each.Equal legs give equal base angles, each \(\frac{180 - 90}{2} = 45^\circ\).
- Bearing of \(R\) from \(P\): the direction \(PQ\) is \(060^\circ\), and \(PR\) is \(45^\circ\) further clockwise, so \(060 + 45 = 105^\circ\).\(R\) lies on the far side of \(PQ\) from north, so add.
- Reverse for the bearing of \(P\) from \(R\): add \(180^\circ\) to get \(285^\circ\).Back bearings differ by \(180^\circ\) (add if under \(180^\circ\), subtract if over).
- Giving \(105^\circ\), which is the bearing of \(R\) from \(P\) - the reverse of what is asked (and an offered option).
- Subtracting \(180^\circ\) and getting a negative, or forgetting three-figure form.
\(\overrightarrow{AB} = 3\mathbf{p} + 2\mathbf{q}\) and \(\overrightarrow{BC} = k\mathbf{p} - 4\mathbf{q}\), where \(\mathbf{p}\) and \(\mathbf{q}\) are not parallel. Given that \(A\), \(B\) and \(C\) are collinear, find \(k\).
- A\(2\)
- B\(6\)
- C\(-3\)
- D\(\dfrac{3}{2}\)
- E\(-6\)
- F\(-4\)
Worked solution
Collinear means \(\overrightarrow{BC} = \lambda\overrightarrow{AB}\): \(-4 = 2\lambda \Rightarrow \lambda = -2\), so \(k = 3\lambda = -6\).
- Collinear \(A\), \(B\), \(C\) means \(\overrightarrow{BC} = \lambda\overrightarrow{AB}\) for some scalar \(\lambda\).They share the point \(B\), so the two vectors must point along the same line.
- Write it out: \(k\mathbf p - 4\mathbf q = \lambda(3\mathbf p + 2\mathbf q)\).Because \(\mathbf p\) and \(\mathbf q\) are not parallel, coefficients can be compared independently.
- Compare \(\mathbf q\): \(-4 = 2\lambda\), so \(\lambda = -2\).Use the equation that contains no unknown besides \(\lambda\).
- Compare \(\mathbf p\): \(k = 3\lambda = -6\).Substitute the value of \(\lambda\).
- Taking \(\lambda = 2\) and answering \(6\) - an offered option.
- Comparing \(\mathbf p\) with \(\mathbf q\) coefficients.
A regular hexagon and a regular octagon share a common side. What is the size of the angle between their two other sides that meet at one end of the common side (the angle outside both polygons)?
- A\(120^\circ\)
- B\(95^\circ\)
- C\(105^\circ\)
- D\(135^\circ\)
- E\(115^\circ\)
- F\(75^\circ\)
Worked solution
Angles at the shared vertex: \(120^\circ\) (hexagon) \(+ 135^\circ\) (octagon) \(+ x = 360^\circ\), so \(x = 105^\circ\).
- Hexagon interior angle: exterior \(= \frac{360}{6} = 60^\circ\), so interior \(= 120^\circ\).Regular polygon: interior \(= 180 - \frac{360}{n}\).
- Octagon interior angle: exterior \(= \frac{360}{8} = 45^\circ\), so interior \(= 135^\circ\).Same formula with \(n = 8\).
- At the shared vertex three angles meet and fill a full turn: \(120 + 135 + x = 360\).The two interior angles plus the gap between the outer sides make a complete revolution.
- Solve: \(x = 360 - 255 = 105^\circ\).That gap is the angle asked for.
- Using \(180^\circ\) instead of \(360^\circ\) at the vertex.
- Using exterior angles in the sum, giving \(255^\circ\) or similar.
How this topic has been examined
Real TMUA questions on Angles, polygons, congruence and vectors, newest first. Links open the official paper at the question and the online marking page.