Angles, polygons and properties of shapes – Year 10 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(180-63-57=\) 60° (1 for method, 1 for the answer with the reason: angles in a triangle add up to \(180^\circ\))
   2. \(180-60=\) 120° (1), angles on a straight line add up to \(180^\circ\), or the exterior angle equals \(63+57=120\) (1). So angle \(QRS=120^\circ\).

   In (a), the final mark needs the reason written in words, not just the number. Check (b): the exterior angle of a triangle equals the sum of the two interior opposite angles, \(63^\circ+57^\circ=120^\circ\), which agrees with \(180^\circ-60^\circ\). Both methods are accepted.

2. Answer to Question 2 (4 marks)

   1. Exterior \(=360\div20=18^\circ\) (1); interior \(=180-18=\) 162° (1) The exterior angles of a polygon add up to \(360^\circ\), and each interior angle is \(180^\circ\) minus the exterior angle. The interior angle is \(162^\circ\).
   2. \(360\div18=\) 20 sides (1) Divide 360 by the exterior angle to get the number of sides: the polygon has 20 sides.
   3. \((7-2)\times180=\) 900° (1) A heptagon has 7 sides and splits into 5 triangles, so the angle sum is \(5\times180=900^\circ\).

3. Answer to Question 3 (4 marks)

   1. \(x+2x+10+3x-20+4x=10x-10\) (1) and the sum of the angles of a quadrilateral is \(360\) (1), so \(10x-10=360\). Collect the \(x\) terms: \(x+2x+3x+4x=10x\), and the numbers: \(10-20=-10\).
   2. \(10x=370\), \(x=37\) (1). Largest angle \(=4x=\) 148° (1). (The angles are 37, 84, 91, 148.) Add 10 to both sides, then divide by 10. Check: \(37+84+91+148=360\). The largest angle is \(148^\circ\).

   In (a), the sum must be stated as \(360^\circ\) for the second mark. A quick check of (b) is that each angle is positive and less than \(180^\circ\), which is possible for a quadrilateral.

4. Answer to Question 4 (4 marks)

   1. \(72+180=\) 252° (1) A back bearing differs from the bearing by \(180^\circ\), so the bearing of P from Q is \(252^\circ\).
   2. \(252-144=\) 108° (1) At Q the direction to P is \(252^\circ\) and the direction to R is \(144^\circ\). The difference between them is angle \(PQR=108^\circ\).
   3. Triangle PQR is isosceles, so \(QPR=(180-108)\div2=36^\circ\) (1). Bearing of R from P \(=72+36=\) 108° (1). Since PQ = QR, the base angles at P and R are equal. Add angle \(QPR\) to the bearing of Q from P. The bearing of R from P is \(108^\circ\).

5. Answer to Question 5 (4 marks)

   1. \(AOC=2\times64=\) 128° (1), the angle at the centre is twice the angle at the circumference (1)
   2. \((180-128)\div2=\) 26° (1) The radii OA and OC are equal, so the two base angles are equal and each is \(26^\circ\).
   3. \(90-26=\) 64° (1), because the tangent is perpendicular to the radius OA (this also agrees with the alternate segment theorem, \(TAC=ABC=64^\circ\)). So angle \(TAC=64^\circ\).

Total : 20 marks