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

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1. Answer to Question 1

   1. Angles on a straight line add up to \(180^\circ\): \(a=180-128=\) 52°
   2. Angles round a point add up to \(360^\circ\): \(d=360-(95+110+80)=360-285=\) 75°
   3. Angles in a triangle add up to \(180^\circ\): \(e=180-(47+68)=180-115=\) 65°
   4. Angles on a straight line: \(f=180-112=\) 68°. (The angle opposite \(112^\circ\) is also \(112^\circ\): vertically opposite angles are equal.) So \(f=68^\circ\), because \(f\) and \(112^\circ\) lie on the same straight line.

2. Answer to Question 2

   1. \(a=\) 64° (corresponding angles are equal). Corresponding angles sit in the same position at each parallel line (the F shape).
   2. \(b=\) 64° (alternate angles are equal, the Z shape). So \(a=b=64^\circ\), because both are matched with the given angle on parallel lines.
   3. The angle vertically opposite \(64^\circ\) is \(64^\circ\), and co-interior angles add up to \(180^\circ\) (the C shape), so \(c=180-64=\) 116°. Check: \(64+116=180\).

3. Answer to Question 3

   1. Exterior \(=360\div5=72^\circ\), interior \(=180-72=\) 108° The exterior and interior angles lie on a straight line, so they add up to \(180^\circ\). Each interior angle of a regular pentagon is \(108^\circ\).
   2. Exterior \(=360\div8=45^\circ\), interior \(=180-45=\) 135° Check with the angle sum: \((8-2)\times180\div8=1080\div8=135^\circ\).
   3. Exterior \(=360\div12=30^\circ\), interior \(=180-30=\) 150° So each interior angle of a regular 12-sided polygon is \(150^\circ\).
   4. Number of sides \(=360\div15=\) 24. Interior \(=180-15=\) 165°. The exterior angles of any polygon add up to \(360^\circ\), so the number of sides is \(360\) divided by one exterior angle. The polygon has 24 sides and each interior angle is \(165^\circ\).

4. Answer to Question 4

   1. \((6-2)\times180=4\times180=\) 720° A hexagon has 6 sides, so it splits into 4 triangles from one corner. The angle sum of a hexagon is \(720^\circ\).
   2. \((10-2)\times180=8\times180=\) 1440° A decagon splits into 8 triangles, each with angles adding up to \(180^\circ\). The angle sum is \(1440^\circ\).
   3. \((15-2)\times180=13\times180=\) 2340° A 15-sided polygon splits into 13 triangles, so its interior angles add up to \(2340^\circ\).
   4. \((n-2)\times180=2880\), so \(n-2=16\) and \(n=\) 18. Divide both sides by 180: \(2880\div180=16\). Then add 2. The polygon has 18 sides. Check: \(16\times180=2880\).

5. Answer to Question 5

   1. A rhombus. All four sides are equal, so it is a rhombus; it is not a square because the angles are not right angles.
   2. An isosceles trapezium. One pair of parallel sides makes it a trapezium, and the equal non-parallel sides make it isosceles.
   3. A kite. Two pairs of equal adjacent sides and exactly one line of symmetry describe a kite.
   4. Opposite angles are equal and co-interior angles add to \(180^\circ\): the angles are 70°, 110°, 70°, 110° (sum \(=360^\circ\)). The angle next to \(70^\circ\) is \(180-70=110^\circ\). The other two angles are opposite these, so they are \(70^\circ\) and \(110^\circ\).

6. Answer to Question 6

   1. \((180-40)\div2=\) 70° each (base angles of an isosceles triangle are equal).
   2. \(180-2\times55=\) 70° The two base angles are equal, so together they make \(110^\circ\), and the apex angle is \(70^\circ\) because angles in a triangle add up to \(180^\circ\).
   3. The third interior angle is \(180-50-65=65^\circ\), so the exterior angle is \(180-65=\) 115°. This equals \(50+65\): an exterior angle equals the sum of the two opposite interior angles.
   4. \(130-55=\) 75°. The exterior angle equals the sum of the two opposite interior angles, so the other interior angle is \(75^\circ\).

7. Answer to Question 7

   1. \((180-110)\div2=\) 35° (isosceles triangle).
   2. Angle \(ADE\) and angle \(ABC\) are corresponding angles on parallel lines, so \(ADE=\) 35°.
   3. The strut bisects the apex angle in an isosceles triangle: \(110\div2=\) 55°.
   4. \(55+35+90=180\) (the strut is vertical, so angle \(AMB=90^\circ\)). The angles do add up. So triangle ABM has angles of \(55^\circ\), \(35^\circ\) and \(90^\circ\).

8. Answer to Question 8

   1. \(3\times120=\) 360°, so they fit. Three hexagons fill the space round the point exactly, with no gap and no overlap.
   2. \(2\times135+90=270+90=\) 360°, so they fit. This is why floors are often tiled with octagons and small squares.
   3. \(3\times108=324^\circ\). The gap is \(360-324=\) 36°, so pentagons alone do not tessellate.
   4. \(360-180=180\) and \(180\div90=2\), so 2 squares: 3 triangles and 2 squares (\(3\times60+2\times90=360\)). So 3 triangles and 2 squares fit round a point, because the angles add up to exactly \(360^\circ\).

9. Answer to Question 9

   1. Exterior \(=360\div8=\) 45°, interior \(=180-45=\) 135°. For any regular polygon the exterior angle is \(360\div n\), and the interior and exterior angles add up to \(180^\circ\).
   2. \(8\times25=\) 200 cm. The perimeter is 8 sides of 25 cm, so the perimeter of the sign is 200 cm.
   3. An octagon has \(8-2=6\) triangles, with angle sum \(6\times180=\) 1080°. From one corner we can draw lines to all corners except itself and its two neighbours, which makes 6 triangles. Their angles add up to \(1080^\circ\).
   4. \(8\times135=1080^\circ\), which matches \((8-2)\times180=1080^\circ\). So the angles of the 6 triangles make exactly the interior angles of the octagon, and the sum is \(1080^\circ\) both ways.

10. Answer to Question 10

    1. Back bearing \(=50+180=\) 230°.
    2. At B, the direction back to A is \(230^\circ\) and the direction to C is \(130^\circ\), so angle \(ABC=230-130=\) 100°.
    3. \((180-100)\div2=\) 40°. The two base angles of the isosceles triangle are equal and add to \(180-100=80^\circ\).
    4. Bearing of C from A \(=50+40=\) 090° (due east). Bearing of A from C \(=90+180=\) 270° (due west).

11. Answer to Question 11

    1. Co-interior angles between parallel lines add up to \(180^\circ\).
    2. \(3x+30=180\), so \(3x=150\) and \(x=\) 50. Check: \(2\times50+10=110\) and \(50+20=70\).
    3. \(2x+10=\) 110° and \(x+20=\) 70°. (Sum \(=180\).)
    4. At each crossing there are 2 angles of \(70^\circ\) (vertically opposite) and 2 of \(110^\circ\), so over 3 crossings: 6 angles of 70°.

12. Answer to Question 12

    1. \(360\div6=\) 60°.
    2. The two other angles are equal (isosceles triangle) and add to \(120^\circ\), so each is \(60^\circ\). All three angles are \(60^\circ\), so the triangle is equilateral.
    3. \(6\times40=\) 240 cm. The window has 6 sides of 40 cm, so the perimeter is 240 cm.
    4. One triangle: \(\tfrac12\times40\times20\sqrt3=400\sqrt3=692.8\) cm². Six triangles: \(6\times692.82=\) 4157 cm² (nearest cm²).

13. Answer to Question 13

    1. \(x+2x+10+3x-10=180\), \(6x=180\), \(x=30\). Angles: 30°, 70°, 80°. The angles in a triangle add up to \(180^\circ\). Check: \(30+70+80=180\).
    2. \(6x+60=360\), \(x=50\). Angles: 50°, 80°, 100°, 130°. The angles in a quadrilateral add up to \(360^\circ\): \(x+x+30+2x+2x+30=6x+60\). Check: \(50+80+100+130=360\).
    3. \(8x+100=540\), \(x=55\). Angles: 55°, 110°, 165°, 100°, 110°. The angles of a pentagon add up to \((5-2)\times180=540^\circ\). Check: \(55+110+165+100+110=540\).
    4. Only the pentagon (it has \(165^\circ\)). The triangle's largest angle is \(80^\circ\) and the quadrilateral's is \(130^\circ\), so only the pentagon has an angle bigger than \(150^\circ\).

14. Answer to Question 14

    1. \(B=D=(360-76-112)\div2=172\div2=\) 86°. The angles of a quadrilateral add up to \(360^\circ\), and in a kite the angles between the unequal sides, \(B\) and \(D\), are equal.
    2. AC is the line of symmetry, so \(BAC=76\div2=\) 38° and \(BCA=112\div2=\) 56°. The line of symmetry cuts angles \(A\) and \(C\) exactly in half.
    3. In triangle ABE: \(180-90-38=\) 52°. In triangle CBE: \(180-90-56=\) 34°. Angle \(AEB\) is \(90^\circ\) because the diagonals of a kite cross at right angles.
    4. \(52+34=86\), which equals angle \(B\). The check works, so the answers are consistent.

15. Answer to Question 15

    1. \(x\) and \(q\) are alternate angles between parallel lines, so \(x=q\). In the same way \(y\) and \(r\) are alternate angles, so \(y=r\).
    2. Angles on a straight line at P: \(x+p+y=180^\circ\). Replace \(x\) by \(q\) and \(y\) by \(r\): \(q+p+r=180^\circ\). So the angles of a triangle add up to \(180^\circ\).
    3. Exterior angle at R \(=180-r\) (straight line). From part (b), \(p+q=180-r\). So the exterior angle \(=p+q\).
    4. Third angle \(r=180-55-75=50^\circ\). Exterior angle \(=180-50=130^\circ=55+75\).

16. Answer to Question 16

    1. Let the exterior angle be \(e\). Then \(180-e=8e\), \(9e=180\), \(e=20\), and \(n=360\div20=\) 18.
    2. Exterior \(=180-156=24^\circ\), so \(n=360\div24=\) 15.
    3. \(180-e=e+100\), so \(2e=80\), \(e=40\), \(n=360\div40=\) 9.
    4. The exterior angle would be \(80^\circ\) and \(360\div80=4.5\), which is not a whole number, so no regular polygon has an interior angle of \(100^\circ\).

17. Answer to Question 17

    1. AB = AC (given), BM = CM (M is the midpoint), AM is common to both. All three pairs of sides are equal, so the triangles are congruent by SSS.
    2. Corresponding angles in congruent triangles are equal, so angle \(ABM=\) angle \(ACM\). The angles AMB and AMC are equal and lie on a straight line, so each is \(180\div2=\) 90°.
    3. Pair (i) is congruent by SSS (the same three lengths). Pair (ii) is not congruent because the included angles differ.
    4. The triangles could be different sizes, for example an equilateral triangle of side 3 cm and one of side 6 cm. Equal angles give similar triangles, not congruent ones.

18. Answer to Question 18

    1. A tangent meets the radius at \(90^\circ\). So in triangles OAT and OBT the angles at A and B are right angles, because each tangent is perpendicular to the radius drawn to its point of contact.
    2. \((180-40)\div2=\) 70°. The two equal tangents make the base angles of triangle TAB equal, so each is \(70^\circ\).
    3. \(OAB=OAT-TAB=90-70=\) 20°. Angle \(OAT\) is split into angle \(OAB\) and angle \(TAB\).
    4. Triangle OAB is isosceles (OA = OB), so \(AOB=180-2\times20=\) 140°. Check in quadrilateral OATB: \(90+90+40+140=360^\circ\). The angles of a quadrilateral add up to \(360^\circ\), so the answer \(AOB=140^\circ\) is confirmed.

19. Answer to Question 19

    1. The angle at the centre is twice the angle at the circumference: \(108\div2=\) 54°.
    2. Opposite angles of a cyclic quadrilateral add to \(180^\circ\): \(180-78=\) 102°.
    3. The angle in a semicircle is \(90^\circ\), so \(ACB=90^\circ\) and \(ABC=180-90-34=\) 56°.
    4. By the alternate segment theorem, the angle in the alternate segment is 52°.

20. Answer to Question 20

    1. \((5-2)\times180\div5=540\div5=\) 108°. The angles of a pentagon add up to \(540^\circ\) and in a regular pentagon all five are equal.
    2. The base angle lies on a straight line with an interior angle: \(180-108=\) 72°. Each extended side makes a straight line, so the angle outside the pentagon is \(180-108=72^\circ\). The triangle is isosceles, so both base angles are \(72^\circ\).
    3. Tip angle \(=180-2\times72=\) 36°. The angles of each point add up to \(180^\circ\), so the tip angle is \(36^\circ\).
    4. \(5\times36=\) 180°. So the five tip angles of the star add up to \(180^\circ\), the same as the angles of a single triangle.