Year 9 angles in polygons worksheet answers to download as PDF

These Year 9 angles in polygons worksheet answers give a full worked solution to all 20 questions of the angles in polygons worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. \((5-2)\times180=3\times180=\) \(540^\circ\) A pentagon splits into 3 triangles from one vertex, so its angles add up to three lots of \(180^\circ\).
2. \((7-2)\times180=5\times180=\) \(900^\circ\) A heptagon has 7 sides and splits into 5 triangles, so the angle sum is \(900^\circ\).
3. \((12-2)\times180=10\times180=\) \(1800^\circ\) Use \(n=12\) in the rule: ten triangles of \(180^\circ\) give \(1800^\circ\).
4. \(1620\div180=9\), so \(n-2=9\) and \(n=\) 11 sides Work backwards: divide the angle sum by 180 to find the number of triangles, then add 2. The polygon has 11 sides.
Answer to Question 2
1. Sum \(=360^\circ\). Known: \(85+102+96=283\). \(x=360-283=\) \(77^\circ\) A quadrilateral has 4 sides, so its angles add up to \(360^\circ\); subtract the three known angles.
2. Sum \(=540^\circ\). Known: \(110+95+120+105=430\). \(x=540-430=\) \(110^\circ\) A pentagon has an angle sum of \(540^\circ\); subtracting the four known angles leaves \(x=110^\circ\).
3. Sum \(=720^\circ\). Known: \(130+125+118+140+112=625\). \(x=720-625=\) \(95^\circ\) A hexagon has an angle sum of \(4\times180=720^\circ\); the missing angle is \(95^\circ\).
4. Sum \(=540^\circ\). Known: \(270\). \(2x=270\), so \(x=\) \(135^\circ\) The three right angles make \(270^\circ\), so the two equal angles share the remaining \(270^\circ\): each one is \(135^\circ\).
Answer to Question 3
1. \((6-2)\times180\div6=720\div6=\) \(120^\circ\) Find the angle sum first, then share it equally between the 6 equal angles.
2. \((8-2)\times180\div8=1080\div8=\) \(135^\circ\) An octagon has an angle sum of \(1080^\circ\), shared equally between 8 angles, so each is \(135^\circ\).
3. \((10-2)\times180\div10=1440\div10=\) \(144^\circ\) Check with the exterior angle: \(360\div10=36\) and \(180-36=144\), which agrees.
4. \((15-2)\times180\div15=2340\div15=\) \(156^\circ\) Check with the exterior angle: \(360\div15=24\) and \(180-24=156\). Each angle is \(156^\circ\).
Answer to Question 4
1. \(360\div5=\) \(72^\circ\) The five equal exterior angles share the full turn of \(360^\circ\), so each is \(72^\circ\).
2. \(360\div9=\) \(40^\circ\) Nine equal exterior angles share \(360^\circ\), so each exterior angle is \(40^\circ\).
3. \(360\div30=\) 12 sides Each turn is \(30^\circ\) and the turns must add up to \(360^\circ\), so there are 12 sides.
4. \(180-30=\) \(150^\circ\) The interior and exterior angles lie on a straight line, so they add up to \(180^\circ\). The interior angle is \(150^\circ\).
Answer to Question 5
1. \(1440\div180=8\), so \(n=8+2=\) 10 The angle sum is 8 lots of \(180^\circ\), so the polygon splits into 8 triangles and has 10 sides.
2. Exterior \(=180-140=40^\circ\). \(360\div40=\) 9 Interior and exterior angles add to \(180^\circ\), and the exterior angles fill \(360^\circ\), so the polygon has 9 sides.
3. \(360\div15=\) 24 The exterior angles add up to \(360^\circ\), so the number of sides is \(360\div15\): 24 sides.
4. \(2340\div180=13\), so \(n=13+2=\) 15 The sum is 13 lots of \(180^\circ\), so the polygon has \(13+2=15\) sides.
Answer to Question 6
1. Sum: \(x+(x+20)+2x+(2x+40)=6x+60=360\), so \(6x=300\) and \(x=50\). The angles are \(50^\circ\), \(70^\circ\), \(100^\circ\), \(140^\circ\) (check: they add to 360). A quadrilateral's angles add up to \(360^\circ\). Collect the \(x\) terms (\(6x\)) and the numbers (60), then solve the equation.
2. Each angle is \(540\div5=108\). \(3x+6=108\), so \(3x=102\) and \(x=\) 34. A pentagon's angles add up to \(540^\circ\), and in a regular pentagon they are all equal, so each is \(108^\circ\). Check: \(3\times34+6=108\).
3. Sum \(=720\). Known: \(400\). The other two share \(320\), so each is \(160^\circ\). A hexagon's angles add up to \(720^\circ\). The four angles of \(100^\circ\) make \(400^\circ\), and \(320\div2=160\), so the two equal angles are each \(160^\circ\).
Answer to Question 7
1. Triangle \(60^\circ\), square \(90^\circ\), pentagon \(540\div5=108^\circ\), hexagon \(720\div6=120^\circ\), octagon \(1080\div8=135^\circ\). Use \((n-2)\times180\div n\) for each shape; the triangle and square angles are already known.
2. \(3\times120=360\), so there is no gap. The angles at the point must add up to a full turn of \(360^\circ\), and they do exactly, so hexagons tile.
3. \(135+135+90=360\), so they fit. Two octagons and one square fill the full turn exactly, so this pattern also leaves no gap.
4. \(360\div108=3.33\ldots\) is not a whole number. Three pentagons make \(324^\circ\), leaving a gap of \(36^\circ\); four would overlap. Pentagon angles cannot make exactly \(360^\circ\) at a point, so regular pentagons alone cannot tile a floor.
5. \(360\div30=12\) tiles along and \(240\div30=8\) tiles across, so \(12\times8=\) 96 tiles. Change the metres to centimetres first (360 cm and 240 cm), then count the rows and columns. Maya needs 96 tiles.
Answer to Question 8
1. \(8\times30=\) 240 cm A regular octagon has 8 equal sides, so multiply the side length by 8. The perimeter is 240 cm.
2. Interior \(=135^\circ\), exterior \(=360\div8=45^\circ\). The exterior angle is \(360\div8=45^\circ\), so the interior angle is \(180-45=135^\circ\).
3. He turns through \(8\times45=\) \(360^\circ\) (one full turn). At each corner he turns through the exterior angle; after 8 corners he faces his starting direction again, a turn of \(360^\circ\).
4. \(14\times42.50=\) £595 Multiply the price of one sign by the number of signs: the total cost is £595.
Answer to Question 9
1. \(6\times6=\) 36 mm A hexagon has 6 equal sides of 6 mm, so the perimeter is 36 mm.
2. \(720\div6=\) \(120^\circ\) The angle sum of a hexagon is \(720^\circ\), shared between 6 equal angles, so each is \(120^\circ\).
3. \(3\times120=360\), so the angles fill the full turn around a point. Three angles of \(120^\circ\) make exactly a full turn, so the cells fit together with no gaps or overlaps.
4. \(3200\times9=\) 28 800 cells Multiply the number of cells in one frame by the number of frames: there are 28 800 cells in the hive.
5. Yes. Interior angle \(108^\circ\) does not divide 360, so pentagons would leave gaps. She is right, because \(360\div108\) is not a whole number, so pentagons would leave gaps between the cells.
Answer to Question 10
1. Pentagon \(108^\circ\), hexagon \(120^\circ\). A pentagon's angle sum is \(540^\circ\) and a hexagon's is \(720^\circ\); divide by 5 and by 6.
2. \(108+120+120=\) \(348^\circ\) One pentagon angle and two hexagon angles meet at each corner, giving a total of \(348^\circ\).
3. It is \(12^\circ\) less than \(360^\circ\), so the shapes cannot lie flat. The gap makes the surface curve into a ball. On a flat surface the angles at a point must add up to \(360^\circ\). The missing \(12^\circ\) forces the panels to bend, which is why the football is round.
4. Sides: \(12\times5+20\times6=60+120=180\). Each edge is counted twice, so edges \(=180\div2=\) 90. Every edge is shared by two panels, so counting the sides of every panel counts each edge twice. The ball has 90 edges.
Answer to Question 11
1. Perimeter \(=5\times2=10\) m, cost \(=10\times12=\) £120 The fence goes all the way round, so first find the perimeter, then multiply by the cost per metre. The fence costs £120.
2. \(108^\circ\) A regular pentagon has an angle sum of \(540^\circ\), so each interior angle is \(540\div5=108^\circ\).
3. Angle \(B=108^\circ\), so each base angle is \((180-108)\div2=\) \(36^\circ\) The angles in triangle \(ABC\) add up to \(180^\circ\), and the two base angles are equal because \(AB=BC\), so angle \(BAC=36^\circ\).
4. Angle \(CAE=108-36=\) \(72^\circ\) Angle \(BAE\) is split into angle \(BAC\) and angle \(CAE\), so subtract: angle \(CAE=72^\circ\).
Answer to Question 12
1. \(360\div40=\) 9 The robot turns through the exterior angle each time, and the turns add up to \(360^\circ\), so it repeats the command 9 times.
2. A regular nonagon (9 sides) The path has 9 equal sides and 9 equal angles, so it is a regular nonagon.
3. \(9\times2=\) 18 m Each of the 9 sides is 2 m long, so the robot travels 18 m.
4. \(360\div72=5\), so a regular pentagon, with interior angle \(180-72=\) \(108^\circ\). Now the exterior angle is \(72^\circ\), giving 5 sides, so the robot draws a regular pentagon with interior angles of \(108^\circ\).
Answer to Question 13
1. Sum \(=540\). Known: \(98+121+104+117=440\). Fifth \(=540-440=\) \(100^\circ\) A pentagon's angles add up to \(540^\circ\). Subtract the four measured angles: the fifth angle is \(100^\circ\).
2. \(40+55+62+48+51=\) 256 m The perimeter is the total of the five side lengths: 256 m.
3. \(256\times7.50=\) £1 920 Multiply the perimeter by the cost per metre: fencing the whole field costs £1 920.
4. \(252\times7.50=\) £1 890 The gate gap means 4 m less fence: \(256-4=252\) m. The new cost is £1 890, which is £30 less.
Answer to Question 14
1. \(360\div12=\) \(30^\circ\) The exterior angles of any polygon add up to \(360^\circ\), and there are 12 equal ones, so each is \(30^\circ\).
2. \(180-30=\) \(150^\circ\) The interior and exterior angles lie on a straight line, so the interior angle is \(150^\circ\).
3. The twelve apex angles fill a full turn, so \(360\div12=\) \(30^\circ\). The twelve apex angles meet at the centre and fill a full turn of \(360^\circ\), so each is \(30^\circ\).
4. \((180-30)\div2=75^\circ\) each. Two base angles meet at each vertex: \(75+75=150\), which equals the interior angle. The angles in each triangle add up to \(180^\circ\), so each base angle is \(75^\circ\). The check confirms the interior angle of \(150^\circ\).
Answer to Question 15
1. Pentagon: sum \(540^\circ\), exterior \(72^\circ\). Hexagon: sum \(720^\circ\), interior \(120^\circ\). Octagon: interior \(135^\circ\), exterior \(45^\circ\). Decagon: exterior \(36^\circ\). Last row: \(360\div24=15\) sides, a 15-gon with sum \(13\times180=2340^\circ\) and interior \(156^\circ\). Use the angle sum \((n-2)\times180\), the interior angle (sum divided by \(n\)) and the exterior angle (\(360\div n\)) in each row.
2. Interior \(=2\times\) exterior and they add to \(180\), so \(3e=180\), \(e=60\), \(n=360\div60=6\): a hexagon. Let the exterior angle be \(e\). Interior + exterior \(=180\), so \(2e+e=180\). The polygon is a regular hexagon.
3. \(6e=180\), so \(e=30\) and \(n=360\div30=12\): a dodecagon. This time \(5e+e=180\), so \(e=30\). A regular polygon with 12 sides is a dodecagon.
4. Exterior \(20^\circ\) gives \(n=360\div20=18\) sides, so such a polygon has 18 sides, not fewer than 15. (The two angles do add to \(180^\circ\), so it is consistent.) An exterior angle of \(20^\circ\) forces exactly 18 sides, so the claim about having fewer than 15 sides is impossible.
Answer to Question 16
1. \(108^\circ\) The pentagon is regular, so every interior angle is \(540\div5=108^\circ\).
2. \((180-108)\div2=\) \(36^\circ\) each. Triangle \(ABC\) has \(AB=BC\), so the base angles are equal, and the angles in a triangle add up to \(180^\circ\).
3. Angle \(EAD=36^\circ\) by congruence, so \(CAD=108-36-36=\) \(36^\circ\). Triangles \(ABC\) and \(AED\) are congruent, so angle \(EAD\) also measures \(36^\circ\); subtract both from \(108^\circ\).
4. \(36+36+36=108\). The three angles at \(A\) are each \(36^\circ\), so together they make the full interior angle of \(108^\circ\). The diagonals split the angle into three equal parts.
5. Angles in triangle \(ACD\): \(180-36=144\), shared equally: \(72^\circ\). In triangle \(ACD\), angle \(CAD=36^\circ\), so the other two equal angles share \(144^\circ\). Angle \(ACD=72^\circ\).
Answer to Question 17
1. Only Ali is correct: \(5\times180=900\). Use \((n-2)\times180\) to test Ali, \(360\div\) exterior angle to test Bea and Cal, and divide by 180 to test Dev.
2. Bea: exterior angle would be \(80^\circ\) and \(360\div80=4.5\) is not a whole number, so no regular polygon has interior angle \(100^\circ\). Cal: it is \(360\div20=18^\circ\). Dev: \(1000\div180=5.56\), which is not a whole number, so no polygon has this angle sum (the sum is always a multiple of \(180\)). For a claim to be true, the number of sides must be a whole number of at least 3. Only Ali's claim passes this test.
3. The smallest polygon has 3 sides, so the largest exterior angle is \(360\div3=\) \(120^\circ\): the equilateral triangle. The fewer sides a regular polygon has, the bigger each exterior angle is. A triangle has the fewest sides, so its exterior angle of \(120^\circ\) is the largest.
Answer to Question 18
1. \(1080\div8=\) \(135^\circ\) An octagon has an angle sum of \(6\times180=1080^\circ\), shared equally between 8 angles, so each is \(135^\circ\).
2. \(135\div2=\) \(67.5^\circ\) Each rail is cut at half the interior angle, so the two cut ends fit together to make \(135^\circ\). The cut angle is \(67.5^\circ\).
3. \(8\times1.5=\) 12 m The frame has 8 equal sides of 1.5 m, so the perimeter is 12 m.
4. \(12\div0.5=\) 24 posts (it is a closed loop, so the number of posts equals the number of gaps). Divide the perimeter by the spacing. Around a closed shape the first and last posts are not doubled, so 24 posts are needed.
5. \(24\times3.50=\) £84 Multiply the number of posts by the price of one post: the posts cost £84.
Answer to Question 19
1. \(5\times180=\) \(900^\circ\) A heptagon has 7 sides, so it splits into 5 triangles: the angles add up to \(900^\circ\).
2. Interior \(=900\div7=128.6^\circ\); exterior \(=360\div7=51.4^\circ\). Share the angle sum between 7 equal angles, and divide \(360^\circ\) by 7 for the exterior angle. Check: \(128.6+51.4=180\).
3. \(7\times2.4=\) 16.8 cm The coin has 7 equal sides of 2.4 cm, so the perimeter is 16.8 cm.
4. \(360\div128.57=2.8\) is not a whole number: two coins leave a gap and three would overlap. For shapes to fit around a point with no gaps, the interior angles must divide exactly into \(360^\circ\). They do not, so seven heptagons cannot fit.
5. Two coins use \(2\times128.57=257.1^\circ\), so the gap is \(360-257.1=\) \(102.9^\circ\). Two heptagon angles leave a space of about \(102.9^\circ\), which is too small for a third coin.
Answer to Question 20
1. Sum \(=(n-2)\times180\). An \(n\)-sided polygon splits into \(n-2\) triangles, each with angles adding up to \(180^\circ\).
2. Sum \(=150(n-1)+90\). There are \(n-1\) angles of \(150^\circ\) and one angle of \(90^\circ\).
3. \(180n-360=150n-150+90=150n-60\). So \(30n=300\) and \(n=\) 10. The two expressions give the same angle sum, so set them equal, expand the brackets and collect the \(n\) terms. The polygon has 10 sides.
4. Sum \(=8\times180=1440\). Also \(9\times150+90=1350+90=1440\). They agree. Both methods give \(1440^\circ\), so the answer \(n=10\) is correct.

How to use these Year 9 angles in polygons worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the angles in polygons lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 9 angles in polygons worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 9 angles in polygons worksheet answers show every step, so nothing is left unexplained.