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GCSE Maths · EdexcelChapter 20 · Pythagoras and trigonometry
GCSE Maths · Pythagoras and Trigonometry · Higher · Lesson 20.2

GCSE Maths: Is it right-angled? Pythagorean triples and problems in context (Higher)

Part of the GCSE Maths pythagoras and trigonometry chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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20.2

Is it right-angled? Pythagorean triples and problems in context

Edexcel 1MA1: G20, G6
Grade 4-6

Pythagoras’ theorem works backwards too. If the squares of the two shorter sides add up to the square of the longest side, then the triangle must have a right angle. If they do not, it has no right angle. This gives us a test for a right angle using only lengths.

ABC7 cm24 cm25 cm
Test: 72 + 242 = 49 + 576 = 625 and 252 = 625. They are equal, so the triangle has a right angle opposite the 25 cm side.

Whole numbers that fit the theorem are called a Pythagorean triple. The most famous is 3, 4, 5 (9 + 16 = 25). Others you should recognise are 5, 12, 13 and 8, 15, 17 and 7, 24, 25. If you multiply every number in a triple by the same amount you get another triple, so 6, 8, 10 and 30, 40, 50 also work. Builders use 3, 4, 5 to check that a corner is square: they measure 3 m along one wall and 4 m along the other and see if the gap between the ends is exactly 5 m.

If the sum of the squares of the two shorter sides is less than the square of the longest side, the biggest angle is more than 90° (obtuse). If it is more, the biggest angle is less than 90°. You do not need this to answer most questions, but it explains why the test says “not right-angled”.

Problems in context. Many real problems hide a right-angled triangle: a ladder against a wall, the diagonal of a rectangle, a ship that sails north then east, a wire holding up a pole. Follow these steps each time:

  1. Draw a sketch and mark the right angle.
  2. Mark the lengths you know, and call the one you want x.
  3. Decide whether x is the hypotenuse (the longest side, opposite the right angle) or a shorter side.
  4. Use Pythagoras: add the squares for the hypotenuse, subtract for a shorter side.
  5. Check that the answer is sensible, then write it with its unit.

Key ideas

  • Test for a right angle: (short)2 + (short)2 = (longest)2.
  • A Pythagorean triple is three whole numbers that fit: 3,4,5; 5,12,13; 8,15,17; 7,24,25.
  • Multiplying a triple by the same number gives another triple.
  • Always draw a sketch for a word problem and mark the right angle.
Worked example 20.2.1
A triangle has sides of length 7 cm, 24 cm and 25 cm. Is it a right-angled triangle? You must show your working.
7 cm24 cm25 cm

Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.

Method. Square all three sides. If (short)2 + (short)2 equals (longest)2, the triangle has a right angle.

  1. Find the longest side. If the triangle is right-angled, the longest side must be the hypotenuse, so it goes on its own on one side of the equation.
    Sides in order: 7, 24, 25. Longest = 25 cm.
  2. Square the two shorter sides and add. This is the left-hand side of Pythagoras’ theorem.
    72 + 242 = 49 + 576 = 625
  3. Square the longest side. This is the right-hand side.
    252 = 625
  4. Compare the two answers. If they are equal, Pythagoras’ theorem works, so the triangle is right-angled. If not, it is not.
    625 = 625. They are equal, so the triangle is right-angled.

Check. Sides that fit Pythagoras’ theorem are called a Pythagorean triple. This triple cannot be made smaller, so it is a basic one. ✓

Watch out. Add the squares of the two shorter sides and compare with the square of the longest side. Do not add the longest side by mistake.

Worked example 20.2.2
A triangle has sides of length 6 cm, 7 cm and 10 cm. Is it a right-angled triangle? You must show your working.
6 cm7 cm10 cm

Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.

Method. Square all three sides. If (short)2 + (short)2 equals (longest)2, the triangle has a right angle.

  1. Find the longest side. If the triangle is right-angled, the longest side must be the hypotenuse, so it goes on its own on one side of the equation.
    Sides in order: 6, 7, 10. Longest = 10 cm.
  2. Square the two shorter sides and add. This is the left-hand side of Pythagoras’ theorem.
    62 + 72 = 36 + 49 = 85
  3. Square the longest side. This is the right-hand side.
    102 = 100
  4. Compare the two answers. If they are equal, Pythagoras’ theorem works, so the triangle is right-angled. If not, it is not.
    85 ≠ 100. They are not equal, so the triangle is not right-angled.

Check. If it were right-angled the longest side would be √(36 + 49) = 9.21954…, but it is 10. ✓ The longest side is longer than that, so the biggest angle is more than 90°.

Watch out. Add the squares of the two shorter sides and compare with the square of the longest side. Do not add the longest side by mistake.

Worked example 20.2.3
A ladder 5.6 m long leans against a vertical wall. The foot of the ladder is 1.4 m from the base of the wall, on horizontal ground. How far up the wall does the ladder reach? Give your answer correct to 2 decimal places.
1.4 mhladder 5.6 m

Key words. The hypotenuse is the longest side of a right-angled triangle; it is always opposite the right angle. Squaring a number means multiplying it by itself. The square root (√) undoes squaring.

Method. To find a shorter side, square the hypotenuse and the known side, subtract, then take the square root.

  1. Draw a sketch. The wall is vertical and the ground is horizontal, so they make a right angle. The ladder is the hypotenuse because it is opposite that right angle.
    Ladder = 5.6 m (hypotenuse). Foot of ladder to the wall = 1.4 m. Height up the wall = h.
  2. Spot the hypotenuse. The hypotenuse is opposite the right angle and is the longest side. This time it is given, and we are looking for one of the two shorter sides.
    Hypotenuse = 5.6 m. The other known side = 1.4 m. The side we want is h.
  3. Write down Pythagoras’ theorem and put the numbers in. The hypotenuse is always the one on its own on the right-hand side: (shorter)2 + (shorter)2 = (hypotenuse)2.
    a2 + b2 = c2
    h2 + 1.42 = 5.62
  4. Square the known numbers. Squaring means multiplying the number by itself.
    1.42 = 1.4 × 1.4 = 1.96
    5.62 = 5.6 × 5.6 = 31.36
    h2 + 1.96 = 31.36
  5. Subtract 1.96 from both sides. We want h2 on its own. Taking 1.96 away from both sides keeps the equation balanced.
    h2 = 31.36 − 1.96 = 29.4
  6. Take the square root. We want h, not h2.
    h = √29.4 = 5.42217…
  7. Round to 2 d.p.. The question tells us how accurate to be.
    h = 5.42 m

Check. Put it back: 5.422 + 1.42 = 29.38 + 1.96 = 31.34, which is close to 5.62 = 31.36. ✓ The answer is shorter than the hypotenuse 5.6.

Watch out. When you are finding a shorter side you subtract. A shorter side must always be smaller than the hypotenuse; if your answer is bigger, you added by mistake.

Worked example 20.2.4
For any whole numbers m and n with m > n, show that m2 − n2, 2mn and m2 + n2 are the sides of a right-angled triangle. Then use m = 3 and n = 2 to make a triple.

Key words. A Pythagorean triple is a set of three whole numbers a, b, c with a2 + b2 = c2. To show that means to give every step of a proof.

Method. The longest side should be m2 + n2. Show that the two shorter sides squared and added equal its square, using algebra.

  1. Square the first short side. Use (p − q)2 = p2 − 2pq + q2 with p = m2 and q = n2.
    (m2 − n2)2 = m4 − 2m2n2 + n4
  2. Square the second short side. Square both the 2 and the mn.
    (2mn)2 = 4m2n2
  3. Add the two squares. Collect like terms: −2m2n2 + 4m2n2 = 2m2n2.
    m4 − 2m2n2 + n4 + 4m2n2 = m4 + 2m2n2 + n4
  4. Square the longest side. Use (p + q)2 = p2 + 2pq + q2.
    (m2 + n2)2 = m4 + 2m2n2 + n4
  5. Compare. The two results are the same, so Pythagoras’ theorem holds and the triangle has a right angle.
    (m2 − n2)2 + (2mn)2 = (m2 + n2)2
  6. Make a triple with m = 3 and n = 2. Substitute into the three expressions.
    32 − 22 = 9 − 4 = 5
    2 × 3 × 2 = 12
    32 + 22 = 9 + 4 = 13
    The triple is 5, 12, 13.

Check. 52 + 122 = 25 + 144 = 169 = 132. ✓

Watch out. Expand brackets carefully: (m2 − n2)2 is not m4 − n4.

Watch out

When testing for a right angle, put the longest side on its own. A common mistake is to add the square of the longest side to another one. Also, “close enough” is not good enough: 6, 7, 10 has 36 + 49 = 85, which is not 100, so there is no right angle.

Exercise 20C  Right-angled triangles in problems

Calculator allowed
1
FIs each triangle right-angled? Show your working. All the lengths are in centimetres.
(a)
3 cm4 cm5 cm
(b)
5 cm12 cm13 cm
(c)
6 cm8 cm11 cm
(d)
9 cm12 cm15 cm
(e)
7 cm8 cm12 cm
(f)
20 cm21 cm29 cm
2
FEach triangle is right-angled and the lengths are in centimetres. Find the missing length x (a whole number).
(a)(3 whole numbers 8, 15, x)
8 cm15 cmx
(b)(9, x, 41)
x9 cm41 cm
(c)(x, 21, 29)
21 cmx29 cm
(d)(12, 16, x)
12 cm16 cmx
3
FA builder marks out the corner of a patio. She measures 0.9 m along one edge and 1.2 m along the other edge. The distance between the two marks is 1.5 m. Is the corner a right angle? You must show your working.
ABC0.9 m1.2 m1.5 m
4
PSA ladder reaches 5.2 m up a vertical wall. The foot of the ladder is 1.5 m from the wall. Work out the length of the ladder, correct to 2 decimal places.
1.5 m5.2 mL
5
PSA boat sails 24 km due west and then 10 km due south. Work out the distance of the boat from its starting point.
24 km10 kmdStartWEnd
6
PSThe diagonals of a rhombus are 16 cm and 12 cm long. Work out the length of one side of the rhombus, x.
x16 cm12 cm
7
PSTwo cyclists leave the same place at the same time. One cycles north at 12 km/h and the other cycles east at 5 km/h. How far apart are they after 2 hours?
24 km10 kmdStartC1C2
8
PSA joiner checks that a door frame is square. The frame is 0.9 m wide and 2.1 m tall, and a diagonal measures 2.3 m. Is the frame exactly square (right-angled)? Give a reason using Pythagoras’ theorem.
ABC0.9 m2.1 mdiagonal 2.3 m
9
PSA vertical pole is 6 m tall. A wire 10 m long runs from the top of the pole to a point on level ground. How far is that point from the foot of the pole?
dpole 6 mwire 10 m
10
RRaj says: “This triangle is not right-angled because 8 + 15 = 23, and 23 is not 16.” Explain why Raj’s method is wrong. Then use the correct method to decide whether the triangle is right-angled.
ABC8 cm15 cm16 cm
11
RChallenge. Jada says: “If I double every length of a right-angled triangle, the new triangle is still right-angled.” Use the triangle 5 cm, 12 cm, 13 cm to show that Jada is right for this triangle (the doubled triangle is drawn). Then explain why this must be true for any right-angled triangle.
ABC10 cm24 cm26 cm
12
RThe sides of a right-angled triangle are x cm, (x + 7) cm and (x + 8) cm, as shown. Form an equation and solve it to find x.
x(x + 7)(x + 8)

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