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GCSE Maths · EdexcelChapter 20 · Pythagoras and trigonometry
GCSE Maths · Pythagoras and Trigonometry · Foundation · Lesson 20.3

GCSE Maths: Pythagoras with coordinates, isosceles and equilateral triangles (Foundation)

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

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20.3

Pythagoras with coordinates, isosceles and equilateral triangles

Edexcel 1MA1: G20, G6, G11
Grade 4-6

Distance between two points. To find the distance between two points on a grid, imagine a right-angled triangle between them. The straight line joining the points is the hypotenuse. The horizontal step is the difference of the x-coordinates and the vertical step is the difference of the y-coordinates. Then use Pythagoras. A minus sign in a step does not matter, because squaring turns it into a positive number.

xy-11234567891011-11234567891011A(1, 2)B(7, 10)
A(1, 2) and B(7, 10): the steps are 7 − 1 = 6 across and 10 − 2 = 8 up, so AB2 = 62 + 82 = 100 and AB = 10.

Isosceles triangles. An isosceles triangle has two equal sides, and the angles at the base are equal. If you draw the line of symmetry from the top down to the base, it hits the base at 90° and cuts it exactly in half. That makes two right-angled triangles that are mirror images of each other. Their hypotenuse is one of the equal sides, one shorter side is half the base and the other shorter side is the height of the triangle. Once you know the height you can find the area: ½ × base × height.

13 cm13 cm10 cmh
The height splits the base into two halves of 5 cm. In each right-angled triangle: h2 + 52 = 132.

Equilateral triangles. An equilateral triangle has three equal sides and three 60° angles. It is a special isosceles triangle, so the same trick works. For a side of length s the half-base is s ÷ 2, so the height h satisfies h2 + (s ÷ 2)2 = s2.

ABC8 cm8 cm60°60°h4 cm4 cm
Equilateral triangle of side 8 cm: h2 + 42 = 82.

Key ideas

  • Distance between points: (x step)2 + (y step)2 = distance2.
  • An isosceles triangle splits into two right-angled triangles: use half the base.
  • Area of the triangle = ½ × whole base × height.
  • Equilateral triangle: height2 = side2 − (½ side)2.
Worked example 20.3.1
Work out the distance between the points A(1, 2) and B(7, 10).
xy-11234567891011-11234567891011A(1, 2)B(7, 10)

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. The distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical steps between them.

  1. Draw a right-angled triangle between the two points. The line from one point to the other is the hypotenuse. The two shorter sides are the horizontal step and the vertical step.
    Horizontal step across, vertical step up or down.
  2. Find the horizontal and vertical steps. Subtract the x-coordinates, then subtract the y-coordinates. Only the size of each step matters, so ignore any minus sign.
    Horizontal: 7 − 1 = 6, so the step is 6
    Vertical: 10 − 2 = 8, so the step is 8
  3. Use Pythagoras’ theorem. The distance d is the hypotenuse.
    d2 = 62 + 82 = 36 + 64 = 100
  4. Take the square root. We want d, not d2.
    d = √100 = 10

Check. 6, 8, 10 fits a Pythagorean triple pattern: 36 + 64 = 100. ✓

Watch out. Subtract the coordinates in the same order for x and y. A negative step is fine: squaring removes the minus sign.

Worked example 20.3.2
The points P(−3, 4) and Q(5, −2) are joined by a straight line. Work out the length of PQ.
xy-4-3-2-1123456-3-2-11234567P(−3, 4)Q(5, −2)

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. The distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the horizontal and vertical steps between them.

  1. Draw a right-angled triangle between the two points. The line from one point to the other is the hypotenuse. The two shorter sides are the horizontal step and the vertical step.
    Horizontal step across, vertical step up or down.
  2. Find the horizontal and vertical steps. Subtract the x-coordinates, then subtract the y-coordinates. Only the size of each step matters, so ignore any minus sign.
    Horizontal: 5 − (−3) = 8, so the step is 8
    Vertical: −2 − 4 = −6, so the step is 6
  3. Use Pythagoras’ theorem. The distance d is the hypotenuse.
    d2 = 82 + 62 = 64 + 36 = 100
  4. Take the square root. We want d, not d2.
    d = √100 = 10

Check. 8, 6, 10 fits a Pythagorean triple pattern: 64 + 36 = 100. ✓

Watch out. Subtract the coordinates in the same order for x and y. A negative step is fine: squaring removes the minus sign.

Worked example 20.3.3
An isosceles triangle has two sides of length 13 cm and a base of 10 cm. Work out the area of the triangle.
13 cm13 cm10 cmh

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. An isosceles triangle has two equal sides.

Method. Split the isosceles triangle down the middle into two right-angled triangles, then use Pythagoras’ theorem.

  1. Draw the line of symmetry from the top to the middle of the base. In an isosceles triangle this line is the perpendicular height. It cuts the base in half and splits the triangle into two identical right-angled triangles.
    Half the base = 10 ÷ 2 = 5 cm
  2. Pick one right-angled triangle. Its hypotenuse is the slanting side, 13 cm. Its shorter sides are the half-base and the height h.
    h2 + 52 = 132
  3. Square and subtract. We are finding a shorter side, so we subtract.
    h2 = 169 − 25 = 144
  4. Take the square root. The height is h, not h2.
    h = √144 = 12 cm
  5. Use area of a triangle = ½ × base × height. The base is the whole base, not the half. Use the exact height from the calculator to avoid rounding errors.
    Area = ½ × 10 × 12 = 60
  6. Write the area with its units. Area is measured in square units. The answer is exact, so no rounding is needed.
    Area = 60 cm2

Check. Another way: two small triangles each have area ½ × 5 × 12 = 30.00; doubling gives 60.00, matching 60. ✓

Watch out. Use half the base in Pythagoras’ theorem, but the whole base in the area formula.

Worked example 20.3.4
An equilateral triangle has sides of length 8 cm. Work out its height, correct to 2 decimal places.
ABC8 cm8 cm60°60°h4 cm4 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. An isosceles triangle has two equal sides.

Method. Split the isosceles triangle down the middle into two right-angled triangles, then use Pythagoras’ theorem.

  1. Draw the line of symmetry from the top to the middle of the base. In an isosceles triangle this line is the perpendicular height. It cuts the base in half and splits the triangle into two identical right-angled triangles.
    Half the base = 8 ÷ 2 = 4 cm
  2. Pick one right-angled triangle. Its hypotenuse is the slanting side, 8 cm. Its shorter sides are the half-base and the height h.
    h2 + 42 = 82
  3. Square and subtract. We are finding a shorter side, so we subtract.
    h2 = 64 − 16 = 48
  4. Take the square root. The height is h, not h2.
    h = √48 = 6.92820…
    h = 6.93 cm

Check. The height (6.93) must be shorter than the slanting side (8). Also 6.932 + 42 = 64.02, close to 64. ✓

Watch out. Use half the base in Pythagoras’ theorem, but the whole base in the area formula.

Watch out

For an isosceles triangle, use half the base in Pythagoras’ theorem, but the whole base in the area formula ½ × base × height. Mixing these up is the most common mistake.

Exercise 20D  Coordinates, isosceles and equilateral triangles

Calculator allowed
1
FWork out the distance between each pair of points. Give your answers correct to 1 decimal place when they are not whole numbers.
(a)A(1, 2) and B(4, 6)
xy-11234567-11234567A(1, 2)B(4, 6)
(b)C(0, 3) and D(5, −9)
xy24681012-10-8-6-4-224C(0, 3)D(5, −9)
(c)P(−4, 2) and Q(3, 5)
xy-5-4-3-2-11234-112345678P(−4, 2)Q(3, 5)
(d)R(−6, −1) and S(2, 4)
xy-7-6-5-4-3-2-1123-2-112345678R(−6, −1)S(2, 4)
2
PSOn a map, each grid square is 1 km wide. A town P is at (2, 3) and a town Q is at (14, 8). Work out the straight-line distance from P to Q.
xy24681012142468101214P(2, 3)Q(14, 8)
3
RA(2, 1), B(6, 4) and C(2, 7) are the vertices of a triangle. Show that triangle ABC is isosceles.
xy-112345678-112345678A(2, 1)B(6, 4)C(2, 7)
4
FEach isosceles triangle has the equal sides and the base shown. Work out the perpendicular height h of each triangle.
(a)Equal sides 10 cm, base 12 cm
10 cm10 cm12 cmh
(b)Equal sides 13 cm, base 10 cm
13 cm13 cm10 cmh
(c)Equal sides 17 cm, base 16 cm
17 cm17 cm16 cmh
(d)Equal sides 6.5 cm, base 5 cm
6.5 cm6.5 cm5 cmh
5
FAn isosceles triangle has two sides of length 10 cm and a base of 12 cm. Work out its area.
10 cm10 cm12 cmh
6
FAn equilateral triangle has sides of length 10 cm. Work out its height. Give your answer correct to 2 decimal places.
ABC10 cm10 cm60°60°h5 cm5 cm
7
PSA roof is supported by an isosceles triangular frame. The two sloping beams are each 5.2 m long and the horizontal span is 9.6 m. Work out the height of the roof frame.
5.2 m5.2 m9.6 mh
8
PSThe cross-section of a tent is an isosceles triangle. The sloping sides are 1.9 m long and the base is 2.4 m wide. Work out the height of the tent. Give your answer correct to 2 decimal places.
1.9 m1.9 m2.4 mh
9
PSA road sign is an equilateral triangle of side 50 cm. Work out the area of the sign. Give your answer correct to 1 decimal place.
ABC50 cm50 cm60°60°h25 cm25 cm
10
RZara says: “The distance between A(1, 2) and B(4, 6) is 3 + 4 = 7.” Explain why Zara is wrong and work out the correct distance.
xy-11234567-11234567A(1, 2)B(4, 6)
11
RIsosceles triangle ABC has AB = AC = 17 cm and BC = 16 cm. Show that its area is 120 cm2.
17 cm17 cm16 cmh

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