Two triangles worth knowing by heart
Tonight's idea
Two right triangles turn up so often that their sides are worth knowing by heart. Each comes from cutting a simple shape in half.
Half a square. Cut a square of side 1 along its diagonal. Each half has angles 45°, 45°, 90° and two legs of length 1. Pythagoras gives the diagonal: \(1^2 + 1^2 = 2\), so it is \(\sqrt2\).
Half an equilateral triangle. Cut an equilateral triangle of side 2 down its line of symmetry. Each half has angles 30°, 60°, 90°. Its hypotenuse is 2 and its short leg is half the base, 1. Pythagoras gives the other leg: \(2^2 - 1^2 = 3\), so it is \(\sqrt3\).
45°-45°-90°: sides \(1 : 1 : \sqrt2\). 30°-60°-90°: sides \(1 : \sqrt3 : 2\), with the 1 opposite the 30° angle.
Any triangle with the same angles is a scaled copy, so multiply all three sides by the same number. Leave the root in your answer: that is exact, and a decimal is only rounded.
Watch one: a ladder at 60°
A 6 m ladder leans against a wall and makes 60° with the ground. How high does it reach?
Ladder, ground and wall make a 30°-60°-90° triangle, with the ladder as hypotenuse. The pattern's hypotenuse is 2, so the scale factor is \(6 \div 2 = 3\).
Foot of the ladder to the wall: \(1 \times 3 = 3\) m. Height: \(\sqrt3 \times 3 = 3\sqrt3 \approx 5.20\) m.
Check with Pythagoras: \(3^2 + (3\sqrt3)^2 = 9 + 27 = 36 = 6^2\).
Where you met Pythagoras: the grade 8 sheet Pythagoras, distances, and the diagonal of a box.
Your turn
- (a) A square has side 9 cm. Give its diagonal exactly, then to one decimal place. (b) A square has diagonal 12 cm. Find its side exactly.
- A triangle has angles 30°, 60°, 90°. (a) Its hypotenuse is 10. Find the other two sides. (b) In another, the longer leg is 6. Find the short leg and the hypotenuse.
- An equilateral triangle has side 6 cm. (a) Find its height. (b) Find its area. (c) Find the area when the side is \(s\).
- A hexagonal nut has six equal sides of 4 mm. A regular hexagon is six equilateral triangles meeting at the centre. A spanner grips two opposite flat sides. How wide must its jaws be? How far apart are opposite corners?
Stretch: both triangles at once
A triangle has angles 45°, 60° and 75°. The side joining the 45° and 60° corners has length \(1 + \sqrt3\). Drop the height from the 75° corner: it splits the triangle into our two special ones. Find the other two sides and the area.
Answers
- (a) \(9\sqrt2 \approx 12.7\) cm. (b) Side \(s\) has \(s\sqrt2 = 12\), so \(s = \dfrac{12}{\sqrt2} = 6\sqrt2 \approx 8.49\) cm. Check: \(6\sqrt2 \times \sqrt2 = 12\).
- (a) Scale factor 5: short leg 5, longer leg \(5\sqrt3 \approx 8.66\). (b) The scale factor \(k\) has \(k\sqrt3 = 6\), so \(k = \dfrac{6}{\sqrt3} = 2\sqrt3\). Short leg \(2\sqrt3 \approx 3.46\), hypotenuse \(4\sqrt3 \approx 6.93\).
- (a) Half of it is a 30°-60°-90° triangle with scale factor 3: height \(3\sqrt3 \approx 5.20\) cm. (b) \(\tfrac12 \times 6 \times 3\sqrt3 = 9\sqrt3 \approx 15.6\) cm². (c) Height \(\dfrac{\sqrt3}{2}s\), so area \(\tfrac12 \times s \times \dfrac{\sqrt3}{2}s = \dfrac{\sqrt3}{4}s^2\). With \(s = 6\) this is \(9\sqrt3\) again.
- Each of the six triangles has side 4 and height \(2\sqrt3\). Flat side to flat side is two heights: \(4\sqrt3 \approx 6.93\) mm, so a 7 mm spanner fits. Corner to corner is two sides: 8 mm.
Stretch. Let the height be \(h\). On the 45° side the foot of the height is \(h\) from the corner. On the 60° side it is \(\dfrac{h}{\sqrt3}\). So \(h + \dfrac{h}{\sqrt3} = 1 + \sqrt3\), which gives \(h = \sqrt3\). The side from the 45° corner is \(\sqrt3 \times \sqrt2 = \sqrt6\). The side from the 60° corner is 2. Area \(= \tfrac12(1 + \sqrt3)\sqrt3 = \dfrac{3 + \sqrt3}{2} \approx 2.37\).