Pythagoras, distances, and the diagonal of a box
Tonight's idea
A right triangle has legs \(a\) and \(b\) and a hypotenuse \(c\), the side opposite the right angle. Pythagoras says \(a^2 + b^2 = c^2\). Tonight: one proof you can see, and two uses.
The proof. Take a square of side \(a + b\) and put four copies of the triangle in its corners. The hole in the middle is a tilted square of side \(c\). (Its corners are 90° because the triangle's two sharp angles add to 90°.) Now find the big square's area two ways. As a square: \((a+b)^2 = a^2 + 2ab + b^2\). As four triangles plus the hole: \(4 \times \tfrac12 ab + c^2 = 2ab + c^2\). Same area, so \(a^2 + 2ab + b^2 = 2ab + c^2\), and \(a^2 + b^2 = c^2\).
Distances. Two points on a grid, \((x_1, y_1)\) and \((x_2, y_2)\). The horizontal gap is \(x_2 - x_1\), the vertical gap is \(y_2 - y_1\), and the line joining them is the hypotenuse:
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Squaring kills signs, so point order does not matter.
Boxes. A box with sides \(a, b, c\). Its floor diagonal is \(f = \sqrt{a^2 + b^2}\). That diagonal and the height \(c\) make a second right triangle, so the long diagonal is \(d = \sqrt{a^2 + b^2 + c^2}\).
The grade 6 sheet Pythagoras and square roots gives the two-square version with no algebra.
Watch one: the longest pencil in a 12 × 4 × 3 cm box
Floor first: \(f^2 = 12^2 + 4^2 = 144 + 16 = 160\). Do not take the root yet. Stand the height on it: \(d^2 = f^2 + 3^2 = 160 + 9 = 169\), so \(d = 13\) cm. A 13 cm pencil fits exactly, corner to opposite corner.
The same in one line: \(d = \sqrt{144 + 16 + 9} = \sqrt{169} = 13\). Keeping squares until the last step avoids messy decimals and rounding errors.
Your turn
- (a) A right triangle has legs 8 and 15. Find the hypotenuse. (b) A right triangle has hypotenuse 25 and one leg 7. Find the other leg.
- Find the distance between (a) \((1, 2)\) and \((7, 10)\); (b) \((-3, 4)\) and \((2, -8)\). Sketch the right triangle you are using in (b).
- A 6.5 m ladder leans against a wall with its foot 2.5 m from the wall. (a) How high up the wall does it reach? (b) The foot slides out to 3.9 m from the wall. How far does the top slide down? (c) The foot moved 1.4 m. Why did the top move less?
- (a) A suitcase measures 75 cm by 55 cm by 25 cm. Will a straight 1 m rod fit inside? What is the longest rod that fits? (b) A cube has side \(s\). Show that a face diagonal is \(s\sqrt2\) and the long diagonal is \(s\sqrt3\).
Stretch: a spider on a cube
A spider sits at one corner of a cube-shaped room of side 2 m and wants the opposite corner. It must walk on the walls, floor or ceiling. Unfold two adjacent faces into a flat rectangle and find the shortest route. Compare with flying straight through and with walking along the edges.
Answers
- (a) \(\sqrt{64 + 225} = \sqrt{289} = 17\). (b) \(\sqrt{625 - 49} = \sqrt{576} = 24\).
- (a) Gaps 6 and 8: \(\sqrt{36 + 64} = 10\). (b) Gaps \(5\) and \(-12\); squares 25 and 144: \(\sqrt{169} = 13\). The triangle has a horizontal leg from \(x = -3\) to \(x = 2\) and a vertical leg from \(y = 4\) down to \(y = -8\).
- (a) \(\sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6\) m. (b) New height \(\sqrt{42.25 - 15.21} = \sqrt{27.04} = 5.2\) m, so the top slides down \(0.8\) m. (c) Pythagoras trades squares, not lengths. Near the top of the wall the ladder is steep, and a small change in height makes a big change in the foot's distance; near the floor it is the other way round.
- (a) \(d^2 = 75^2 + 55^2 + 25^2 = 5625 + 3025 + 625 = 9275\), so \(d \approx 96.3\) cm. A 1 m rod does not fit; 96.3 cm is the longest. (b) Face diagonal \(\sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt2\). Long diagonal \(\sqrt{s^2 + s^2 + s^2} = s\sqrt3\).
Stretch. Two adjacent faces unfold into a 2 m by 4 m rectangle, and the two corners are its opposite corners: \(\sqrt{2^2 + 4^2} = \sqrt{20} \approx 4.47\) m. Flying straight through: \(2\sqrt3 \approx 3.46\) m. Along three edges: 6 m. The walking route is straight on the unfolded paper, which is why it is shortest; on the cube it bends once, at the shared edge's midpoint.