9.9 Further Applications

The Baudhayana–Pythagoras theorem is among the handiest tools in all of geometry. Here is a classic riverside puzzle in the same spirit as the old Sanskrit verse-problems.

The Bending Bamboo > A bamboo shoot rises straight out of a pond, standing 2 hand-spans above the water. A gust of wind bends it over until its tip just brushes the surface, 8 hand-spans away from where it grew. How deep is the pond at the base of the bamboo?

Worked Example

Find the depth of the pond.

Let x be the length of bamboo below the surface, that is the depth we want. Because the shoot rises 2 spans above the water, the whole bamboo has length x + 2. When the wind bends it down, the bamboo (length x+2) becomes the hypotenuse, the depth x is one leg, and the horizontal reach 8 is the other leg. By the theorem: 8^2 + x^2 = (x+2)^2. Expand the right side: 64 + x^2 = x^2 + 4x + 4. Cancel x^2 from both sides: 64 = 4x + 4 \;\Rightarrow\; 4x = 60 \;\Rightarrow\; x = 15. The pond is 15 spans deep. (What looked like “too little information” turned out to be plenty!)

Figure it Out: Applications

Practice

  1. Find the diagonal of a square of side 8 cm.
  2. Find the missing side in each right triangle:  (i) legs 9 and 14  (ii) leg 6, hypotenuse 10  (iii) leg 9, hypotenuse 41  (iv) leg 9, hypotenuse \sqrt{130}  (v) leg 16, hypotenuse 20  (vi) leg 5, hypotenuse \sqrt{74}.
  3. Find the side of a rhombus whose diagonals are 18 and 80 units.
  4. Is the hypotenuse always the longest side of a right triangle? Justify your answer.
  5. True or False: every Baudhayana triple is either a primitive triple or a scaled copy of a primitive triple.
  6. Give 5 examples of rectangles whose sides and diagonals are all whole numbers.
  7. Construct a square whose area equals the difference of the areas of squares of sides 5 and 13.
  8. On a square dot grid, can you draw a square of area (a) 2, (b) 3, (c) 4, (d) 5 square units? If the grid runs on forever, which integer areas are achievable?
  9. Find the area of an equilateral triangle of side 8. (Hint: an altitude bisects the base; get the height from the theorem.)
  1. Diagonal = \sqrt{8^2 + 8^2} = \sqrt{128} = 8\sqrt2 \approx \mathbf{11.31} cm.
    1. \sqrt{9^2+14^2} = \sqrt{277} \approx \mathbf{16.6}.  (ii) \sqrt{10^2-6^2} = \sqrt{64} = \mathbf{8}.  (iii) \sqrt{41^2-9^2} = \sqrt{1600} = \mathbf{40}.  (iv) \sqrt{130-9^2} = \sqrt{49} = \mathbf{7}.  (v) \sqrt{20^2-16^2} = \sqrt{144} = \mathbf{12}.  (vi) \sqrt{74-5^2} = \sqrt{49} = \mathbf{7}.
  3. A rhombus’s diagonals bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs \tfrac{18}{2}=9 and \tfrac{80}{2}=40: side = \sqrt{9^2+40^2} = \sqrt{1681} = \mathbf{41} units.
  4. Yes. Since c^2 = a^2 + b^2 with a, b > 0, we have c^2 > a^2 and c^2 > b^2, so c exceeds each leg. The hypotenuse (opposite the right angle, the biggest angle) is always the longest side.
  5. True. Every triple either has no common factor (primitive) or, after dividing out its common factor, becomes a primitive triple that scales back up.
  6. Use any triple as (length, breadth, diagonal): (3,4,5), (6,8,10), (5,12,13), (8,15,17), (9,12,15). Each gives a rectangle with whole-number sides and whole-number diagonal.
  7. The wanted square has area 13^2 - 5^2 = 169 - 25 = 144, so its side is \sqrt{144} = \mathbf{12}. Make a right triangle with hypotenuse 13 and one leg 5; the other leg is \sqrt{169-25} = \sqrt{144} = 12, and the square on that leg has area 144.
  8. Tilted squares on the grid have area a^2 + b^2 for whole numbers a, b (the legs of the tilt). So 2 (1^2+1^2), 4 (2^2+0^2), and 5 (2^2+1^2) are possible; 3 is not (no two squares sum to 3). In general the reachable integer areas are exactly those of the form a^2 + b^2.
  9. Drop an altitude: it splits the base into halves of 4, and the height is \sqrt{8^2 - 4^2} = \sqrt{48} = 4\sqrt3. Area = \tfrac12 \times 8 \times 4\sqrt3 = 16\sqrt3 \approx \mathbf{27.71} sq. units.