ABSTRACT
Thus d = √144 = 12 cm i.e. EF = 12 cm
Problem 2. Two aircraft leave an airfield at the same time. One travels due north at an average speed of 300 km/h and the other due west at an average speed of 220 km/h. Calculate their distance apart after 4 hours
Figure 22.3 After 4 hours, the first aircraft has travelled 4 × 300 = 1,200 km, due north, and the second aircraft has travelled 4 × 220 = 880 km due west, as shown in Fig. 22.3. Distance apart after 4 hour = BC. From Pythagoras’ theorem:
BC2 = 12002 + 8802 = 1440000 + 774400 and
BC = √2214400 Hence distance apart after 4 hours = 1488 km
Now try the following exercise
Exercise 84 Further problems on the theorem of Pythagoras
1. In a triangle CDE, D = 90◦ , CD = 14.83 mm and CE = 28.31 mm. Determine the length of DE. [24.11 mm]
2. Triangle PQR is isosceles, Q being a right angle. If the hypotenuse is 38.47 cm find (a) the lengths of sides PQ and QR, and (b) the value of ∠QPR. [(a) 27.20 cm each (b) 45◦]
3. A man cycles 24 km due south and then 20 km due east. Another man, starting at the same time as the first man, cycles 32 km due east and then 7 km due south. Find the distance between the two men. [20.81 km]
4. A ladder 3.5 m long is placed against a perpendicular wall with its foot 1.0 m from the wall. How far up the wall (to the nearest centimetre) does the ladder reach? If the foot of the ladder is now moved 30 cm further away from the wall, how far does the top of the ladder fall? [3.35 m, 10 cm]
5. Two ships leave a port at the same time. One travels due west at 18.4 km/h and the other due south at 27.6 km/h. Calculate how far apart the two ships are after 4 hours.
