Question

22.P and Q start walking in opposite directions P covers 3 km and Q covers 4 km. Then P turns right and walks 6 km and Q turns left and walks 8 km. How far cach is from the starting point?(use Pythagoras theorem)​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer:

Therefore P is 6.7 km away from starting point.

Therefore Q is 8.9 km away from starting point.

Step-by-step explanation:

P and Q walks in opposite directions. So let's say P walks towards North(N), and Q walks towards South(S).

So when the turn turn left/right they turn towards east/west making it a 90 degree angle.

The path will be like in figure 1 (attached), where, A is P's ending position, B is where P takes a right, C is where Q takes a left, and D is Q's ending position and O is where both of them started walking

Join A and O, and D and O with imaginary lines (Distance of each of them from starting point). Now we have 2 right angled triangles ABO and DCO.

AB - 6km, BO - 3km, OC - 4km, DC - 8km. The figure is like this now (refer 2nd attached).

Now, according to Pythagoras theorem, in a right angled triangle, square of its Hypotenuse will be equal to the sum of the squares of its base and altitude.   $hypotenuse^{2} = base^{2} + altitude^{2}$

Consider Triangle ABO, P's route.

Hypotenuse - AO, Base - AB = 6km, Altitude - BO = 3km

$=> AO^{2} = 6^{2} + 3^{2}\\=> AO = \sqrt{36 + 9} \\=> AO = \sqrt{45} \\=> AO = 6.7 km$

Therefore P is 6.7 km away from starting point.

Same way...

Consider Triangle DCO, Q's route.

Hypotenuse - DO, Base - DC = 8km, Altitude - OC = 4km

$=> DO^{2} = 8^{2} + 4^{2}\\=> DO = \sqrt{64 + 16} \\=> DO = \sqrt{80} \\=> DO = 8.9 km$

Therefore Q is 8.9 km away from starting point.