Question

A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city wants to construct three paths from the corner point opposite to the longest side such that these paths divide the longest side into four equal segments. Determine the sum of the squares of the lengths of the three paths. 

(a) 4000 m 

(b) 4800 m 

(c) 5600 m 

(d) 6400 m

Answer 1

(c) 5600 m

CE, CD and CF are the required paths such that the longest side of the park AB is divided into four equal segments. 

AE = ED = DF = FB = 20 m. 

Let AC = b, BC = a 

Then, by Apollonius theorem in ΔACD 

AC² + CD² = 2(CE² + 20²) 

⇒ \(\frac{1}{2}\) (b² + CD²) = CE² + 20²             …(i) 

Similarly in ΔCDB, (CB² + CD²) = 2(CF² + 20²) 

⇒ CF² + 20² = \(\frac{1}{2}\)(a² + CD²)            …(ii) 

Adding (i) and (ii), 

CE² + CF² + 2 x 20² = \(\frac{1}{2}\) (a² + b² + 2 x CD²)

⇒ CE² + CF² = \(\frac{1}{2}\) (80² + 2 x 40²) - 2 x 20² 

(∵ AC² + CB² = AB² ⇒ a² + b² = 80²)

(∵ CD = 40, ∵ line joining the vertex at the right ∠ to the mid-point of the hypotenuse is half the hypotenuse) 

⇒ CE² + CF² = \(\frac{1}{2}\) (6400 + 3200) - 800 = 4000. 

Now CE² + CF² + CD² = 4000 + 402 = 5600.