Aju starts from his home and walks x meters in north direction. After that he turns right and walks another x meters. He meets his friends and then reaches back home by the shortest possible path. The area enclosed by the path taken by Aju for the whole journey is 50 m2. What is x?
10m
Let the home be at A.
In the figure below, AC is the shortest path from C to AArea of ∆ABC = 50 m2 (given)
In ∆ABC and ∆ADC
AB = AD (sides of a square)
BC = DC (sides of a square)
AC = AC (common)
By S-S-S congruence condition.
∆ABC ≅ ∆ADC
∴ Area(∆ABC) = Area(∆ADC)
So, Area(ABCD) = 2 x Area(∆ABC)
= 2 x 50 m2 = 100 m2
∴ x × x = 100 m2 ⇒ x2 = 100 m2 ⇒ x = 10 m
Alternate:
Area of triangle = b× h2 = x22 = 50
⇒x2 = 100
⇒x = 10 m