Question

20. Consider the length of a triangular park as in the figure
shown below. AC is 24 m. The joint D is the middle point
of length AC and the length of DB is 9 m. Vikram jogs
from point A to point B and then from point B to point C.
What is the total distance he covers while jogging? If he
returns to the starting point A from C, what is the total
distance covered by him?​

Answer 1

answer:54m.

explanation:

Given,

AC=24m, BD=9m

and D is the mid point of AC

so, AD=DC=12m .

This leaves us with two triangles namely,

∆ADB and ∆CDB.

Using pythogoras rule for ∆ADB we get,

AB^2 =AD^2 + BD^2

=>12^2 + 9^2 = 144 + 81 =225

So AD^2 = 225

then AD= 15m

Similarly perform this for ∆CDB , we get CB = 15m

As it was asked total diatance travelled, that means he is asking about perimeter of that particular triangle.

So perimeter of ∆ABC = AB+BC+AC

=>15 +15 +24=54m.