pole c is 200 ft.

pole d is 150 ft.

Dist b/w these poles is 250 ft.

Assume grain is place at dist x from pole c.

so that now the dist b/w c and d is x + (250-x)

now since the speed of the bird is same and they both reach at same time. The Dist travelled by them will be same. i.e. the hypotenues will be same.

So 200^2 + x^2 150^2 + (250-x)^2

solving this x 90 ft.

Hence Dist b/w c-pole and grain is 90 ft.

pole c is 200 ft.

pole d is 150 ft.

Dist b/w these poles is 250 ft.

Assume grain is place at dist x from pole c.

so that now the dist b/w c and d is x + (250-x)

now since the speed of the bird is same and they both reach at same time. The Dist travelled by them will be same. i.e. the hypotenues will be same.

So 200^2 + x^2 150^2 + (250-x)^2

solving this x 90 ft.

Hence Dist b/w c-pole and grain is 90 ft.