(35/7)*4 = 20 Ft.
As the sun is far enough away, the rays of light are effectively parallel. This produces similar triangles with the ratio of sides the same in each case. As the shadow of the post is 12 ft and the shadow of the tree is 24 ft, the sides of the triangle of the tree are double that of the post. Assuming the post is parallel to the tree, the tree's height is twice the height of the post → tree = 12 ft × 2 = 24ft high.
The height of the tree is in direct proportion to the pole and its shadow
A 1 foot shadow I think.
It works out as 12 feet and 4 inches in height
If the shadow of a 6-ft person is 4-ft long, then in this place at this moment, all shadowsare 2/3 the length of the vertical object that casts them.The 9-ft shadow therefore 2/3 the height of the tree. The height is (9)/(2/3) = (9 x 3/2) = 13.5-ft.-----------------------------------------(9/4)*6=13.5 ft.
Ratio of object to its shadow is the same. So if T is the height of the tree, then T/21 = 4/6 So T = 21*4/6 = 84/6 = 14 feet
As the sun is far enough away, the rays of light are effectively parallel. This produces similar triangles with the ratio of sides the same in each case. As the shadow of the post is 12 ft and the shadow of the tree is 24 ft, the sides of the triangle of the tree are double that of the post. Assuming the post is parallel to the tree, the tree's height is twice the height of the post → tree = 12 ft × 2 = 24ft high.
The height of the tree is in direct proportion to the pole and its shadow
40 ft
50 feet
A 1 foot shadow I think.
It is 90 feet in height
It works out as 12 feet and 4 inches in height
Ten is to two as 40 is to x, yielding: 200ft.
It depends on the time of day because the angle of the sun will determine the shadow length
That depends on the height of the yardstick whose height has not been given.
The tangent ratios are equal so let the height of the tree be x: x/33 = 7/6 Multiply both sides 33: x = 231/6 = 38.5 Height of tree = 38.5 units of measurement.