Question

Joshua is 1.45 meters tall. At 2 p.m., he measures the length of a tree's shadow to be 31.65 meters. He stands 26.2 meters away from the tree, so that the tip of his shadow meets the tip of the tree's shadow. Find the height of the tree to the nearest hundredth of a meter .

Answer 1

Answer:

The height of the tree=8.42 m

Step-by-step explanation:

We are given that

Height of Joshua, h=1.45 m

Length of tree's shadow, L=31.65 m

Distance between tree and Joshua=26.2 m

We have to find the height of the tree.

BC=26.2 m

BD=31.65m

CD=BD-BC

CD=31.65-26.2=5.45 m

EC=1.45 m

All right triangles are similar .When two triangles are similar then the ratio of their corresponding sides are equal.

$\triangle ABD\sim \triangle ECD$

$\frac{AB}{EC}=\frac{BD}{CD}$

Substitute the values

$\frac{AB}{1.45}=\frac{31.65}{5.45}$

$AB=\frac{31.65\times 1.45}{5.45}$

$AB=8.42m$

Hence, the height of the tree=8.42 m