In the afternoon, the person (who is 1.8 m tall) casts a shadow that is 10 m. The distance along the ground from the person (H) to the tree (G) is 24 m, and the distance from the tree (G) to the building (F) is 150 m. Calculate the height of the tree and the building. Round answers to the nearest tenth of a meter and show all your work.

In the afternoon the person who is 18 m tall casts a shadow that is 10 m The distance along the ground from the person H to the tree G is 24 m and the distance class=

Respuesta :

Answer:

The height of the tree is 6.12 meters and the height of the building is 33.12 meters.

Step-by-step explanation:

Since the person is 1.8 meters tall, HC = 1.8

And since their shadow is 10 meters long, HD = 10.

We are also given that GH is 24 meters and that FG is 150 meters.

Height of the Tree:

The height of the tree is given by GB.

Again, since m∠BGD = m∠CHD = 90°, ∠BGD ≅ ∠CHD.

Likewise, ∠D ≅ ∠D. So, by AA-Similarity:

[tex]\displaystyle \Delta BGD\sim \Delta CHD[/tex]

Corresponding parts of similar triangles are in proportion. Therefore:

[tex]\displaystyle \frac{GB}{GD}=\frac{HC}{HD}[/tex]

Note that:

[tex]GD=GH+HD[/tex]

Find GD:

[tex]GD=(24)+(10)=34[/tex]

Substitute the known values into the proportion:

[tex]\displaystyle \frac{GB}{34}=\frac{1.8}{10}[/tex]

Cross-multiply:

[tex]10GB=61.2[/tex]

Therefore:

[tex]GB=6.12\text{ meters}[/tex]

The height of the tree is 6.12 meters.

Height of the Building:

The height of the building is given by FA.

Since m∠AFD = m∠CHD = 90°, ∠AFD ≅ ∠CHD.

∠D ≅ ∠D. So, by AA-Similarity:

[tex]\Delta AFD\sim \Delta CHD[/tex]

Corresponding parts of similar triangles are in proportion. Therefore:

[tex]\displaystyle \frac{FA}{FD}=\frac{HC}{HD}[/tex]

Note that:

[tex]FD=FG+GH+HD[/tex]

Find FD:

[tex]FD=(150)+(24)+(10)=184[/tex]

Substitute the known values into the proportion:

[tex]\displaystyle \frac{FA}{184}=\frac{1.8}{10}[/tex]

Cross-multiply:

[tex]10FA=331.2[/tex]

Therefore:

[tex]FA=33.12\text{ meters}[/tex]

The height of the building is 33.12 meters.