<h2><u>Answers:</u></h2>
These problems can be solved by the Thales’s Theorem, which states:
<em>Two triangles are similar when they have equal angles and proportional sides </em>
In addition, i<u>f two triangles are similar, their angles are similar as well</u>. This means that the relation between two sides of the big triangle is equal to the relation between two sides of the small triangle, as follows:
Knowing this, lets’s begin with the answers:
1. For this first problem, see the <u>first figure attached</u>. There are shown the two triangles, and we are told both are similar, this means (according the prior explanation above) that we can use the Thale’s Theorem to find
.
Therefore, we can establish a relation between two sides of the big triangle which is equal to the relation between two sides of the small triangle; in this case let’s use
and
for the first triangle and
and
for the second:
![\frac{C}{B}=\frac{c}{b}](https://tex.z-dn.net/?f=%5Cfrac%7BC%7D%7BB%7D%3D%5Cfrac%7Bc%7D%7Bb%7D)
Now we find
:
Finally, the value of
is 12 units.
2. For this problem, see <u>the second figure attached</u>. In order to find the values of the segments BE and EC, we will use two sides of each triangle (ABE and DCE) according to the Thale’s Theorem:
<u>For the segment BE:</u>
![\frac{BE}{10}=\frac{x+3}{4}](https://tex.z-dn.net/?f=%5Cfrac%7BBE%7D%7B10%7D%3D%5Cfrac%7Bx%2B3%7D%7B4%7D)
![BE=\frac{(x+3)(10)}{4}](https://tex.z-dn.net/?f=BE%3D%5Cfrac%7B%28x%2B3%29%2810%29%7D%7B4%7D)
Simplifying:
>>>>>This is the length of the segment BE
<u>For the segment EC:</u>
![\frac{EC}{4}=\frac{2x+10}{10}](https://tex.z-dn.net/?f=%5Cfrac%7BEC%7D%7B4%7D%3D%5Cfrac%7B2x%2B10%7D%7B10%7D)
![EC=\frac{(2x+10)(4)}{10}](https://tex.z-dn.net/?f=EC%3D%5Cfrac%7B%282x%2B10%29%284%29%7D%7B10%7D)
Simplifying:
>>>>>This is the length of the segment EC
3. For this problem, see the <u>third figure attached:</u>
![\frac{12 inches}{8 inches}=\frac{x}{7 inches}](https://tex.z-dn.net/?f=%5Cfrac%7B12%20inches%7D%7B8%20inches%7D%3D%5Cfrac%7Bx%7D%7B7%20inches%7D)
Solving for
:
![x=\frac{(12 inches)(7 inches)}{8 inches}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%2812%20inches%29%287%20inches%29%7D%7B8%20inches%7D)
Finally:
>>>>>This is the height of the smaller sail
4. For this problem, see the <u>fourth figure attached.</u>
![\frac{80 inches}{45 inches}=\frac{x}{40 inches}](https://tex.z-dn.net/?f=%5Cfrac%7B80%20inches%7D%7B45%20inches%7D%3D%5Cfrac%7Bx%7D%7B40%20inches%7D)
Solving for
:
![x=\frac{(80 inches)(40 inches)}{45 inches}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%2880%20inches%29%2840%20inches%29%7D%7B45%20inches%7D)
>>>>>This is the height from the floor to the son's hand