# Math Help - Geometry Word Problem

1. ## Geometry Word Problem

Okay, so I need some help with these two:

*8. In the triangle ABC, D and E are the midpoints of AC and BC. The segments AE and BD intersect at F. Show that regions a2 and a4 have equal areas.

(Picture Below)

9. A quarter (a 25 cent piece) is 3/4 inch in diameter and when placed 7 feet from the eye will just block out the disc of the moon. If the diameter of the moon is 2160 miles, how far is the moon from the earth?

2. Hello, A Beautiful Mind!

The second one involves similar triangles.

9. A quarter (a 25-cent piece) is 3/4 inch in diameter and,
when placed 7 feet from the eye, will just block out the disc of the moon.
If the diameter of the moon is 2160 miles, how far is the moon from the earth?
Code:
              A
-           *
:          /|\
:         / | \
:        /  |  \
:       /   |84 \
d      /    |    \
:    B*-----+-----*C
:    /      ¾      \
:   /               \
:  /                 \
-D*-------------------*E
: - - - 2160  - - - :

The eye is at $A.$
The coin is: . $BC = \tfrac{3}{4}$ inch.
The altitude of $\Delta ABC$ is 84 inches.

The diameter of the moon is: . $DE \:=\:2160$ miles.
The distance to the moon is $d$ miles.

Since $\Delta ABC \sim \Delta ADE$, we have: . $\frac{d}{2160\text{ miles}} \;=\;\frac{84\text{ inches}}{\frac{3}{4}\text{ inch}}$

Therefore: . $d \;=\;\frac{(2160)(64)}{\frac{3}{4}} \;=\;241,\!920$ miles.

3. In triangle ABC D and E are mid points of AC and BC, Hence DE is parallel to AB.Hence area ADB = area AEB. In that A2 is common. So area A1 = A3. In triangle ABC BD bisects AC. Therefore area ABD = area DBC.Or A1 + A2 = A3 + A4. But A1 = A3. Hence A2 = A4.