7) In an old episode of the TV program McGyver, a planeload of gold was being transported from the SovietUnion to the United States during WWU. The place crashed in the Arctic region. To prevent the "badguys" from getting the gold, the pilot and copilot transferred the gold into a cave by stacking it on a door of the crashed plane and dragging the "sled" into the cave. You were led to believe in the episode that
they accomplished the move in one trip. The gold was stacked neatly in the shape of a cube, measuring
about 1 meter on a side. Calculate the weight, in tons, of one cubic meter of gold. The density of gold is
19.4 g/cc. Would it have been possible for the two pilots to accomplish this feat? Would a plane of WWII
vintage be able to carry this much gold? (cc = cubic centimeter, 1cm = 1mL)

Here we need to remember that:

density = mass/volume

Only by using this equation, we will find that the mass of gold is 19,400 kg

In this case, we know that the gold is in the shape of a cube with side length of one meter.

Remember that for a cube of side length L, the volume is:

V = L^3

Then in this case, the volume of the cube of gold will be:

V = (1m)^3  = 1m^3

And we know that the density of the gold is:

density = 19.4 g/cc

cc = cm^3

density = 19.4 g/cm^3

Notice that we have different units of volume, so we would want to write the volume in cm^3

We know that:

1m^3 = (1×10^6)cm^3

Then the volume of the cube of gold is just:

V = (1×10^6)cm^3 = 1,000,000 cm^3

Now let's return to the density equation:

density = mass/volume

Solving for the mass, we get:

density×volume = mass

Now we can replace the two things in the left side:

(19.4 g/cm^3)×(1,000,000 cm^3) = mass = 19,400,000 g

Now remember that:

1000k = 1kg

Then we can rewrite:

mass = 19,400,000 g = (19,400,000/1000)kg = 19,400 kg

So the cube of gold has a mass of  19,400 kg, there is no way that two persons could carry this much gold into a cave.

if you want to learn more, you can read:

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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