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melisa1 [442]
3 years ago
10

The triangles are similar. The area of the larger triangle is 1200 cm?

Mathematics
2 answers:
Wittaler [7]3 years ago
7 0

Answer:

I would say 75cm^2.

borishaifa [10]3 years ago
4 0

Answer:

75 cm^2

Step-by-step explanation:

We have the lengths of two corresponding sides. The scale factor from the large triangle to the small triangle is 16/64 = 1/4. Each side of the smaller triangle is 1/4 times the length of the corresponding side of the large triangle.

For the areas, the scale factor is the square of the scale factor of the lengths. Area scale factor = 1/4^2 = 1/16.

The area of the small triangle is 1/16 the area of the large triangle.

area of small triangle = (area of large triangle) * (area scale factor)

area of small triangle = 1200 cm^2 * 1/6 = 1200 cm^2/16

area of small triangle = (1200 cm^2)/16 = 75 cm^2

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Step-by-step explanation:

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How would I find a? What formula would I use?
xenn [34]

Answer:

  You can use either of the following to find "a":

  • Pythagorean theorem
  • Law of Cosines

Step-by-step explanation:

It looks like you have an isosceles trapezoid with one base 12.6 ft and a height of 15 ft.

I find it reasonably convenient to find the length of x using the sine of the 70° angle:

  x = (15 ft)/sin(70°)

  x ≈ 15.96 ft

That is not what you asked, but this value is sufficiently different from what is marked on your diagram, that I thought it might be helpful.

__

Consider the diagram below. The relation between DE and AE can be written as ...

  DE/AE = tan(70°)

  AE = DE/tan(70°) = DE·tan(20°)

  AE = 15·tan(20°) ≈ 5.459554

Then the length EC is ...

  EC = AC - AE

  EC = 6.3 - DE·tan(20°) ≈ 0.840446

Now, we can find DC using the Pythagorean theorem:

  DC² = DE² + EC²

  DC = √(15² +0.840446²) ≈ 15.023527

  a ≈ 15.02 ft

_____

You can also make use of the Law of Cosines and the lengths x=AD and AC to find "a". (Do not round intermediate values from calculations.)

  DC² = AD² + AC² - 2·AD·AC·cos(A)

  a² = x² +6.3² -2·6.3x·cos(70°) ≈ 225.70635

  a = √225.70635 ≈ 15.0235 . . . feet

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