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PtichkaEL [24]
3 years ago
9

HELPPPP JUST 2 QUESTIONS BASED ON PYTHAGOREAN THEOREM AND IM CONFUSED HELPPPP​

Mathematics
1 answer:
Anna007 [38]3 years ago
4 0

9514 1404 393

Answer:

  3. no; does not have the correct ratio to other sides

  4. (1; right), (2; obtuse), (3; right)

Step-by-step explanation:

3. There are a couple of ways you can go at this.

a) the ratio of sides of a 30°-60°-90° right triangle (half an equilateral triangle) is 1 : √3 : 2. Here, the ratio of sides is ...

  (√30/2) : √15 : √30 = 1 : √2 : 2 . . . . MQ is <em>not</em> the height

b) the height is perpendicular to the base, so if MQ were the height, the sides of the triangle would satisfy the Pythagorean theorem:

  LQ² +MQ² = LM²

  (√30/2)² +(√15)² = (√30)²

  30/4 +15 = 30

  22.5 = 30 . . . . NOT TRUE

The length MQ cannot be the height of ∆LMN. (It is too short.)

__

4. To see if these lengths form a right triangle, you can test them in the Pythagorean theorem relation. In the attached we have reformulated the equation so it gives a value of 0 if the triangle is a right triangle:

  c² = a² +b² . . . . . Pythagorean theorem

  c² -a² -b² = 0 . . . . rewritten

If the value of c² -a² -b² is not zero, then the side lengths do not form a right triangle. The attached shows this math performed by a graphing calculator. The results are ...

  1. right triangle
  2. not a right triangle — long side is too long, so the triangle is obtuse
  3. right triangle

_____

<em>Additional comment</em>

The attached table demonstrates an artifact of computer arithmetic. Not all numbers can be represented exactly in a computer, so a difference that is expected to be zero using exact arithmetic may be slightly different from zero when computed by a computer or calculator. Here, we see a difference of about 6×10^-14 when zero is expected. On most calculators (with 10, or 12 displayed digits), this would be displayed as 0.

The same calculation done "by hand" gives ...

  (√425)² -5² -20² = 425 -25 -400 = 0 . . . Triangle 1 is a <em>right triangle</em>

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