Every four years is a leap year. If a year is randomly chosen, what is the probability that it will NOT be a leap year?
1 answer:
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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.)
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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 ...
- right triangle
- not a right triangle — long side is too long, so the triangle is obtuse
- right triangle
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<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>
