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8090 [49]
2 years ago
5

PLEASE HELP!!!!!

Mathematics
2 answers:
Debora [2.8K]2 years ago
8 0

The correct answer is; "△DEF is not a right triangle because no two sides are perpendicular." - I took the test.

Alenkasestr [34]2 years ago
7 0
<span>â–łDEF is not a right triangle because no two sides are perpendicular. There are two ways to determine if DEF is a right triangle. You can use the pythagorean theorem to see of a^2 + b^2 = c^2, or you can calculate the slope of each pair of lines to see if their product is -1. I'll demonstrate both methods, but given the available options, I suspect that the product of the slopes is the one you need. Slope method. Slope of DE = (1-(-1))/(-4-3) = 2/-7 = -2/7 Slope of EF = (-1-(-4))/(3-(-1)) = 3/4 Slope of DF = (1-(-4))/(-4-(-1)) = 5/-3 = -5/3 Since we're looking for -1 as a product and we have 2 negative slopes and 1 positive slope, we only need to check DE against EF and EF against DF. So -2/7 * 3/4 = -6/28 = -3/14. This isn't -1, so not perpendicular. 3/4 * -5/3 = -15/12 = -5/4. Still not -1, so not perpendicular. Since no sides of perpendicular to any other sides, DEF is not a right triangle. Now for the pythagorean theorem method. First, let's take a look at the length of the 3 sides of DEF. Actually, we just need to look at the square of the lengths of each side. So square of length DE = (-4 - 3)^2 + (1 - (-1))^2 = -7^2 + 2^2 = 49 + 4 = 53. square of length EF = (3-(-1))^2 + (-1-(-4))^2 = 4^2 + 3^2 = 16+9 = 25 square of length DF = (-4-(-1))^2 + (1-(-4))^2 = -3^2 + 5^2 = 9 + 25 = 34 The longest side is DE, so lets add up the squares of the other 2 sides to see if they equal the square of DE. 25 + 34 = 59. And 59 is not equal to 53. So we know that triangle DEF is NOT a right triangle. Now let's look at the options and see what's correct. â–łDEF is not a right triangle because no two sides are perpendicular. * This is a true statement. So it's the correct answer. The remaining 3 other options all claim that one side is perpendicular to another and as such are incorrect.</span>
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