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2 years ago

6

help please i been stuck for 2 hours........... the triangle above is a right-angled isosceles triangle. what would the measurem

ent of x have to be, given that the hypotenuse is 8cm and legs a and b have the same length? please explain your answer

1 answer:

Likurg_2 [28]2 years ago

8 0

Answer:

4√2 or 5.66

Step-by-step explanation:

x² + x² = 8²

=> 2x² = 64

=> x² = 32

=> x = √32 = 4√2 or 5.66

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One equation for this would be

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We start by finding the slope between the two points:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{-12-\frac{-19}{5}}{-2-\frac{6}{5}}
\\
\\=(-12+\frac{19}{5}) \div (-2-\frac{6}{5})
\\
\\=(\frac{-60}{5}+\frac{19}{5}) \div (\frac{-10}{5}-\frac{6}{5})
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\\=\frac{-41}{5} \div \frac{-16}{5}=\frac{-41}{5} \times \frac{-5}{16}=\frac{41}{16}$

A line parallel to this one will have the same slope. We will use point-slope form to write our equation:

$y-y_1=m(x-x_1)
\\
\\y-\frac{-19}{5}=\frac{41}{16}(x-\frac{6}{5})
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\\y+\frac{19}{5}=\frac{41}{16}x- \frac{41}{16} \times \frac{6}{5}
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\\y+\frac{19}{5}=\frac{41}{16}x-\frac{246}{80}
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\\y+\frac{304}{80}=\frac{41}{16}x-\frac{246}{80}
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\\y=\frac{41}{16}x-\frac{246}{80}-\frac{304}{80}
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\\y=\frac{41}{16}x-\frac{550}{80}
\\
\\y=\frac{41}{16}x-\frac{55}{8}$

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