Question

Suppose that the endpoints of one leg of a 45°-45°-90° triangle are (0, 3) and (4, –1). What is the length of the hypotenuse?

Answer 1

HELP Please

Answer 2

4 – 0 = 4
-1 -3= -4
4 sq + -4 sq = c sq
16 + 16 = c sq
√32 = c
The distance between the two points is √(16+16) = √32
So, the hypotenuse is √32*√2 = √64 = 8
is this right?

Answer 3

I thought distance was same as hypotenuse. Why do you have to multiply by sq root of 2

Answer 4

This was also given. Hint: Remember how the sides relate in a 45°-45°-90° triangle.

Answer 5

I kept getting √32 as my answer.

Answer 6

using theformula of distance we can write:
\[d = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}} = \sqrt {16 + 16} = \sqrt {32} = 4\sqrt 2 \]

Answer 7

Ok - I did that and that is what I thought was the hypotenuse

Answer 8

yes! the length of the hypotenuse is \[\sqrt {32} = 4\sqrt 2 \]

Answer 9

so the length of each side is:
x=4

Answer 10

That is not a choice. Answers include 8 units, 4 units, 10 units or 2 units.
I was told to multiply by 2 and answer was 8 but without an explanation. I don't understand why

Answer 11

I think that your exercise asks for the value of x. So if
\[x\sqrt 2 = 32 = 4\sqrt 2 \]
then we get:
\[x = 4\]

Answer 12

Question is what is length of hypotenuse so I am lost

Answer 13

oops..I have made an error, hereare the right steps:

\[\begin{gathered}
x\sqrt 2 = \sqrt {32} = 4\sqrt 2 \hfill \\
x = 4 \hfill \\
\end{gathered} \]
yes, I know, nevertheless if I look at your drawing:

I understand that we have to determine the value of x

Answer 14

namely in your exercise there are 2 parts:
the first one asks for the hypotenuse, using the coordinates which you have provided
the second part asks for the value of x, using the length of the hypotenuse computed in the first part

Answer 15

Oh, ok. Question is misleading or confusing.

Answer 16

Thank you