# Thread: Find the length of this triangle?

1. ## Find the length of this triangle?

This is on page 860 #18 on the OG Book.

Note: Figure not drawn to scale

In the figure below, if the legs of triangle ABC are parallel to the axes, what could be the lengths of the sides of triangle ABC?

a) 2, 5 and sqrt 29
b)2, 5, 7
c)3, 3, and 3sqrt2
d)3, 4,5
e)4,5, and sqrt 41

2. Hey there fabxx,

Given that the triangle is a right triangle, side lengths can be found by Pythagoras.

a) 2, 5 and sqrt 29
$\displaystyle \sqrt {2^2 + 5^2} = \sqrt {29}$
a. is possible.

b)2, 5, 7
$\displaystyle \sqrt {2^2 + 5^2} = \sqrt {29} \neq {7}$
b. is not possible.

c)3, 3, and 3sqrt2
$\displaystyle \sqrt {3^2 + 3^2} = \sqrt {18} = \sqrt {9} \ * \sqrt {2} = 3 \sqrt{2}$
c. is possible.

d)3, 4,5
$\displaystyle \sqrt {3^2 + 4^2} = \sqrt {25} = 5$
d. is possible

e)4,5, and sqrt 41
$\displaystyle \sqrt {4^2 + 5^2} = \sqrt {41}$
By Pythagoras, this appears possible, but based on the information on the graph (point above triangle given as (4, 10)), no two side lengths can be greater than (or equal to) 4.
e. is not possible

Trust this helps...

3. Originally Posted by MakeANote
Hey there fabxx,

d)3, 4,5
$\displaystyle \sqrt {3^2 + 4^2} = \sqrt {25} = 5$
d. is possible

Trust this helps...

Hi thanks for replying. How do you know that it's $\displaystyle \sqrt {3^2 + 4^2} = \sqrt {25} = 5$ but not the other way around? For example $\displaystyle \sqrt {3^2 + 5^2} = ?$

4. Originally Posted by fabxx
Hi thanks for replying. How do you know that it's $\displaystyle \sqrt {3^2 + 4^2} = \sqrt {25} = 5$ but not the other way around? For example $\displaystyle \sqrt {3^2 + 5^2} = ?$

You know that 5 should be the hypotenuse because it is the longest side of the triangle.

5. As icemanfan indicated, the longest side of the right triangle is the hypotenuse (always opposite the right-angle).

Something to note, however, that the question is not complete. Pythagoras was used to simply remove answers that could not be considered.

The points given in the diagram are also significant.

Two points located on the line given on your diagram are (0,0) and (4, 10). The side lengths must also be able to be represented on this line.

One possible triangle (but not given) would have side lengths 4 and 10 units in size (if one side of the triangle were touching the x-axis).

Hence, another triangle must obey the side-lengths by ratio, because the triangles would be similar (sharing the same angle sizes, and the sides share a common ratio).

4:10 = 2:5

This means that a triangle with side-lengths of 2 and 5 would use this line.

This means that the only two possible answers are a. and b.

However, b. does not conform to the rules of a right-triangle. Hence, the only answer possible given is a) 2, 5 and $\displaystyle \sqrt {29}$.

(This is the solution).
Again, I trust this helps.

6. Originally Posted by MakeANote
Hey there fabxx,

Given that the triangle is a right triangle, side lengths can be found by Pythagoras.

a) 2, 5 and sqrt 29
$\displaystyle \sqrt {2^2 + 5^2} = \sqrt {29}$
a. is possible.

b)2, 5, 7
$\displaystyle \sqrt {2^2 + 5^2} = \sqrt {29} \neq {7}$
b. is not possible.

c)3, 3, and 3sqrt2
$\displaystyle \sqrt {3^2 + 3^2} = \sqrt {18} = \sqrt {9} \ * \sqrt {2} = 3 \sqrt{2}$
c. is possible.

d)3, 4,5
$\displaystyle \sqrt {3^2 + 4^2} = \sqrt {25} = 5$
d. is possible

e)4,5, and sqrt 41
$\displaystyle \sqrt {4^2 + 5^2} = \sqrt {41}$
By Pythagoras, this appears possible, but based on the information on the graph (point above triangle given as (4, 10)), no two side lengths can be greater than (or equal to) 4.
e. is not possible

Trust this helps...

a,c ,and d are all possible. How do you decide which is the correct answer?

7. If all that was necessary was that the triangle be a right triangle, you would be correct. But the triangle must also fit the slope of the line, which is 5/2. And only a) fits that slope.