b Complete the table with appropriate lengths and evaluate the ratios.
C
c What can you say about the ratios looking at the table you made in the previous part?
A
a See solution.
B
b See solution.
C
c The ratio of the hypotenuse to a leg of an isosceles right triangle is always sqrt(2).
Practice makes perfect
a We are asked to draw three different isoscelesright triangles that have whole-number side lengths. Let's name them ABC, MNP and XYZ and label the leg lengths. The right angles will be located at vertices A, M, and X respectively.
Next, we will find the exact length of the hypotenuse of each of the drawn triangles. To do this, we will use the Pythagorean Theorem. Let's start with BC. The length of this segment will be represented by a.
Next we will take a square root of both sides of the equation. Notice that, since a represents the side length, we will consider only the positive case when taking a square root of a^2.
The hypotenuse in â–ł ABC has a length of 2sqrt(2). We can find the lengths of NP and YZ in the same way. Let b and c represent the lengths of NP and YZ.
Triangle
The Pythagorean Theorem
Simplify
â–ł ABC
2^2+ 2^2=a^2
a=2sqrt(2)
â–ł MNP
3^2+ 3^2=b^2
b=3sqrt(2)
â–ł XYZ
1^2+ 1^2=c^2
c=sqrt(2)
Let's add these side lengths to our picture.
b In this part, we are asked to complete the given table using appropriate side lengths.
Triangle
Length
Ratio
ABC
BC
2sqrt(2)
AB
2
BC/AB
2sqrt(2)/2=sqrt(2)
MNP
NP
3sqrt(2)
MN
3
NP/MN
3sqrt(2)/3=sqrt(2)
XYZ
YZ
sqrt(2)
XY
1
YZ/XY
sqrt(2)/1=sqrt(2)
c Looking at the table, we can see that the ratio of the hypotenuse to a leg of an isosceles right triangle in each case is sqrt(2). Therefore, we can assume that this ratio is constantly equal to sqrt(2) for all isosceles right triangles.
