So you are given a triangle ABC and you need to build another one DEF that meets the following:
[tex]\begin{gathered} AB=DE \\ m\angle E=90^{\circ} \\ EF=BC \end{gathered}[/tex]
First of all we should find the lengths of sides AB and BC. For this purpose we can use the coordinates of points A, B and C. The length of AB is the distance between A and B and the length of BC is the distance between B and C. The distance between two generic points (a,b) and (c,d) is given by:
[tex]\sqrt[]{(a-c)^2+(b-d)^2}[/tex]
Then the length of AB is:
[tex]AB=\sqrt[]{(1-1)^2+(6-1)^2}=\sqrt[]{0+5^2}=5[/tex]
And that of BC is:
[tex]BC=\sqrt[]{(1-5)^2+(1-1)^2}=\sqrt[]{4^2}=4[/tex]
Then the triangle DEF must meet these three conditions:
[tex]\begin{gathered} DE=5 \\ EF=4 \\ m\angle E=90^{\circ} \end{gathered}[/tex]
Since there is no rules about its position we can draw it anywhere. For example you can choose E=(-4,1). Then if D=(-4,6) we have that the length of DE is 5:
[tex]DE=\sqrt[]{(-4-(-4))^2+(6-1)^2}=\sqrt[]{0+5^2}=5[/tex]
And if we take F=(0,1) we get EF=4:
[tex]EF=\sqrt[]{(-4-0)^2+(1-1)^2}=\sqrt[]{16}=4[/tex]
Then a possibility for triangle DEF is:
As you can see it also meets the condition that the measure of E is 90°. And that would be part A.
In part B we have to use the pythagorean theorem to state a relation between the sides of DEF. For a right triangle with legs a and b the theorem states that its hypotenuse h is given by:
[tex]h^2=a^2+b^2[/tex]
We can do the same for DEF. Its legs are DE and EF whereas its hypotenuse is DF so we get:
[tex]DF^2=DE^2+EF^2[/tex]
And that's the equation requested in part B.
