Question

D(5, 7), E(4, 3), and F(8, 2) form the vertices of a triangle. What is m∠DEF?

Answer 1

Answer:

m∠DEF is 90°

Step-by-step explanation:

It is given that  D(5, 7), E(4, 3), and F(8, 2) form the vertices of a triangle. Thus, using the distance formula, we have

$DE=\sqrt{(3-7)^2+(4-5)^2}$

⇒$DE=\sqrt{16+1}$

⇒$DE=\sqrt{17}$

and $EF=\sqrt{(2-3)^2+(8-4)^2}$

⇒$EF=\sqrt{1+16}$

⇒$EF=\sqrt{17}$

Also,  $DF=\sqrt{(2-7)^2+(8-6)^2}$

⇒$DF=\sqrt{25+9}$

⇒$DF=\sqrt{34}$

Now, $(DF)^2=(DE)^2+(EF)^2$

⇒$(\sqrt{34})^2=(\sqrt{17})^2+(\sqrt{17})^2$

⇒$34=34$

Thus, Pythagoras theorem holds.

Hence, m∠DEF is 90°⇒ΔDEF is right angled triangle which is right angled at E.

Answer 2

DE^2 = 1 + 16 = 17. 
EF^2 = 16 + 1 = 17. 
DF^2 = 9 + 25 = 34. 
Since (DE)^2 + (EF)^2 = (DF)^2, 
 angle E is 90 degrees.
 DF must be the hypotenuse 
see attachment below