Respuesta :

Answer:  Choice D) right

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Explanation:

If you apply the distance formula, you should find that triangle DEF has the following side lengths

  • DE = 5
  • EF = 5
  • FD = sqrt(50)

Those distances are exact. The sqrt(50) approximates to 7.071

Since 7.071 is the longest side, we'll make c = sqrt(50)

Then we have a = 5 and b = 5

Note that a^2+b^2 = 5^2+5^2 = 50 while c^2 = (sqrt(50))^2 = 50

Both a^2+b^2 and c^2 are equal to 50.

That means a^2+b^2 = c^2 is a true equation for the a,b,c values mentioned.

By the converse of the pythagorean theorem, we have shown that triangle DEF is a right triangle.

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Side notes:

  • Triangle DEF is also isosceles because a = b = 5.
  • If a^2+b^2 > c^2, then the triangle is acute
  • If a^2+b^2 < c^2, then the triangle is obtuse
  • If a = b = c, then we have an equiangular triangle
Ver imagen jimthompson5910
bhadelsimron19

Answer:

right

Step-by-step explanation:

we should find the distance of

DE

EF

FD

for DE

D(2,3) = (x1,y1)

E(5,7) = (x2,y2)

disatnce = [tex]\sqrt{(x2 - x1)^2 + (y2 - y1)^2}[/tex]

=[tex]\sqrt{(5-2)^2 + (7-3)^2}[/tex]

=[tex]\sqrt{3^2 + 4^2}[/tex]

=[tex]\sqrt{16+9}[/tex]

=[tex]\sqrt{25}[/tex]

=5 units

for EF

E(5,7) = (x1 , y1)

F(9,4) = (x2 , y2)

distance =[tex]\sqrt{(x2 - x1)^2 + (y2 - y1)^2}[/tex]

=[tex]\sqrt{(9-5)^2 + (4-7)^2}[/tex]

=[tex]\sqrt{(4)^2 + (-3)^2}[/tex]

=[tex]\sqrt{16 + 9}[/tex]

=[tex]\sqrt{25}[/tex]

=5 units

for DF

D(2,3) = (x1 , y1)

F(9,4) = (x2 , y2)

diatance =[tex]\sqrt{(x2 - x1)^2 + (y2 - y1)^2}[/tex]

=[tex]\sqrt{(9-2)^2 + (4-3)^2}[/tex]

=[tex]\sqrt{7^2 + 1^2}[/tex]

=[tex]\sqrt{49 + 1}[/tex]

=[tex]\sqrt{50}[/tex]

=[tex]5\sqrt{2}[/tex]

It is a right angle triangle .

to prove right angle triangle

sum of square of two smaller sides = sum of square of longest side

5^2 + 5^2 =[tex](5\sqrt{2})^2[/tex]

25 + 25 = 52 *2

50 + 50

hence proved.

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