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Segment EA is an altitude of triangle DEF. Find the area of the triangle. Triangle DEF with altitude AE is shown. Point D is at negative 2, 1. Point E is at 2, 5. Point F is at 6, 1. Point A is at 2, 1. 14.5 15 15.5 16

Respuesta :

Answer:

16

Step-by-step explanation:

A=(1/2)bh

A=(1/2)(df)(ea)

A=(1/2)(8)(4)

Answer:

(D) 16

Step-by-step explanation:

It is given that Segment EA is an altitude of triangle DEF. Point D(-2,1), E(2,5), F(6,1) and A(2,1).

Now, using the distance formula, we have

[tex]DF=\sqrt{(1-1)^2+(6+2)^2}[/tex]

[tex]DF=\sqrt{64}[/tex]

[tex]DF=8 units[/tex]

And [tex]EA=\sqrt{(1-5)^2+(2-2)^2}[/tex]

[tex]EA=\sqrt{16}[/tex]

[tex]EA=4 units[/tex]

Thus, the area of the triangle is=[tex]\frac{1}{2}{\times}DF{\times}SA[/tex]

=[tex]\frac{1}{2}{\times}8{\times}4[/tex]

=[tex]16 sq units[/tex]

Thus, the area of the triangle is 16 sq units.

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