Question

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

Answer 1

16

Explanation:

A=(1/2)bh

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

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

Answer 2

(D) 16

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

`DF=√((1-1)^2+(6+2)^2)`

`DF=√(64)`

`DF=8 units`

And
`EA=√((1-5)^2+(2-2)^2)`

`EA=√(16)`

`EA=4 units`

Thus, the area of the triangle is=
`(1)/(2){*}DF{*}SA`

=
`(1)/(2){*}8{*}4`

=
`16 sq units`

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