Can someone help me solve this problem?

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Answer 1 · 2018-03-05T00:58:43

The fourth answer down is correct:

$A = \frac{1}{2} b h = \frac{1}{2} e f sin D$ $A=1/2bh=1/2efsinD$

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The small triangle to the left of the altitude $h$ $h$ is a right angle triangle. We can write:

$sin D = \frac{Opposite}{Hypotenuse} = \frac{h}{f}$ $sinD=("Opposite")/("Hypotenuse")=h/f$

We can solve for $h$ $h$ :

$h = f sin D$ $h=fsinD$

In the formula for the Area of the triangle $Delta DEF$ , we can plug in $e$ for base $b$ , and $fsinD$ for $h$ :

$A=1/2bh=1/2efsinD$