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2 years ago

Find the area of the triangle with vertices: q(3,-4,-5), r(4,-1,-4), s(3,-5,-6).

Mathematics

1 answer:

maw [93]2 years ago

3 0

Answer:

(√6)/2 square units

Step-by-step explanation:

The area of a triangle is half the magnitude of the cross product of the vectors representing adjacent sides.

QR = (4-3, -1-(-4), -4-(-5)) = (1, 3, 1)

QS = (3 -3, -5-(-4), -6-(-5)) = (0, -1, -1)

The cross product is the determinant ...

$\text{det}\left|\begin{array}{ccc}i&j&k\\1&3&1\\0&-1&-1\end{array}\right|=-2i+j-k$

The magnitude of this is ...

|QR × QS| = √((-2)² +1² +(-1)²) = √6

The area of the triangle is half this value:

Area = (1/2)√6 . . . . square units

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A = 374.123 ft^2

Step-by-step explanation:

First, lets calculate the perimeter:

Perimeter (p) = side length (s) * number of sides (n)

$p = s * n$

$p = 12 * 6$

$p = 72$

Next, lets find the apothem, which is the shortest length from any side to the middle. It's like the radius in a circle, but more complicated.

Apothem (a) = side length (s) / ( 2 * tan(180/number of sides (n)) )

$a = \frac{s}{2 * tan (\frac{180}{n} )}$

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$a = \frac{12}{2 * \frac{\sqrt{3} }{3}}$

$a = \frac{12}{\frac{2\sqrt{3} }{3}}$

$a = \frac{12*3}{2\sqrt{3}}$

$a = \frac{6*3}{\sqrt{3}}$

$a = \frac{18}{\sqrt{3}}$

Now, finally, to find the area of a regular polygon, we use the following equation:

Area (A) = ( apothem (a) * perimeter (p) ) / 2

$A = \frac{a * p}{2}$

$A = \frac{\frac{18}{\sqrt{3} } * 72}{2}$

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