Answer to Question #222328 in Analytic Geometry for hanes

Question #222328

Calculate the area of the triangle through the points (5,3,2),(-2,7,-1) and (4,-2,6)


1
Expert's answer
2021-08-18T05:47:52-0400

Solution.

We can use Heron formula which states

The area of a triangle with sides a,b,c is equal to

S=p(pa)(pb)(pc)S=\sqrt{p(p-a)(p-b)(p-c)}


 where p=a+b+c2.p=\frac{a+b+c}{2}.

Using the formula to find the distance between two points A(x1,x2,x3),B(y1,y2,y3)A(x_1,x_2,x_3), B(y_1,y_2,y_3)

which is

AB=(x1y1)2+(x2y2)2+(x3y3)2AB=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}


we can calculate the length of sides between the three points given

let say A(5,3,2), B(-2,7,-1) and C(4,-2,6).

After that, we substitute to Heron formula.

AB=(5+2)2+(37)2+(2+1)2=49+16+9=74.BC=(24)2+(7+2)2+(16)2=36+81+49=166.AC=(54)2+(3+2)2+(26)2=1+25+16=42.p=a+b+c2=12(8.6+12.88+6.48)=13.98S=(13.98)(13.988.6)(13.9812.88)(13.986.48)=24.9square unitsAnswer. 24.9 square units.AB=\sqrt{(5+2)^2+(3-7)^2+(2+1)^2}=\sqrt{49+16+9}= \sqrt{74}.\newline BC=\sqrt{(-2-4)^2+(7+2)^2+(-1-6)^2}=\sqrt{36+81+49}= \sqrt{166}.\newline AC=\sqrt{(5-4)^2+(3+2)^2+(2-6)^2}=\sqrt{1+25+16}= \sqrt{42}.\newline p = \frac{ a + b + c}{ 2} = \frac{ 1}{ 2} ( 8.6 + 12.88 + 6.48 ) = 13.98\newline S=\sqrt{(13.98)(13.98 - 8.6)(13.98 - 12.88)(13.98 - 6.48) }=24.9 \text{square units} \newline \text{Answer. 24.9 square units.}


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