Q. 43 - Calculus

Question

Suppose a triangle has side lengths a, b and c. The semi-perimeter of the triangle is defined to be $s=\frac{1}{2}(a+b+c)$. We can use the side lengths of the triangle to calculate its area by applying Heron’s formula:

$\mathrm{Area}=\sqrt{s(s-a)(s-b)(s-c)}$

Use Heron’s formula to compute the areas of the triangles determined by the points P, Q and R in Exercises 42–44:

P, Q and R from Exercise 37

Step-by-Step Solution

Answer

The areas of the triangles determined by the points P, Q and R is $\mathrm{approximately} 14.58.$.

Step 1. Given Information

Suppose a triangle has side lengths a, b and c. The semi-perimeter of the triangle is defined to be $s=\frac{1}{2}(a+b+c)$. We can use the side lengths of the triangle to calculate its area by applying Heron’s formula:

$\mathrm{Area}=\sqrt{s(s-a)(s-b)(s-c)}$

Use Heron’s formula to compute the areas of the triangles determined by the points P, Q and R in Exercises 42–44:

P, Q and R from Exercise 37

Step 2. The points from the Exercise 37 are P ( 4 ,   3 , − 2 ) ,   Q ( 0 ,   0 ,   3 ) ,   R ( − 1 ,   3 ,   6 ) .

The herons formula $\mathrm{Area}=\sqrt{s(s-a)(s-b)(s-c)}$

Where $s=\frac{1}{2}(a+b+c)$

$$\begin{gathered} a=\mathrm{Distance} \mathrm{between} QR \\ b=\mathrm{Distance} \mathrm{between} PR \\ c=\mathrm{Distance} \mathrm{between} PQ \end{gathered}$$

Step 3. Now finding the value of a.

$$\begin{gathered} a=QR=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2} \\ a=QR=\sqrt{{\left\{\right(-1)-0\}}^2+{(3-0)}^2+{(6-3)}^2} \\ a=QR=\sqrt{{(-1)}^2+{\left(3\right)}^2+{\left(3\right)}^2} \\ a=QR=\sqrt{1+9+9} \\ a=QR=\sqrt{19} \\ a=QR\approx 4.36 \end{gathered}$$

Step 4. Now finding the value of b.

$$\begin{gathered} b=PR=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2} \\ b=PR=\sqrt{{\left\{\right(-1)-4\}}^2+{(3-3)}^2+{\{6-(-2\left)\right\}}^2} \\ b=PR=\sqrt{{(-5)}^2+{\left(0\right)}^2+{\left(8\right)}^2} \\ b=PR=\sqrt{25+0+64} \\ b=PR=\sqrt{89} \\ b=PR\approx 9.43 \end{gathered}$$

Step 5. Now finding the value of c.

$$\begin{gathered} c=PQ=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2} \\ c=PQ=\sqrt{{(0-4\}}^2+{(0-3)}^2+{\{3-(-2\left)\right\}}^2} \\ c=PQ=\sqrt{{(-4)}^2+{(-3)}^2+{(-5)}^2} \\ c=PQ=\sqrt{16+9+25} \\ c=PQ=\sqrt{50} \\ c=PQ\approx 7.07 \end{gathered}$$

Step 6. Firstly finding the value of s .

$$\begin{gathered} s=\frac{a+b+c}{2} \\ s=\frac{4.36+9.43+7.07}{2} \\ s=\frac{20.86}{2} \\ s=10.43 \end{gathered}$$

Step 7. Now finding the area.

$$\begin{gathered} \mathrm{Area}=\sqrt{s(s-a)(s-b)(s-c)} \\ \mathrm{Area}=\sqrt{10.43(10.43-4.36)(10.43-9.43)(10.43-7.07)} \\ \mathrm{Area}=\sqrt{10.43\times 6.07\times 1\times 3.36} \\ \mathrm{Area}=\sqrt{212.72} \\ \mathrm{Area}\approx 14.58 \end{gathered}$$