Q. 42 - 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 36

Step-by-Step Solution

Answer

The areas of the triangles determined by the points P, Q and R is approximately $22.34$.

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 36

Step 2. The points from the Exercise 36 are P ( 1 ,   4 ,   6 ) ,   Q ( − 3 ,   5 ,   0 ) ,   R ( 3 ,   2 , − 1 ) .

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.

$$\begin{gathered} a=QR=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2} \\ a=QR=\sqrt{{\{3-(-3\left)\right\}}^2+{(2-5)}^2+{(-1-0)}^2} \\ a=QR=\sqrt{{\left(6\right)}^2+{(-3)}^2+{(-1)}^2} \\ a=QR=\sqrt{36+9+1} \\ a=QR=\sqrt{46} \\ a=QR\approx 6.78 \end{gathered}$$

Step 4. Now finding the value of b.

$$\begin{gathered} b=PR=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2} \\ b=PR=\sqrt{{(3-1)}^2+{(2-4)}^2+{(-1-6)}^2} \\ b=PR=\sqrt{{\left(2\right)}^2+{(-2)}^2+{(-7)}^2} \\ b=PR=\sqrt{4+4+49} \\ b=PR=\sqrt{57} \\ b=PR\approx 7.55 \end{gathered}$$

Step 5. Now finding the value of c.

$$\begin{gathered} c=PQ=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2} \\ c=PQ=\sqrt{{(-3-1)}^2+{(5-4)}^2+{(0-6)}^2} \\ c=PQ=\sqrt{{(-4)}^2+{\left(1\right)}^2+{(-6)}^2} \\ c=PQ=\sqrt{16+1+36} \\ c=PQ=\sqrt{53} \\ c=PQ\approx 7.28 \end{gathered}$$

Step 6. Firstly finding the value of s .

$$\begin{gathered} s=\frac{a+b+c}{2} \\ s=\frac{6.78+7.55+7.28}{2} \\ s=\frac{21.61}{2} \\ s=10.805 \end{gathered}$$

Step 7. Now finding the area.

$$\begin{gathered} \mathrm{Area}=\sqrt{s(s-a)(s-b)(s-c)} \\ \mathrm{Area}=\sqrt{10.805(10.805-6.78)(10.805-7.55)(10.805-7.28)} \\ \mathrm{Area}=\sqrt{10.805\times 4.025\times 3.255\times 3.525} \\ \mathrm{Area}=\sqrt{499} \\ \mathrm{Area}\approx 22.34 \end{gathered}$$