Q. 44 - 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 41

Step-by-Step Solution

Answer

The areas of the triangles determined by the points P, Q and R is $\mathrm{approximately} 25.99$.

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 41

Step 2. The points from the Exercise 41 are P ( 1 ,   6 ) ,   Q ( 0 , − 3 ) ,   R ( − 5 ,   4 ) .

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 .

$$\begin{gathered} a=QR=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2} \\ a=QR=\sqrt{{(-5-0)}^2+{\{4-(-3\left)\right\}}^2} \\ a=QR=\sqrt{{(-5)}^2+{\left(7\right)}^2} \\ a=QR=\sqrt{25+49} \\ a=QR=\sqrt{74} \\ a=QR\approx 8.60 \end{gathered}$$

Step 4. Now finding the value of b.

$$\begin{gathered} b=PR=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2} \\ b=PR=\sqrt{{(-5-1)}^2+{(4-6)}^2} \\ b=PR=\sqrt{{(-6)}^2+{(-2)}^2} \\ b=PR=\sqrt{36+4} \\ b=PR=\sqrt{40} \\ b=PR\approx 6.32 \end{gathered}$$

Step 5. Now finding the value of c.

$$\begin{gathered} c=PQ=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2} \\ c=PQ=\sqrt{{(0-1)}^2+{(-3-6)}^2} \\ c=PQ=\sqrt{{(-1)}^2+{(-9)}^2} \\ c=PQ=\sqrt{1+81} \\ c=PQ=\sqrt{82} \\ c=PQ\approx 9.06 \end{gathered}$$

Step 6. Firstly finding the value of s .

$$\begin{gathered} s=\frac{a+b+c}{2} \\ s=\frac{8.60+6.32+9.06}{2} \\ s=\frac{23.98}{2} \\ s=11.99 \end{gathered}$$

Step 7. Now finding the area.

$$\begin{gathered} \mathrm{Area}=\sqrt{s(s-a)(s-b)(s-c)} \\ \mathrm{Area}=\sqrt{11.99(11.99-8.60)(11.99-6.32)(11.99-9.06)} \\ \mathrm{Area}=\sqrt{11.99\times 3.39\times 5.67\times 2.93} \\ \mathrm{Area}=\sqrt{675.26} \\ \mathrm{Area}\approx 25.99 \end{gathered}$$