Q. 46 - Calculus

Question

Use the results of Exercise 45 to find the areas of the quadrilaterals $PQRS$ specified in Exercises 46–49.

$PQ=6, QR=7, RS=8, SP=9, PR=10$

Step-by-Step Solution

Answer

The areas of the quadrilaterals $PQRS$ is $54.85$.

Step 1. Given Information

Using the results of Exercise 45 we have to find the areas of the quadrilaterals $PQRS$:

$PQ=6, QR=7, RS=8, SP=9, PR=10$

Step 2. The Construction of quadrilateral by the given measurement is

To find the area of quadrilateral we firstly finding the area of $∆PQR \mathrm{and} ∆PSR$ then add them.

Step 3. Now finding the area of ∆ P Q R .

The area of $∆PQR=\sqrt{s(s-a)(s-b)(s-c)}$

where $s=\frac{a+b+c}{2}$

From the diagram $a=6, b=7, c=10$

So $s=\frac{6+7+10}{2}=\frac{23}{2}=11.5$

Step 4. Putting the value of s in the formula of area.

$$\begin{gathered} \mathrm{Area} ∆PQR=\sqrt{s(s-a)(s-b)(s-c)} \\ \mathrm{Area} ∆PQR=\sqrt{11.5(11.5-6)(11.5-7)(11.5-10)} \\ \mathrm{Area} ∆PQR=\sqrt{11.5\times 5.5\times 4.5\times 1.5} \\ \mathrm{Area} ∆PQR\approx \sqrt{426.94} \\ \mathrm{Area} ∆PQR\approx 20.66 \end{gathered}$$

Step 5. Now finding the area of ∆ P S R

From the diagram $a=9, b=8, c=10$

So 

$$\begin{gathered} s=\frac{9+8+10}{2} \\ s=\frac{27}{2} \\ s=13.5 \end{gathered}$$

Step 6. Putting the value of s in the formula of area.

$$\begin{gathered} \mathrm{Area} ∆PQR=\sqrt{s(s-a)(s-b)(s-c)} \\ \mathrm{Area} ∆PQR=\sqrt{13.5(13.5-9)(13.5-8)(13.5-10)} \\ \mathrm{Area} ∆PQR=\sqrt{13.5\times 4.5\times 5.5\times 3.5} \\ \mathrm{Area} ∆PQR\approx \sqrt{1169.44} \\ \mathrm{Area} ∆PQR\approx 34.19 \end{gathered}$$

Step 7. Now finding the area of quadrilateral P Q R S

$$\begin{gathered} \mathrm{Area} \mathrm{of} PQRS=\mathrm{Area} \mathrm{of} ∆PQR+\mathrm{Area} \mathrm{of} ∆PSR \\ \mathrm{Area} \mathrm{of} PQRS=20.66+34.19 \\ \mathrm{Area} \mathrm{of} PQRS=54.85 \end{gathered}$$