Q. 49 - Calculus

Question

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

$PQ=7, QR=8, RS=8, SP=9, \angle R={60}^{◦}$

Step-by-Step Solution

Answer

The areas of the quadrilaterals $PQRS$ is $12\sqrt{5}+16\sqrt{3}$.

Step 1. Given Information

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

$PQ=7, QR=8, RS=8, SP=9, \angle R={60}^{◦}$

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

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

Step 3. Firstly finding the area of ∆ Q R S .

From the diagram we can see that $a=8, b=8, \angle C={60}^o$

$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆SPQ=\frac{1}{2}\times a\times b\times \sin C \\ \mathrm{Area} \mathrm{of} ∆QRS=\frac{1}{2}\times 8\times 8\times \sin {60}^o \\ \mathrm{Area} \mathrm{of} ∆QRS=4\times 8\times \frac{\sqrt{3}}{2} \\ \mathrm{Area} \mathrm{of} ∆QRS=2\times 8\times \sqrt{3} \\ \mathrm{Area} \mathrm{of} ∆QRS=16\sqrt{3} \end{gathered}$$

Step 4. Now finding the area of ∆ S P Q .

Before finding the area of $∆SPQ$ we firstly find the value of diagonal.

$$\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ \mathrm{where} b=8, c=8, A={60}^o \\ {a}^2={\left(8\right)}^2+{\left(8\right)}^2-2\times 8\times 8\times \cos {60}^o \\ {a}^2=64+64-2\times 8\times 8\times \frac{1}{2} \\ {a}^2=64+64-64 \\ {a}^2=64 \\ a=\pm 8 \end{gathered}$$

Step 5. Now finding the area of ∆ Q R S using the formula Area   of   ∆ Q R S = s ( s - a ) ( s - b ) ( s - c ) .

$$\begin{gathered} s=\frac{a+b+c}{2} \\ s=\frac{7+9+8}{2} \\ s=\frac{24}{2} \\ s=12 \end{gathered}$$

$$\begin{gathered} \mathrm{Area} \mathrm{of} ∆QRS=\sqrt{12(12-7)(12-8)(12-9)} \\ \mathrm{Area} \mathrm{of} ∆QRS=\sqrt{4\times 3\times 5\times 4\times 3} \\ \mathrm{Area} \mathrm{of} ∆QRS=4\times 3\sqrt{5} \\ \mathrm{Area} \mathrm{of} ∆QRS=12\sqrt{5} \end{gathered}$$

Step 6. Now finding the area of quadrilateral P Q R S

$$\begin{gathered} \mathrm{Area} \mathrm{of} PQRS=\mathrm{Area} \mathrm{of} ∆SPQ+\mathrm{Area} \mathrm{of} ∆QRS \\ \mathrm{Area} \mathrm{of} PQRS=12\sqrt{5}+16\sqrt{3} \end{gathered}$$