Question

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

role="math" localid="1649740722250" $PQ=6,QR=7,RS=8,SP=9,QS=10$

Short answer

The areas of the quadrilaterals $PQRS$is $\frac{5}{4}\left(\sqrt{455}+3\sqrt{55}\right)$.

Step-by-step solution

Step 1. Given Information

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

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

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

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

Step 3. Now finding the area of ∆QPS.

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

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

From the diagramrole="math" localid="1649739691369" $a=6,b=9,c=10$

Sorole="math" localid="1649739746415" $s=\frac{6+9+10}{2}=\frac{25}{2}$

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

$$\begin{gathered} \mathrm{Area}∆QPS=\sqrt{s(s-a)(s-b)(s-c)} \\ \mathrm{Area}∆QPS=\sqrt{\frac{25}{2}\left(\frac{25}{2}-6\right)\left(\frac{25}{2}-9\right)\left(\frac{25}{2}-10\right)} \\ \mathrm{Area}∆QPS=\sqrt{\frac{25}{2}\left(\frac{25-12}{2}\right)\left(\frac{25-18}{2}\right)\left(\frac{25-20}{2}\right)} \\ \mathrm{Area}∆QPS=\sqrt{\frac{25}{2}\times \frac{13}{2}\times \frac{7}{2}\times \frac{5}{2}} \\ \mathrm{Area}∆QPS=\frac{5}{2\times 2}\sqrt{13\times 7\times 5} \\ \mathrm{Area}∆QPS=\frac{5}{4}\sqrt{455} \end{gathered}$$

Step 5. Now finding the area of ∆QRS

From the diagram$a=7,b=8,c=10$

So

$$\begin{gathered} s=\frac{7+8+10}{2} \\ s=\frac{25}{2} \end{gathered}$$

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

$$\begin{gathered} \mathrm{Area}∆QPS=\sqrt{s(s-a)(s-b)(s-c)} \\ \mathrm{Area}∆QPS=\sqrt{\frac{25}{2}\left(\frac{25}{2}-7\right)\left(\frac{25}{2}-8\right)\left(\frac{25}{2}-10\right)} \\ \mathrm{Area}∆QPS=\sqrt{\frac{25}{2}\left(\frac{25-14}{2}\right)\left(\frac{25-16}{2}\right)\left(\frac{25-20}{2}\right)} \\ \mathrm{Area}∆QPS=\sqrt{\frac{25}{2}\times \frac{11}{2}\times \frac{9}{2}\times \frac{5}{2}} \\ \mathrm{Area}∆QPS=\frac{5\times 3}{2\times 2}\sqrt{55} \\ \mathrm{Area}∆QPS=\frac{15}{4}\sqrt{55} \end{gathered}$$

Step 7. Now finding the area of quadrilateral PQRS

$$\begin{gathered} \mathrm{Area}\mathrm{of}PQRS=\mathrm{Area}\mathrm{of}∆QPS+\mathrm{Area}\mathrm{of}∆QPS \\ \mathrm{Area}\mathrm{of}PQRS=\frac{5}{4}\sqrt{455}+\frac{15}{4}\sqrt{55} \\ \mathrm{Area}\mathrm{of}PQRS=\frac{5}{4}\left(\sqrt{455}+3\sqrt{55}\right) \end{gathered}$$