Question

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

role="math" localid="1649738595160" $PQ=6,QR=7,RS=8,SP=9,PR=10$

Short answer

The areas of the quadrilaterals $PQRS$is localid="1649862103327" $54.85$.

Step-by-step solution

Step 1. Given Information

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

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

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 ∆PQR.

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 ∆PSR

From the diagram$a=9,b=8,c=10$

So

localid="1649861896902" $$\begin{gathered} s=\frac{9+8+10}{2} \\ s=\frac{27}{2} \\ s=13.5 \end{gathered}$$

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 PQRS

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