Hello again, Outragexl10!

Now I see that you've solved many of them already.

Oh well . . . maybe I can offer a different approach.

5) .[6 points] . In circle $\displaystyle V,\:RT$ is a diameter.

a) Find $\displaystyle m\angle T$ and $\displaystyle m\angle R.$

b) If $\displaystyle ST = 22$, find the length of the radius to the nearest tenth. (a) Since $\displaystyle \Delta RST$ is inscribed in a semicircle, $\displaystyle \angle S = 90^o.$

Hence, $\displaystyle \angle T$ and $\displaystyle \angle R$ are complementary: . $\displaystyle (x^2 + 30) + (5x + 24) \:=\:90$

. . which simplifies to:. . $\displaystyle x^2 + 5x - 36 \:=\:0$

. . which factors: .$\displaystyle (x - 4)(x + 9) \:=\:0$

. . and has the positive root: .$\displaystyle x\,=\,4$

Therefore: .$\displaystyle \begin{array}{ccccccc}\angle T & = & 5x + 24 & = & 5(4)+24 & = & 44^o \\ \angle R & = & x^2 + 30 & = & 4^2 + 30 & = & 46^o\end{array}$

(b) Let $\displaystyle r = VR = VT$. .Then: .$\displaystyle RT = 2r$

In right triangle $\displaystyle RST$, we have: .$\displaystyle \sin46^o \:=\:\frac{22}{2r}\quad\Rightarrow\quad r \:=\:\frac{11}{\sin46^o}$

Therefore: .$\displaystyle r \;=\;15.2917995 \;\approx\;15.3$

7) .[4 points] .$\displaystyle PQ$ is a common external tangent of circle $\displaystyle R$ and circle $\displaystyle S$.

The radius of circle $\displaystyle S$ is 12 feet and the radius of circle $\displaystyle R$ is 5 feet. .$\displaystyle TU = 8$ ft.

Find the area of quadrilateral $\displaystyle PQRS$.

Note that radii $\displaystyle SP$ and $\displaystyle RQ$ are perpendicular to tangent $\displaystyle PQ$.

Note also that: $\displaystyle ST = SP = 12,\;RU = RQ = 5$

. . Hence: .$\displaystyle SR \:=\:12 + 8 + 5 \:=\:25$

Invert the quadrilateral. Code:

S *
| * 25
7 | *
| *
V + - - - - - - - - - - - * R
| x |
5 | | 5
| |
* - - - - - - - - - - - *
P Q

Through $\displaystyle R$ draw a line parallel to $\displaystyle PQ$, meeting $\displaystyle SP$ at $\displaystyle V$.

. . Let $\displaystyle x = VR$

In right triangle $\displaystyle SVR$, we have: .$\displaystyle x^2 + 7^2 \:=\:25^2$

. . Then: .$\displaystyle x^2 + 49 \:=\:625\quad\Rightarrow\quad x^2 \:=\:576\quad\Rightarrow\quad x \:=\:24$

Quadrilateral $\displaystyle PQRS$ is a trapezoid with parallel bases 5 and 12

. . and height 24.

Therefore, its area is: .$\displaystyle a \;=\;\frac{24}{2}(5 + 12) \;=\;204$ ft².