## SCIE 4101 Review Packet #3 Solutions October 13, 2014

Posted by Ms. Miller in UTA.
1. x-axis: $U'(-2,1), T'(2,-4), A'(3,1)$
y-axis: $U'(2,-1), T'(-2,4), A'(-3,-1)$
2. $H'(5,2), A'(8,-1), N'(6,-11), D'(2,-3)$
3. $M'(-3,2.5), A'(4,-2.5), V'(-3,-2.5)$
4. $M'(-5,-6), A'(5,8), V'(5,-6)$
5.a. First we use the Pythagorean Theorem to find the missing side: $b=\sqrt{13^2-5^2}=12$.
$P=2(5)+2(12)=34$
$A=(5)(12)=60$
b. Again, we use the Pyth. Theorem to find the missing side: $b=\sqrt{17^2-8^2}=15$.
$P=8+15+17=40$
$A=\frac{1}{2}(15)(8)=60$
6.a. $A=\pi(16^2)=256\pi \approx 804.25$ (use the $\pi$ key on your calculator, since it is more accurate)
b. $A=\pi(\frac{21}{2})^2=110.25\pi \approx 346.36$
7.a. Since we know the perimeter, we can find the missing side: $64-2(25)=14$. To find x, we take half of the base and use Pyth. Theorem: $x=\sqrt{25^2-7^2}=24$
b.
 $(x+4)(x-10)=240$ $x^2-6x-40=240$ $x^2-6x-280=0$ Solve by factoring Solve by quadratic formula $(x-20)(x+14)=0$ $a=1, b=-6, c=-280$ $x=\frac{6\pm\sqrt{6^2-4(1)(-280)}}{2(1)}$ $x=\frac{6\pm\sqrt{36+1120}}{2}$ $x=\frac{6\pm\sqrt{1156}}{2}$ $x=\frac{6\pm 34}{2}$ $x=3\pm 17$ $x=20,-14$

Since length must be positive, our answer is $x=20$.

c.
 $\pi(\frac{y}{2})^2=121\pi$ $\pi(\frac{y^2}{4})=121\pi$ $\frac{y^2}{4}=121$ $y^2=484$ $\sqrt{y^2}=\sqrt{484}$ $y=\pm 22$

Again, since length must be positive, our answer is $y=22$.

d. Because the legs of the trapezoid are congruent, we can split the trapezoid into a rectangle and two right triangles. The base of one of the triangles will be $\frac{15-9}{2}=3$. After that, we can use Pythagorean Theorem to find the height. Since the hypotenuse is 5, we have a 3-4-5 triangle, so the height is 4.
8.a. To find the leg from the hypotenuse, divide by $\sqrt{2}$: $\frac{60\sqrt{2}}{\sqrt{2}}=60$
b. The long leg is $7\sqrt{3}$, and the hypotenuse is $2(7)=14$
c. To go from the long leg to the short leg, divide by $\sqrt{3}$: $\frac{4\sqrt{3}}{\sqrt{3}}=4$, and the hypotenuse is $2(4)=8$.
9.a. The sum of the angles of a hexagon is $(6-2)180=720\textdegree$, and each angle of a regular hexagon is $\frac{720}{6}=120\textdegree$. Therefore, $7x+15=120$, and $x=15$.
b. The sum of the angles of a pentagon is $(5-2)180=540\textdegree$, and each angle of a regular pentagon is $\frac{540}{5}=108\textdegree$ (and one-half would be $54\textdegree$). Therefore,

 $x^2-3x=54$ $x^2-3x-54=0$ $(x-9)(x+6)=0$ $x=9,-6$

Angles must be positive, but either value of x would work.

10.a. $P=2(8)+2(20)=56$
$L=(56)(12)=672$
$S=672+2(8)(20)=992$
$V=(8)(20)(12)=1920$
b. $S=(8\pi)(9)+2\pi(4^2)=104\pi$
$V=\pi(4^2)(9)=144\pi$
c. $S=4\pi 8^2$
$V=\frac{4}{3}\pi (8^3)=\frac{4}{3}(512)\pi=\frac{2048}{3}\pi$
d. To find the slant height, consider the right triangle formed by the height, half the side length, and the slant height: $l=\sqrt{8^2+6^2}=10$. $S=\frac{1}{2}(48)(10)+12^2=240+144=384$
$V=\frac{1}{3}(12^2)(8)=384$
11. Surface area: $S=(2P)(2h)+2(2l)(2w)=4Ph+8lw=4(Ph+2lw)$. Therefore the total surface are would become $4(150)=600\ \mathrm{cm^2}$.
Volume: $V=(2l)(2w)(2h)=8lwh$. Therefore, the new volume would be $8(225)=1800\ \mathrm{cm^3}$

Notice that the scale factor for the area is the square of the length scale factor, and the scale factor for the volume is the cube of the length. This holds true for any increase in the sides of a shape.

12.
 $4\pi(r^2)=180\pi$ $r^2=\frac{180\pi}{4\pi}$ $r^2=45$ $r=\sqrt{45}=3\sqrt{5}$
13. $\frac{180-176}{180}=\frac{4}{180}=2.\overline{2}\%$
14. Since the lines are parallel, the alternate interior angles are congruent:

 $6x-21=5x+4$ $x=25$

Now that we know x, we can find the angle:

 $\alpha=180-(5(25)+4)$ $\alpha=51\textdegree$
15. These are two similar triangles, so we can set up a proportion:

 $\frac{5}{2}=\frac{x}{224}$ $2x=1120$ $x=560\ \mathrm{feet}$
16.
 2. Angle addition postulate 3. Transitive prop. of = 4. Subtraction prop. of =
17.
 1. ∠1 and ∠2 are supplementary. 4. Def. of supplementary 5. $m\angle 1+m\angle 2=m\angle 1+m\angle 3$ 6. $m\angle 2=m\angle 3$ 7. Def. of congruence 8. $a \parallel b$
18.
 1. $C$ is the midpoint of $\overline{BD} and \overline{AE}$. 1. Given 2. Def. of midpoint 4. $\bigtriangleup ABC \cong \bigtriangleup EDC$ 5. $\angle A \cong \angle E$

Because the triangles are congruent, this also means that $\overline{AB} \cong \overline{DE}$.

19.
a. A quadrilateral is a parallelogram if opposite sides are congruent. This means we need to show $\overline{RA} \cong \overline{GU}$ and $\overline{RG} \cong \overline{AU}$:

$RA=\sqrt{(4-(-6))^2+(1-(-4))^2}=\sqrt{10^2+5^2}=\sqrt{125}$
$GU=\sqrt{8-(-2)^2+(9-4)^2}=\sqrt{10^2+5^2}=\sqrt{125}$
Since $RA=GU$, $\overline{RA} \cong \overline{GU}$.

$RG=\sqrt{-2-(-6))^2+(4-(-4))^2}=\sqrt{4^2+8^2}=\sqrt{80}$
$AU=\sqrt{(8-4)^2+(9-1)^2}=\sqrt{4^2+8^2}=\sqrt{80}$
Since $RA=AU$, $\overline{RG} \cong \overline{AU}$.

Therefore, since opposite sides are congruent, the quadrilateral is a parallelogram.

b. If opposite sides of a quadrilateral are parallel, then it is a parallelogram. Therefore, we need to show that $\overline{RA}\parallel\overline{GU}$ and $\overline{RG}\parallel\overline{AU}$ by finding their slopes:

$\overline{RA}:\ m=\frac{1-(-4)}{4-(-6)}=\frac{5}{10}=\frac{1}{2}$
$\overline{GU}:\ m=\frac{9-4}{8-(-2)}=\frac{5}{10}=\frac{1}{2}$
Since the slopes are the same, $\overline{RA}\parallel\overline{GU}$.

$\overline{RG}:\ m=\frac{4-(-4)}{-2-(-6)}=\frac{8}{4}=2$
$\overline{AU}:\ m=\frac{9-1}{8-4}=\frac{8}{4}=2$
Since the slopes are the same, $\overline{RG}\parallel\overline{AU}$

Because the opposite sides are parallel, the quadrilateral is a parallelogram.

20.a.
b.
21. You can solve these either using the formula $(S=\pi r^2(\frac{\mathrm{arc}}{360}))$ or as a proportion $(\frac{S}{\pi r^2}=\frac{\mathrm{arc}}{360})$.
a. $S=\pi(6^2)(\frac{114}{360})=11.4\pi\ \mathrm{ft^2}\approx 35.8\ \mathrm{ft^2}$
b. $S=\pi(8^2)(\frac{26}{360})=\frac{208}{45}\pi\ \mathrm{in^2}=4.6\overline{2}\pi\ \mathrm{in^2}\approx 14.5\ \mathrm{in^2}$
22. Again, you can solve these either using the formula or as a proportion.
a. $L=2\pi(15)(\frac{80}{360})=\frac{20}{3}\pi=6.\overline{6}\pi$
b. $L=2\pi(18)(\frac{170}{360})=17\pi$
23. We can use the area of the sector to find the radius, which we will then double to get the diameter:
$\pi r^2(\frac{81}{360})=90\pi$
$0.225\pi r^2=90\pi$
$\frac{0.225\pi r^2}{0.225\pi}=\frac{90\pi}{0.225\pi}$
$r^2=400$
$\sqrt{r^2}=\sqrt{400}$
$r=20$
Therefore, the diameter is $2(20)=40$.