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  1. #1
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    almost done with angles and circles

    Find the measure of angle JLN to the nearest degree.

    18
    20
    36
    80
    POINT VALUE: 3 points
    Givenandare tangents of circle T, find the measure of arc PQR.
    54
    126
    306
    not enough information
    POINT VALUE: 3 points
    ABCD is an inscribed quadrilateral.

    a.) Find the values of x and y. Show all work.
    b.) Find mB and mD.

    WORTH 4 points (Not yet scored by your instructor)
    In circle L,is a tangent andis a diameter. Find the measure of arc MN. Show all work.



    WORTH 4 points (Not yet scored by your instructor)

    In circle V,is a diameter.

    a.) Find mSTR and mSRT. Show all work.
    b.) If ST = 22, find the length of the radius to the nearest tenths place. Show all work.

    WORTH 6 points (Not yet scored by your instructor)

    the next page is...
    is a common external tangent of circle A and circle B. EC = 36, EA = 27, DB = 6. Find HC.
    10
    4
    20
    25
    POINT VALUE: 4 points
    is a common external tangent of circle R and circle S. The radius of circle S is 12 feet and the radius of circle R is 5 feet. TU = 8 ft. Find the area of quadrilateral PQRS.

    WORTH 4 points (Not yet scored by your instructor)
    For circle K and circle L, is a common internal tangent. If JK = 3, LM = 9 and JM = 20, find KL. Show all work.

    WORTH 4 points (Not yet scored by your instructor)
    In circle A, is a tangent. Find AD if CD = 8 and BD = 20.
    21
    22
    29
    37
    POINT VALUE: 4 points
    Write the equation of the circle with center (-2, -3) and tangent to the line 2x + 5y = 10. (x – 2)2+(y – 3)2 = 29
    (x – 2)2+(y – 3)2 = 5
    (x + 2)2+(y + 3)2 = 29
    (x + 2)2+(y + 3)2 = 5
    none of these
    POINT VALUE: 4 points and the last page is... Describe the locus of points that are 9 cm from point B. a segment 18 cm long
    a circle with a radius of 9 cm
    a sphere with a radius of 9 cm
    a circle with an area of 81 sq cm
    POINT VALUE: 1 points Describe the locus of points whose coordinates satisfy both statements:
    x2 + (y – 2)2 = 25
    y < 0
    the part of the circle with center (0, 2) and radius 5, on or below the x-axis
    the part of the circle with center (0, 2) and radius 5, below the x-axis
    the part of the circle with center (0, -2) and radius 5, on or to the left of the y-axis
    the part of the circle with center (0, 2) and radius 5, to the left of the y-axis
    POINT VALUE: 1 points Describe the locus of points whose coordinates satisfy both statements:
    x2 = 49
    x2 + y2 = 85

    WORTH 2 points (Not yet scored by your instructor) Given a cylinder with a surface area of 60 units squared and a height of 7 units. Find the volume of a sphere that has the same size radius as the given cylinder. Show all work.
    WORTH 3 points (Not yet scored by your instructor) Find the surface area of the frustum. Show all work.

    WORTH 4 points (Not yet scored by your instructor) In a cone with a slant height of 6 feet, the slant height forms a 42 degree angle with the radius. Find the volume of the cone. Show all work.
    WORTH 4 points (Not yet scored by your instructor) A sphere with center A has a surface area of 144 units squared. Find the volume of a sphere with center B whose radius is twice as large as the sphere with center A. What is the ratio of the volumes of the two spheres?
    WORTH 3 points (Not yet scored by your instructor)





    The stuff in red i have already been helped with, and for those that have helped me so far, thank you VERY much!


    i need to get this done very very soon... i have an online class and I've asked people for help but i haven't had much luck... the instructions the class gives me doesn't even tell me what to do... can i get some help PLEASE... i have a 76 now and with these i will be able to attain a B, and thats all i want...

    i have a few of these left and then my exam... so i want to do good on these...

    edit... let me just say I'm not looking for answers but a point in the right direction... i don't like to cheat...
    Last edited by Outragexl10; July 21st 2007 at 01:11 PM.
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  2. #2
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    Quote Originally Posted by Outragexl10 View Post
    [FONT=Helvetica,Arial][SIZE=2] ...
    A sphere with center A has a surface area of 144π units squared. Find the volume of a sphere with center B whose radius is twice as large as the sphere with center A. What is the ratio of the volumes of the two spheres?
    ...
    Hello,

    the surface area of a sphere is calculated by: a_{surface}=4 \pi r^2 . Plug in the value you know:

    144 \pi = 4 \pi r^2~\Longrightarrow~ r = 6

    The volume of a sphere is calculated by: V_{sphere} = \frac{4}{3}\pi r^3 . Plug in the value of the radius:

    V_{sphere} = \frac{4}{3}\pi \cdot 6^3 = 288 \pi \text{  cubic units}

    If the second radius is twice as large as the first one you have: r_2 = 2 \cdot r_1 . Calculate the volume:

    V_{sphere} = \frac{4}{3}\pi r_2^3 = \frac{4}{3}\pi (2 \cdot r_1)^3 = 8 \cdot \frac{4}{3}\pi r_1^3 . That means:

    \frac{V_1}{V_2} = \frac{1}{8}
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  3. #3
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    Quote Originally Posted by Outragexl10 View Post
    ...
    In a cone with a slant height of 6 feet, the slant height forms a 42 degree angle with the radius. Find the volume of the cone. Show all work.
    ...
    Hello,

    make a rough sketch of the situation.

    Let s be the slant, h be the height of the con and r be the radius. s, h and r form a right triangle.

    Therefore:
    1. r = s \cdot \cos(42^\circ) \approx 4.459 '
    2. h = s \cdot \sin(42^\circ) \approx 4.015 '

    The volume of a cone is calculated by: V_{cone} = \frac{1}{3} \pi \cdot r^2 \cdot h . Plug in the values you know:

    V_{cone} = \frac{1}{3} \pi \cdot (s \cdot \cos(42^\circ))^2 \cdot s \cdot \sin(42^\circ) = \frac{1}{3} \pi \cdot s^3 \cdot \cos(42^\circ) \cdot \sin(42^\circ) \approx 112.478 \text{ cft}
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  4. #4
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    Quote Originally Posted by Outragexl10 View Post
    ...
    Describe the locus of points whose coordinates satisfy both statements:
    x2 = 49
    x2 + y2 = 85
    ...
    Hello,

    the first equation describes 2 (two!) straight lines parallel to the y-axis. The second equation describes a circle with the centre at the origin and the radius \sqrt{85}. Plug in the value of x of the 1rst equation into the 2nd equation. You'll get 2 solutions. That means the locus consists of 4 points:

    P_1(-7, 6), P_2(-7, -6), P_3(7, 6), P_4(7, -6)
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  5. #5
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    Quote Originally Posted by Outragexl10 View Post
    ...
    Describe the locus of points whose coordinates satisfy both statements:
    x2 + (y 2)2 = 25
    y < 0
    - the part of the circle with center (0, 2) and radius 5, on or below the x-axis
    - the part of the circle with center (0, 2) and radius 5, below the x-axis
    - the part of the circle with center (0, -2) and radius 5, on or to the left of the y-axis
    - the part of the circle with center (0, 2) and radius 5, to the left of the y-axis
    ...
    Hello,

    make a sketch. Then you'll see easily that only answer #2 could be right.
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    Hello, Outragexl10!

    For the first two, you need to be familiar with this theorem:

    If two tangents, or a tangent and a secant, or two secants are drawn to a circle,
    . . the included angle is one-half the difference of the two intercepted arcs.



    Find the measure of \angle JLN to the nearest degree.


     18\qquad  20 \qquad  36 \qquad  80

    The theorem says: . 4x \:=\:\frac{1}{2}\left[(x^2 + 8x) - 20\right]

    which simplifies to: . x^2\,=\,20\quad\Rightarrow\quad x\,=\,2\sqrt{5}

    Therefore: . \angle JLN \:=\:4x\:=\:4(2\sqrt{5}) \:\approx\:\boxed{18^o}



    Given PS and RS are tangents of circle T, find the measure of arc PQR.



    54 \qquad  126 \qquad  306 \qquad \text{not enough information}

    We see that: . \overline{PQR} \:=\:360^o - \overline{PR} \:=\:360 - 6x^2


    The theorem says: . 42x \:=\:\frac{1}{2}\left[(360 - 6x^2) - 6x^2\right]

    This simplifies to: . x^2+ 7x - 30 \:=\:0

    . . which factors: . (x - 3)(x + 10) \:=\:0

    . . and has the positive root: . x \,=\,3


    Therefore: . \overline{PQR} \:=\:360 - 6x^2 \:=\:360 - 6\cdot3^2 \:=\:\boxed{306^o}

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    Quote Originally Posted by Outragexl10 View Post
    ABCD is an inscribed quadrilateral.

    a.) Find the values of x and y. Show all work.
    b.) Find mB and mD.
    You said you don't want the straight answers, so i will tell you how to do this on your own, or give you hints.

    Recall the theorem.

    Theorem: Opposite angles of a cyclic quadrilateral are supplementary (add up to 180 degrees or pi radians).

    thus for a) we have a system of equations:

    (3y - 5) + 2x = 180 .......................(1)
    (3x + y) + 60 = 180 ........................(2)




    For b), it is easy once you found x and y.

    m \angle B = 3y - 5

    just plug in the value for y you got in part (a)

    m \angle D = 2x

    just plug in the value for x you got in part (a)
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    Quote Originally Posted by Outragexl10 View Post


    WORTH 4 points (Not yet scored by your instructor)
    In circle V,is a diameter.

    a.) Find mSTR and mSRT. Show all work.



    Recall the theorem,

    Theorem: Angles inscribed in a semi-circle measure 90 degrees.

    From that theorem, we know that we are dealing with a right angled triangle, with the right angle located at
    \angle RST

    since the angles in a triangle add up to 180 degrees, we can find x as follows.

    \angle RST + \angle TRS + \angle RTS = 180

    \Rightarrow 90 + (x^2 + 30) + (5x + 24) = 180

    if you simplify the above, you get a quadratic equation, i'll assume you don't have problems with quadratic equations (unless you tell me otherwise).

    once you find x, just plug in it's values to find the two angles that are desired.

    remember: m \angle STR = 5x + 24 and m \angle SRT = x^2 + 30


    b.) If ST = 22, find the length of the radius to the nearest tenths place. Show all work.
    Again, recall we are working with a right triangle, so we think of trig ratios and Pythagoras' theorem when we are dealing with such triangles.

    Note that the radius is \frac {1}{2} RT

    we can find RT as follows:

    \sin \left( \angle SRT \right) = \frac {ST}{RT}

    you found \angle SRT in the previous question, and ST is given, so your only unknown is RT, just solve for it. when you get the answer, divide it by 2 to get the radius
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    Quote Originally Posted by Outragexl10 View Post
    is a common external tangent of circle A and circle B. EC = 36, EA = 27, DB = 6. Find HC.
    10
    4
    20
    25
    Recall, Theorem: A tangent to a circle makes an angle of 90 degrees with the radius of the circle.

    so \angle AEC = \angle BDC = 90

    Thus you should notice two things:
    (1) we are working with right-triangles (think of Pythagoras' theorem and trig ratios)

    (2) We have SIMILAR triangles here, so the angles of \triangle AEC are congruent to those of \triangle BDC and the sides of \triangle AEC are proportional to the sides of \triangle BDC

    Now we get to work.

    We want HC, we should note immediately that HC = BC - BH = BC - BD, since BH and BD are the radii of the smaller circle so they are equal. BD is given, so HC = BC - 6

    so our objective is to find BC, then we can easily find HC

    Note, since we have similar triangles, BC is proportional to AC, and we can use Pythagoras' theorem to find AC.

    By Pythagoras, (AC)^2 = (EC)^2 + (AE)^2

    EC and AE are given, so finding AC is no big deal

    Now, since we have similar triangles:

    \frac { \mbox {Hypotenuse of big triangle }}{ \mbox {Base of big triangle}} = \frac { \mbox {Hypotenuse of small triangle}}{ \mbox {Base of small triangle}}

    \Rightarrow \frac {AC}{AE} = \frac {BC}{BD}

    You found AC already, and AE and BD were given. now just solve for BC

    when you get BC, subtract 6 from it to get HC
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    Quote Originally Posted by Jhevon View Post
    You said you don't want the straight answers, so i will tell you how to do this on your own, or give you hints.

    Recall the theorem.

    Theorem: Opposite angles of a cyclic quadrilateral are supplementary (add up to 180 degrees or pi radians).

    thus for a) we have a system of equations:

    (3y - 5) + 2x = 180 .......................(1)
    (3x + y) + 60 = 180 ........................(2)




    For b), it is easy once you found x and y.

    m \angle B = 3y - 5

    just plug in the value for y you got in part (a)

    m \angle D = 2x

    just plug in the value for x you got in part (a)
    i still don't understand this one fully... can you go through it please...
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    Quote Originally Posted by Outragexl10 View Post
    is a common external tangent of circle R and circle S. The radius of circle S is 12 feet and the radius of circle R is 5 feet. TU = 8 ft. Find the area of quadrilateral PQRS.
    See the attached diagram.

    Step 1: Draw a line from R to cut PS at a point I called V, such that \angle QRV = 90^{ \circ}

    Remember from my last post, that tangents intersect with radii at 90 degrees. So \angle SPQ = \angle RQP = 90

    So now you see we can separate the quadrilateral into two shapes that we can find the area of easily: A rectangle and a right-triangle (think of Pythagoras' theorem and trig ratios--always! when you hear right-triangle, that's what you think about).

    Also note that SR = ST + TU + UR = 12 + 8 + 5 = 25

    since RV \parallel PQ, we have that PV = QR, so SV = 12 - 5 = 7

    Now we know two sides of the right-triangle, and two sides of the quadrilateral. The missing sides of the quadrilateral are equal to the missing side of the triangle, so let's use Pythagoras' theorem to find the missing side of the triangle:

    By Pythagoras, (RV)^2 = (RS)^2 - (SV)^2. we know RS and SV, so finding RV is no big deal.

    Now we realize that RV = PQ, and we have everything we need!

    \mbox { area} PQRS = \mbox { area} PQRV + \mbox { area} \triangle RSV

    (area of a rectangle is length times with, area of a triangle is 1/2 * base * height)

    \Rightarrow \mbox { area} PQRS = PQ \cdot QR + \frac {1}{2} RV \cdot SV

    and now we know all the unknowns, so just solve for the area
    Attached Thumbnails Attached Thumbnails almost done with angles and circles-geometry.jpg  
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    Quote Originally Posted by Outragexl10 View Post
    i still don't understand this one fully... can you go through it please...
    i there any particular part you don't get? do you get how i set up the system of equations? if you add the opposite angles of the quad, you get 180, that's all i did.

    Quote Originally Posted by Jhevon View Post
    Theorem: Opposite angles of a cyclic quadrilateral are supplementary (add up to 180 degrees or pi radians).

    thus for a) we have a system of equations:

    \color {red} \angle B + \angle D = 180
    \color {red} \angle A + \angle C = 180

    (3y - 5) + 2x = 180 .......................(1)
    (3x + y) + 60 = 180 .......................(2)
    Ok, let me simplify a bit and maybe you will see it. the equations boil down to

    2x + 3y = 185 ..........................(1)
    3x + y = 120 ............................(2)

    Can you continue now? Please explain which concept you don't get, so I can address it directly.
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    Quote Originally Posted by Outragexl10 View Post
    WORTH 4 points (Not yet scored by your instructor)
    For circle K and circle L, is a common internal tangent. If JK = 3, LM = 9 and JM = 20, find KL. Show all work.
    Ok, again, based on the tangent-radii theorem i mentioned several times before in this thread, we have \angle KJM = \angle LMJ = 90^{ \circ}

    So we are dealing with right-triangles (think of Pythagoras' theorem and trig ratios)

    see the diagram to see what we have. do you think you can figure out how to find KL based on the diagram?
    Attached Thumbnails Attached Thumbnails almost done with angles and circles-tridouble.gif  
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    Quote Originally Posted by Jhevon View Post
    i there any particular part you don't get? do you get how i set up the system of equations? if you add the opposite angles of the quad, you get 180, that's all i did.

    Ok, let me simplify a bit and maybe you will see it. the equations boil down to

    2x + 3y = 185 ..........................(1)
    3x + y = 120 ............................(2)

    Can you continue now? Please explain which concept you don't get, so I can address it directly.
    How to find x and y from those two...for some reason its not coming to mind...
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    Quote Originally Posted by Outragexl10 View Post
    WORTH 4 points (Not yet scored by your instructor)
    In circle A, is a tangent. Find AD if CD = 8 and BD = 20.

    21
    22
    29
    37
    Again, by the same theorem about intersecting tangents and radii, we have \angle ABD = 90^{ \circ}

    thus we have a right-triangle (think of Pythagoras' theorem and trig ratios)

    By Pythagoras, (AD)^2 = (BD)^2 + (AB)^2

    but AB = AC since both have the radius of the circle for their length

    so, (AD)^2 = (BD)^2 + (AC)^2

    since AD = AC + CD, we get

    (AD)^2 = (BD)^2 + (AD - CD)^2

    expand the last set of brackets, we get:

    (AD)^2 = (BD)^2 + (AD)^2 - 2 AD \cdot CD + (CD)^2

    \Rightarrow 2 AD \cdot CD = (BD)^2 + (CD)^2

    \Rightarrow AD = \frac {(BD)^2 + (CD)^2}{2 CD}

    and i think you can take it from here
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