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Math Help - geometry hw need help please thanks

  1. #1
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    geometry hw need help please thanks

    1. the base angles of an isosceles trapezoid are each 60 degrees and one of the diagonals bisects the base angle. if the median is 15 inches long, find the area of the trapezoid

    2. A circles has its center ate (2, -3). the equation tangent to the circle is
    4y-3x=7. Find the equation of the circle

    3. The altitude to the hypotenuse of a right triangle is 8 and the median to the hypotenuse is 10. find the perimeter of the triangle and the area of the part of the circumscribed circle which lies outside the triangle
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    Quote Originally Posted by anime_mania View Post
    3. The altitude to the hypotenuse of a right triangle is 8 and the median to the hypotenuse is 10. find the perimeter of the triangle and the area of the part of the circumscribed circle which lies outside the triangle
    Consider a right triangle with smaller sides,
    a,b
    And hypotenuse,
    c.

    The length of the median upon the hypotenuse is 10.
    By this is a special case, the length of median is half the length of hypotenuse. Thus c=20.

    The altitude upon the hypotenuse is 8. Thus,
    ab=8c
    ab=8(20)=160.

    But by the Pythagorean theorem we also know that,
    a^2+b^2=20^2=400.

    Thus,
    \left\{ \begin{array}{c}ab=160\\ a^2+b^2=400 \end{array} \right\}.
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    Quote Originally Posted by anime_mania View Post
    1. the base angles of an isosceles trapezoid are each 60 degrees and one of the diagonals bisects the base angle. if the median is 15 inches long, find the area of the trapezoid
    I drew a red altitutude upon the diagnol.
    The blue is half the length of diagnol (height of isoseles triangle).

    The red can be found from the blue, because it is a 30-60-90 triangle.

    The blue can be from from the base b side again a 30-60-90 triangle.

    Thus, you have an equation involving a,b.

    And another equation from the conditions of the problem.
    a+b=30.

    IF you want I can do it out. I am lazy.
    Attached Thumbnails Attached Thumbnails geometry hw need help please thanks-picture3.gif  
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    thanks so much again mr.hacker but i dont know how you got ab=8c
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    2. A circles has its center ate (2, -3). the equation tangent to the circle is
    4y-3x=7. Find the equation of the circle
    The radius of the circle is the distance from the center (2,-3) to the line.

    The equation of the line in Standard Form is:
    -3x+4y-7=0
    You can also write, (multiply by negative).
    3x-4y+7=0

    The distance to point (2,-3) is:
    \frac{|3(2)-4(-3)+7|}{\sqrt{3^2+4^2}}=\frac{|25|}{\sqrt{25}}=\fra  c{25}{5}=5

    Thus, the equation of circle center at (2,-3) radius r=5 is,
    (x-2)^2+(y+3)^2=5^2=25
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  6. #6
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    Quote Originally Posted by anime_mania View Post
    thanks so much again mr.hacker but i dont know how you got ab=8c
    Ah! I was hoping you figure that out thyself.

    The area of that triangle is base times height. Now there are 2 bases and 2 height.

    1)The legs a,b are a base and height with each other.

    2)The altitude 8 and hypotenuse c are bases and heights with each other.

    3)Area is well-defined. Thus the area using a,b is the same as using the area 8,c. ( A=\frac{1}{2}bh). And there are 2 ways of getting the area.

    4)Thus, \frac{1}{2}ab=\frac{1}{2} (8c).
    ab=8c.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    I drew a red altitutude upon the diagnol.
    The blue is half the length of diagnol (height of isoseles triangle).

    The red can be found from the blue, because it is a 30-60-90 triangle.

    The blue can be from from the base b side again a 30-60-90 triangle.

    Thus, you have an equation involving a,b.

    And another equation from the conditions of the problem.
    a+b=30.

    IF you want I can do it out. I am lazy.
    yes, yes, yes, please work it out thanks and im still wonderin where how that median would fit into the problem
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  8. #8
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    Quote Originally Posted by anime_mania View Post

    yes, yes, yes, please work it out thanks and im still wonderin where how that median would fit into the problem
    I am sure you understand how to get the angles.

    1)Basically, base angles are 60 each. A diagnol is drawn it bisects it. The triangle that we form need to add up to 180 thus the remaining angle is 90 degrees.

    2)The other angles (top angles of trapezoid) need to add up 120. (Because 120+120+60+60=360 in a quadralateral). To the remaining top right angle is 30.

    3)Upper triangle is an isoseles triangle (opposite angles are equal). Thus the altitutde (red) is a perpendicular bisector of the base.

    4)The remaing angle on top is found from the triangle property (x+y+z=180).

    5)Let a be the top side of the trapezoid. And b be the bottom side of trapezoid (the base).

    6)The lower triangle is a 30-60-90 Triangle. That means the side opposite 60 (the one having some blue) is,
    \frac{b\sqrt{3}}{2}.

    7)The blue side is half the side in #6 because of reason in #3. Thus, \mbox{BLUE}=\frac{b\sqrt{3}}{4}

    8)Look at the upper triangle. The hypotenuse is a. Because we have 30-60-90 triangle that means,
    \frac{a\sqrt{3}}{2}=\mbox{BLUE}

    9)Thus #8 and #7 together say that,
    \frac{a\sqrt{3}}{2}=\frac{b\sqrt{3}}{4}
    Some cancelling,
    2a=b
    Meaning the bottom is twice as long.

    10)The median says \frac{a+b}{2}=15
    Thus, a+b=30

    11)Substitute #9 into #10 to get,
    a+2a=3a=30 thus, a=10
    Thus b=20.
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  9. #9
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    cAn you also help me on these too

    1.In triangle ABC, D is a point on AB and E is a point on AC so that DE ll BC. If BC=20 and the area of trapezoid DBCE is one-fourth of the area of triangle ABX, find DE.

    2. Prove using coordinate geometry, that the diagonals of a rhombus are perpendicular ( my teacher was like use variables because just using coordintaes doesnt work to me so if you can do it thanks a lot)

    3. also i am still not sure on how to find the area of the circle because i see no circle for the question about "The altitude to the hypotenuse of a right triangle is 8 and the median to the hypotenuse is 10. find the perimeter of the triangle and the area of the part of the circumscribed circle which lies outside the triangle"

    Thanks so much you are a life saver
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  10. #10
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    Quote Originally Posted by anime_mania View Post

    3. also i am still not sure on how to find the area of the circle because i see no circle for the question about "The altitude to the hypotenuse of a right triangle is 8 and the median to the hypotenuse is 10. find the perimeter of the triangle and the area of the part of the circumscribed circle which lies outside the triangle"
    When you circumscribe that right triangle. The diameter is the length of the hypotenuse! Since you know the diameter you can find the radius and then the area of circle \pi r^2.
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