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Thread: geometry question

  1. #1
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    geometry question

    Question: A triangle is bounded by the three lines


    x +y +1 =0, y=2x-1, y=k
    where k is a positive integer. For what values of k is the area of the triangle less than 2008?
    I found out the point of intersections which are
    (0, -1)
    (-(1+k). k)
    $(\frac{1+k}{2} ,k)$




    what do I do from here
    any help will be appreciate
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  2. #2
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    Re: geometry question

    note there are two possible values for $k$ (a sketch shows an upper triangle and a congruent lower triangle)

    $x+y+1=0$ intersects $y=2x-1$ at $(0,-1)$

    height of the upper triangle is $(k+1)$

    base of the upper triangle is $b = \dfrac{k+1}{2} - [-(k+1)] = \dfrac{3(k+1)}{2}$

    $\dfrac{1}{2} \cdot \dfrac{3(k+1)}{2} \cdot k < 2800$

    solve the quadratic inequality
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  3. #3
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    Re: geometry question

    The lines x+ y+ 1= 0, and y= k intersect where x+ k+ 1= 0 or x= -k- 1. That is the point (-k- 1, k).
    The lines y= 2x- 1 and y= k intersect where 2x- 1= k or x= (k+1)/2. That is the point ((k+1)/2, k).
    The lines x+ y+ 1= 0 and y= 2x- 1 intersect where x+ 2x-1+ 1= 3x= 0. That is the point (0, -1)

    Choose the horizontal line y= k as base. The distance from (1- k, k) to (k+1, k) is k+ 1- (1- k)= 2k. That is the length of the base.
    The altitude is the distance from the third vertex, (0, -1), to the base, measured along the line perpendicular to the base. Since the base is y= k, the perpendicular is x= 0. That crosses y= k at (0, k) which has distance from (0, -1) k- (-1)= k+ 2 and that is the "height" of the triangle. I presume you know that the area of a trangle is "1/2 base times height".
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    Re: geometry question

    Quote Originally Posted by HallsofIvy View Post
    The lines x+ y+ 1= 0, and y= k intersect where x+ k+ 1= 0 or x= -k- 1. That is the point (-k- 1, k).
    The lines y= 2x- 1 and y= k intersect where 2x- 1= k or x= (k+1)/2. That is the point ((k+1)/2, k).
    The lines x+ y+ 1= 0 and y= 2x- 1 intersect where x+ 2x-1+ 1= 3x= 0. That is the point (0, -1)

    Choose the horizontal line y= k as base. The distance from (1- k, k) to (k+1, k) is k+ 1- (1- k)= 2k. That is the length of the base.
    The altitude is the distance from the third vertex, (0, -1), to the base, measured along the line perpendicular to the base. Since the base is y= k, the perpendicular is x= 0. That crosses y= k at (0, k) which has distance from (0, -1) k- (-1)= k+ 2 and that is the "height" of the triangle. I presume you know that the area of a trangle is "1/2 base times height".
    Quote Originally Posted by skeeter View Post
    note there are two possible values for $k$ (a sketch shows an upper triangle and a congruent lower triangle)

    $x+y+1=0$ intersects $y=2x-1$ at $(0,-1)$

    height of the upper triangle is $(k+1)$

    base of the upper triangle is $b = \dfrac{k+1}{2} - [-(k+1)] = \dfrac{3(k+1)}{2}$

    $\dfrac{1}{2} \cdot \dfrac{3(k+1)}{2} \cdot k < 2800$

    solve the quadratic inequality
    Thank you both mathematicians a lot
    I raise a problem with this question

    the answer is $1\leqslant x\leqslant 50$
    but applying the advice you gave me
    I did it this

    $\frac{1}{2}\cdot \frac{3)(k+1)}{2}\cdot (k+1)<2008 $
    $\frac{3}{4}(k+1)^2 <2008$
    how do I get from the above to the answer $1 \leqslant x\leqslant 50$
    I believe that the answer may be wrong but I wanted a second opinion.

    I uploaded the worksheet it is question number 19C1 Coordinate Geometry Hard.pdf
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  5. #5
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    Re: geometry question

    I would set:

    3\le\frac{3}{4}(k+1)^2<2008

    Multiply through \frac{4}{3}:

    4\le (k+1)^2<\frac{8032}{3}

    This implies (keeping in mind that k is a natural number):

    2\le k+1\le\left\lfloor\sqrt{\frac{8032}{3}} \right\rfloor<\sqrt{\frac{8032}{3}}

    2\le k+1\le51

    Subtract through by 1:

    1\le k\le50
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    Re: geometry question

    Quote Originally Posted by MarkFL View Post
    I would set:

    3\le\frac{3}{4}(k+1)^2<2008

    Multiply through \frac{4}{3}:

    4\le (k+1)^2<\frac{8032}{3}

    This implies (keeping in mind that k is a natural number):

    2\le k+1\le\left\lfloor\sqrt{\frac{8032}{3}} \right\rfloor<\sqrt{\frac{8032}{3}}

    2\le k+1\le51

    Subtract through by 1:

    1\le k\le50
    Dear Mr Mark thank you but i want to ask some questions
    firstly where did you get the number 3 from 3\le\frac{3}{4}(k+1)^2<2008

    secondly, the answer uses a less than or equal to sign. I wanted to know is there a reason for this. This is because i dont want to make a mistake by leaving it out since i can lose marks
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  7. #7
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    Re: geometry question

    The problem, as you stated in your first post, required that k be a positive integer. The smallest positive integer is 1 and, if k= 1, \frac{3}{4}(1+ 1)^2= 3. That says that 3\le \frac{3}{4}(k+ 1)^2. The other inequality, that \frac{3}{4}(k+ 1)^2< 2008 follows from the condition that the area is less than 2008. Once you have 3\le k+ 1<\sqrt{\frac{8032}{3}}, knowing that \sqrt{\frac{8032}{3}}= 51.7425, and that k is an integer, it follows that 3\le k+ 1\le 51.
    Last edited by HallsofIvy; Mar 14th 2017 at 06:35 PM.
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  8. #8
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    Re: geometry question

    Quote Originally Posted by HallsofIvy View Post
    The problem, as you stated in your first post, required that k be a positive integer. The smallest positive integer is 1 and, if k= 1, \frac{3}{4}(1+ 1)^2= 3. That says that 3\le \frac{3}{4}(k+ 1)^2. The other inequality, that \frac{3}{4}(k+ 1)^2< 2008 follows from the condition that the area is less than 2008. Once you have 3\le k+ 1<\sqrt{\frac{8032}{3}}, knowing that \sqrt{\frac{8032}{3}}= 51.7425, and that k is an integer, it follows that 3\le k+ 1\le 51.
    thank you very much
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