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    • Thread Starter

    I'm really stuck with this question and keep getting it wrong! Could somebody take me through how you would attempt it step by step?

    1. A triangle has vertices (5,0), (q,0) and (10,-1) where q > 5.
    Given that the area of the triangle is 4, find the value of q.

    Have you drawn a diagram? What have you tried?

    what topic is this?

    Recall the equation for the area of a triangle \frac{1}{2}BH.

    Points (5,0) and (q,0) lie one the same line. You could pose B=q-5, and given that q>5, there is no need for an absolute value.

    Points (5,0) and (q,0) being on the same line (the horizontal axis), H has to be vertical in order to be perpendicular to B. H=|\Delta y|=|-1-0|=1.

    Now that we know the equations for B and H, we can solve for q:

    \frac{1}{2}(q-5)(1)=4 \\ q-5=8 \\ q = 13
    • Thread Starter

    (Original post by samantham999)
    what topic is this?
    Core 1
    • Thread Starter

    (Original post by Kvothe the Arcane)
    Have you drawn a diagram? What have you tried?
    Yes I did, but it turns out I was over complicating the question haha. I've got the answer now
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