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Math Help - Needed some hints to complete my solutions on geometry

  1. #1
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    Needed some hints to complete my solutions on geometry

    Drawing 1:
    parallelogram
    d1=40
    d2=74
    Find the area of the parallelogram of the drawing 1.
    I got a= 51 and b=sqrt(937)
    I am not able to find the area because I don't know the angles.
    The answer is 1224 units^2

    Drawing 2:
    d1=14
    d2=48
    Find the height (h) of the rhombus
    The answer is 13,44 units
    Attached Thumbnails Attached Thumbnails Needed some hints to complete my solutions on geometry-mata1.jpg   Needed some hints to complete my solutions on geometry-mata2.jpg  
    Last edited by totalnewbie; September 13th 2005 at 05:19 AM.
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  2. #2
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    Hi,

    Ok, your drawing is not easy to work with since d1 < d2 which is not as it is in your drawings

    In both figures, I will name the parrallelogram ABCD (clockwise, A at the bottom left) with AC the diagonal D1 and BD the diagonal D2. Fianlly, E is the intersection of d1 and d2.

    You know that d1 and d2 intersect at their middle. So m(AE)=m(EC) and m(BE)=m(ED).

    Let's beggin with the second parallelogram. We got m(AB)=m(BC)=m(CD)=m(DA). So in particular, the triangle ABD is isosceles with m(AB)=m(DA). And since AE cuts the side BD at its middle so it is also perpendicular to BD (because ABD is isosceles). So, the triangle ABE is right-angled triangle. We have m(AE)=7 and m(BE)=24 so m(AB)=sqrt(49+24^2) = 25. Let's find the measure of the angle BAE so sin(BAE)=24/25 so m(BAE)=73.74 degrees. Since AE bisects the angle BAD so m(BAE)=m(EAD)=73.74 degrees.

    Ok, I'll name the foot of your height F (So the height is BF). So the angle BAF measures 180-2*73.74 = 32.52 degrees. Now, we are almost done. The triangle BAF is right-angled so sin(BAF) = m(BF)/25 so sin(32.52)=m(BF)/25 so m(BF)=13.44. Done !

    __________________________________________________ ____________

    For the first one, since you have a and b, it is easy. You don't need the angles. Heron formula for the area of a triangle is sufficient. Use it 1 time : to find the area of the triangle ABD. So ABD has sides 74, sqrt(937) and 51 so the p=semiperimeter=(74+sqrt(937)+51)/2=77.805
    Area(ABD)=sqrt(p*(p-74)*(p-sqrt(937))*(p-51))=612. And since the triangle BDC has the same length of sides so its area is also 612 and so the area of the parallelogram is Area(BDC)+Area(ABD)=2*612=1224.

    Done !

    If you don't understand how to find a and b just reply and ask for that.
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  3. #3
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    First of all Hemza, thank you very much for your helpfulness.

    Isosceles trapezium 1:
    the longer base is 44.
    both sides which are aslant (i don't know what is the name of it in english) are 17.
    the diagonal is 39.
    what is the area of the trapezium ?
    I have been trying to solve it. The diagonal is something for but I can't think out how to use this advantage.

    Isosceles trapezium 2:
    bases are 10 and 26.
    diagonals are perpendicular (across) with both sides which are aslant (i don't know what is the name of it in english)
    what is the area of the trapezium ?
    Last edited by totalnewbie; September 13th 2005 at 09:19 AM.
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  4. #4
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    This is not as easy as the ones before (it will take you some patience) :

    1) I will name my trapezium ABCD (clockwise) again with A at the left bottom and the larger base at the bottom (AD). I draw the diagonal AC.

    So m(AC)=39, m(AD)=44 and m(AB)=m(CD)=17.
    I take the triangle ACD (I have all lengths of the sides) and find its area by Heron formula : p=1/2 (39+44+17)=50 so Area(ACD)=...=330.
    Here, it is magnificiant because we can use the usual formula for the Area of a triangle. A(ACD) = b*h/2 where we can take the base being the side AD so out height CH (H is the foot of the height) will be also the height of the trapezium. Cool !

    so A(ACD)=330=b*h/2. We know b=44 so h=15. So m(CH)=15 and the triangle CHD is right-angled so by Pytagorus we can find m(HD)=sqrt(CD^2-CH^2) = 8 (beautiful numbers in this kind of calculations with sqrt usually are a sign that you are on the right track, I say usually).

    So, since the trapezium is isosceles, we have a symetry and then m(BC)=m(AD)-2*m(HD)=44-2*8=28

    A(trap. ABCD)=(BC+AD)*CH/2=(28+44)*15/2=540.

    __________________________________________________ ______________-
    2) I give the answer to this one also but you should try it on your own using the tricks in number 1).

    (try....)

    Ok, I name again ABCD the same way and draw a height CH the same way. Because of the symetry of the isosceles trapezium, m(HD)=(m(AD)-m(BC))/2=(26-10)/2=8 and then m(AH)=26-8=18.

    We have that the triangle ACD is right-angled at C (diagonals perpendicular to the sides....... in the asumptions). So CH^2 = AH*HD (I hope you know this formula, I think you are the one I worked it with not long ago). This formula can only be used on right-angled triangles. So CH=sqrt(8*18)=12.

    Now, A(trap. ABCD)=(BC+AD)*CH/2=(10+26)*12/2=216.
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