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    (Original post by Mehrdad jafari)
    Is the height of the triangle ABC not 8 as 16sin(30)=8 ? which would result in the height of the other triangle being 2root7
    I found that bit wierd because:

    if you do sin 30 = opp/ 16 , opp = 8cm
    Then via pythag, you get 8 root 3 as the height

    However, if you do sohcohtoa twice, both sides become 8cm.

    When I checked pythag, 8 root 3 worked so I stuck with that
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    (Original post by rede121)
    I found that bit wierd because:

    if you do sin 30 = opp/ 16 , opp = 8cm
    Then via pythag, you get 8 root 3 as the height

    However, if you do sohcohtoa twice, both sides become 8cm.

    When I checked pythag, 8 root 3 worked so I stuck with that
    You cannot use Pythagoras as you don't have the length of the side AB but if you did work it out, then AB= 16cos(30)=8root3 but this is the length of the side Ab and not the height .

    Because the other triangle has two equal sides, one of which is the height of the triangle we have just found out, then the height of the right hand side triangle would be h=root(82-62)= 2root7 . Therefore the area of the right hand side triangle would be 1/2*(6)(2root7)=3root7
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    (Original post by Mehrdad jafari)
    You cannot use Pythagoras as you don't have the length of the side AB but if you did work it out, then AB= 16cos(30)=8root3 but this is the length of the side Ab and not the height .

    Because the other triangle has two equal sides, one of which is the height of the triangle we have just found out, then the height of the right hand side triangle would be h=root(82-62)= 2root7 . Therefore the area of the right hand side triangle would be 1/2*(6)(2root7)=3root7
    There are so many way you can do this questions, so bizarre.

    I found length AB by doing 16cos60 = 8
    Then found length BC using pythag = 8 root 3

    So the area of the right angled triangle (left triangle) would be 0.5 x 8 x 8 root 3 = 32 root 3

    The right hand side triangle is isosceles so you would need to split that in half, to get two right angles, then find the height via pythag again. then you can use the 1/2 x base x height formula

    I am intrigued to see what the answer is for this...
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    (Original post by rede121)
    There are so many way you can do this questions, so bizarre.

    I found length AB by doing 16cos60 = 8
    Then found length BC using pythag = 8 root 3

    So the area of the right angled triangle (left triangle) would be 0.5 x 8 x 8 root 3 = 32 root 3

    The right hand side triangle is isosceles so you would need to split that in half, to get two right angles, then find the height via pythag again. then you can use the 1/2 x base x height formula

    I am intrigued to see what the answer is for this...
    (Original post by Mehrdad jafari)
    You cannot use Pythagoras as you don't have the length of the side AB but if you did work it out, then AB= 16cos(30)=8root3 but this is the length of the side Ab and not the height .

    Because the other triangle has two equal sides, one of which is the height of the triangle we have just found out, then the height of the right hand side triangle would be h=root(82-62)= 2root7 . Therefore the area of the right hand side triangle would be 1/2*(6)(2root7)=3root7
    Isn't the height 8 and the base 8rt3
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    (Original post by swagmister)
    Isn't the height 8 and the base 8rt3
    There are many way to go about this question but the simplest way would be the best way .



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    16sin30 =8 height
    16cos30 = 8rt3 base
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    Oh right yeah then the 8rt3 of abc isn't relevant if they need to find area of bcd but you don't need to half it do you?
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    (Original post by swagmister)
    16sin30 =8 height
    16cos30 = 8rt3 base
    I think it depends on what you are using as your angle, I used 60 degree angle at C to work out AB using 16cos60 = 8 (base) and height was found via pythag
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    (Original post by swagmister)
    Oh right yeah then the 8rt3 of abc isn't relevant if she needs to find area of bcd but you don't need to half it do you?
    You could work that out to find the height of the triangle ABC using the Pythagoras but that would involve another step in calculation which is not really preferred.
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    (Original post by rede121)
    I think it depends on what you are using as your angle, I used 60 degree angle at C to work out AB using 16cos60 = 8 (base) and height was found via pythag
    Yeah, you are right . I used the sine rule there
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    (Original post by rede121)
    I think it depends on what you are using as your angle, I used 60 degree angle at C to work out AB using 16cos60 = 8 (base) and height was found via pythag
    It still gives you 8 as the height though as a and o would be different depending on the angle
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    Isn't the area of bcd 12rt7?
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    (Original post by swagmister)
    Isn't the area of bcd 12rt7?
    Oh sorry! It should be 6root7 as Area= 1/2(6)(2root7)
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    (Original post by Mehrdad jafari)
    Oh sorry! It should be 6root7 as Area= 1/2(6)(2root7)
    But there are 2 triangles in bcd so it's like the area of square
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    (Original post by swagmister)
    But there are 2 triangles in bcd so it's like the area of square
    That's totally correct, my mistake! I'm glad I don't need to go back to GCSE
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    (Original post by Mehrdad jafari)
    That's totally correct, my mistake! I'm glad I don't need to go back to GCSE
    Hahah
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    (Original post by swagmister)
    It still gives you 8 as the height though as a and o would be different depending on the angle
    Ok, I typed it up, does this make more sense? This question is going to be the bane of my life
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    (Original post by rede121)
    Ok, I typed it up, does this make more sense? This question is going to be the bane of my life
    Haha yeah on the first one you drew you wrote the height as 8rt3 that's why I was confused
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    (Original post by swagmister)
    Haha yeah on the first one you drew you wrote the height as 8rt3 that's why I was confused
    my bad
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    the answer i got was 12 root 7.
    BC is 8, so the perpendicular height of BCD is 2 root 7.
    A=0.5bh
    =0.5 x 2root7 x 12
    =12root7
 
 
 
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