isosceles triangles (1 Viewer)

lucymowat

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noooooo my dad figured it out
you just had to divide it into 2 right-angled triangles :(
 

BECBEC1992

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Dammit it took me foreva on that question and I couldnt do it
 

bmn

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I done that, but I didn't label them as right angles... Wonder if I'll get the mark...
 

meilz92

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wtf? i thought we had to divide it into 4 isoceles triangles?

im so conufsed. that was the most pathetic excuse for a HSC question ever.
 

imoO

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I walked out astounded that I couldn't draw some triangles
 

lucymowat

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wtf? i thought we had to divide it into 4 isoceles triangles?
first u had to divide it into right angled ones and then into 2 isosceles in each ... that was how it related to the first question.... dammit wish i'd thought of it
 

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this is what i did


not exactly mathematical but neither was the question
 

lolliepop2

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That's what i did too. Do you think we will get marks for it though?
 

johony

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a ridiculously easy mark in hindsight, but at the time i probably overcomplicated the answer and i've probably lost a mark.
 

mitchwong650

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I can't believe so many people couldn't do this question!

what do you think c) i) to c) iv) were for?

simply construct a line from the top corner to the hypotenuse, and you've got two of the triangles like the one right above it, then construct line bisectors in both and voila.

Here:

 

Aquawhite

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The post above is correct and is what the examiners will be looking for... I don't know if they'll award the others that have 4 isosceles triangles... but they would be looking for this.

Shame I didn't do it XD
 

Finx

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I can't believe so many people couldn't do this question!

what do you think c) i) to c) iv) were for?

simply construct a line from the top corner to the hypotenuse, and you've got two of the triangles like the one right above it, then construct line bisectors in both and voila.

Here:

Would we still have received the mark if we didn't put the right angles in?
 

timmy_tummy

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I can't believe so many people couldn't do this question!

what do you think c) i) to c) iv) were for?

simply construct a line from the top corner to the hypotenuse, and you've got two of the triangles like the one right above it, then construct line bisectors in both and voila.

Here:

YES THAT'S WHAT I DID! =)
they obviously wanted you to use the bits from above...
 

HTB

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errr....u only need the right angles when u divide it the first time down the middle

the other line goes across the middle, dividing the two sides into equal parts (lines intersecting parallel lines in the same ratio)

then its the same idea as the previous part, the other triangles arent necessarily right angled
 

mitchwong650

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errr....u only need the right angles when u divide it the first time down the middle

the other line goes across the middle, dividing the two sides into equal parts (lines intersecting parallel lines in the same ratio)

then its the same idea as the previous part, the other triangles arent necessarily right angled
This makes sense, as long as you make it clear that "E" and "F" are the midpoints you should be fine,

you must have a right angle for "AC" though, in order to make it relate to parts c) i) to c) iv)

so i suppose you don't HAVE to have right angles in the smaller triangles
 

cpt. fantastic

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I can't believe so many people couldn't do this question!

what do you think c) i) to c) iv) were for?

simply construct a line from the top corner to the hypotenuse, and you've got two of the triangles like the one right above it, then construct line bisectors in both and voila.

Here:

thats not 4 isosceles triangles thats 4 isosceles trianges and 2 right angle triangles
 

lyounamu

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That question has to be the most ridiculous question ever.

Thank God I didn't do 2 unit this year haha
 

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