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Ah yes, Surds. My favourite. Or something.
I've been trying to do these basically all evening, so any help would be appreciated muchos.
Right, I have 4 right-angled triangles here, and need to find the perimeter and area of each. I'll list the sides given after each question. I think you have to use pythagoras to work out the last side before finding the area and perimeter.
Any help with putting the square root sign in instead of writing it would be great also :D
1: 6 (a), squareroot24(o)
2: squareroot20 (h), 4 (a)
3: sqaureroot6 - 1 (a), squareroot6 + 1 (o)
4: 4squareroot5, 10, 10 (Isocoles, so two of the longest length?)
Hope that sort of makes sense. I've answers in front of me, but obviously if I don't understand it, and therefore don't do workings, there's no point.
So yes, thankyou in advance :)
many goats
I've been trying to do these basically all evening, so any help would be appreciated muchos.
Right, I have 4 right-angled triangles here, and need to find the perimeter and area of each. I'll list the sides given after each question. I think you have to use pythagoras to work out the last side before finding the area and perimeter.
Any help with putting the square root sign in instead of writing it would be great also :D
1: 6 (a), squareroot24(o)
2: squareroot20 (h), 4 (a)
3: sqaureroot6 - 1 (a), squareroot6 + 1 (o)
4: 4squareroot5, 10, 10 (Isocoles, so two of the longest length?)
Hope that sort of makes sense. I've answers in front of me, but obviously if I don't understand it, and therefore don't do workings, there's no point.
So yes, thankyou in advance :)
many goats
Sine I can't work out SsxPro's stuff (as in, I CBA), and I haven't done any maths for a while, i WANT to do this.
And for the record, aa "V" is a square root sign on this post.
many goats wrote:
> 1: 6 (a), squareroot24(o)
Well, h = V(36 + 24) = V60 = V4 V15 = 2 V3V5 (Ouch)
Area = 1/2.o.a
=1/2.6.V24
=1/2.V36.V24
=1/2. V864 (wow, big number)
=1/2. V16V54
=1/2.4V54
=2V54
=2V9V6
=6V6
Perimeter = V36 + V24 + V60
=V120
=V4V30
=2V30
I'd finish them, but I'm missing Lost argh.
And for the record, aa "V" is a square root sign on this post.
many goats wrote:
> 1: 6 (a), squareroot24(o)
Well, h = V(36 + 24) = V60 = V4 V15 = 2 V3V5 (Ouch)
Area = 1/2.o.a
=1/2.6.V24
=1/2.V36.V24
=1/2. V864 (wow, big number)
=1/2. V16V54
=1/2.4V54
=2V54
=2V9V6
=6V6
Perimeter = V36 + V24 + V60
=V120
=V4V30
=2V30
I'd finish them, but I'm missing Lost argh.
many goats wrote:
> 1: 6 (a), squareroot24(o)
Is that as in 6 is the adjacent side and \sqrt{24} the opposite? If that's the case, assuming this is a right angle triangle you are looking at, can't you calculate the area with "1/2 x base x height"? Then to calculate the perimeter, use pythagorus to find the third side (in this case its \sqrt{36 + 24}=2\sqrt{3}\sqrt{5}, then add that to the other two sides. Each answer has some nasty square roots in, but assuming its a right angle triangle, that should be correct. Also don't forget that \sqrt{24}=2 x \sqrt{2x3} if that helps simplify.
> 2: squareroot20 (h), 4 (a)
Should be the same, but be warned that here you have the largest side, (h), so when using pythag, the missing side is \sqrt{20-16}=2.
> 3: sqaureroot6 - 1 (a), squareroot6 + 1 (o)
Some nice cancellation going on there, but essentially the same again.
> 4: 4squareroot5, 10, 10 (Isocoles, so two of the longest length?)
Perimeter is clear, then perhaps just split into two right angle triangles to work out the area.
Have fun!
> 1: 6 (a), squareroot24(o)
Is that as in 6 is the adjacent side and \sqrt{24} the opposite? If that's the case, assuming this is a right angle triangle you are looking at, can't you calculate the area with "1/2 x base x height"? Then to calculate the perimeter, use pythagorus to find the third side (in this case its \sqrt{36 + 24}=2\sqrt{3}\sqrt{5}, then add that to the other two sides. Each answer has some nasty square roots in, but assuming its a right angle triangle, that should be correct. Also don't forget that \sqrt{24}=2 x \sqrt{2x3} if that helps simplify.
> 2: squareroot20 (h), 4 (a)
Should be the same, but be warned that here you have the largest side, (h), so when using pythag, the missing side is \sqrt{20-16}=2.
> 3: sqaureroot6 - 1 (a), squareroot6 + 1 (o)
Some nice cancellation going on there, but essentially the same again.
> 4: 4squareroot5, 10, 10 (Isocoles, so two of the longest length?)
Perimeter is clear, then perhaps just split into two right angle triangles to work out the area.
Have fun!
many goats wrote:
> And Tomm, I know what pythagoras is, dammit. It's hardly the most
> challenging bit of the damn question, is it?
The surds are hardly that challenging either tbh.
> And Tomm, I know what pythagoras is, dammit. It's hardly the most
> challenging bit of the damn question, is it?
The surds are hardly that challenging either tbh.
I hate surds, managed to get an A-level without being able to do them.
I could do them if i read how, then snapped, and for that one lesson, i would be able to, but its just one of those things you forget.
It may be because its useless...
Im glad i dont have to do that any more.
I could do them if i read how, then snapped, and for that one lesson, i would be able to, but its just one of those things you forget.
It may be because its useless...
Im glad i dont have to do that any more.
munn wrote:
> You haven't said which sides are which though (hypotenuse etc), so I
> can't work them out.
True.
I'll have another go, and I'll edit the original post.
And Tomm, I know what pythagoras is, dammit. It's hardly the most challenging bit of the damn question, is it?
> You haven't said which sides are which though (hypotenuse etc), so I
> can't work them out.
True.
I'll have another go, and I'll edit the original post.
And Tomm, I know what pythagoras is, dammit. It's hardly the most challenging bit of the damn question, is it?
a2 + b2 = c2
Where c is the hypotenuse. It's only squaring, so its not like you'll have to do much to the surds...
Edit for number 4. 4route(5) is the same as route(16x5) incase you didnt know...
Where c is the hypotenuse. It's only squaring, so its not like you'll have to do much to the surds...
Edit for number 4. 4route(5) is the same as route(16x5) incase you didnt know...
You haven't said which sides are which though (hypotenuse etc), so I can't work them out.
Ah yes, Surds. My favourite. Or something.
I've been trying to do these basically all evening, so any help would be appreciated muchos.
Right, I have 4 right-angled triangles here, and need to find the perimeter and area of each. I'll list the sides given after each question. I think you have to use pythagoras to work out the last side before finding the area and perimeter.
Any help with putting the square root sign in instead of writing it would be great also :D
1: 6 (a), squareroot24(o)
2: squareroot20 (h), 4 (a)
3: sqaureroot6 - 1 (a), squareroot6 + 1 (o)
4: 4squareroot5, 10, 10 (Isocoles, so two of the longest length?)
Hope that sort of makes sense. I've answers in front of me, but obviously if I don't understand it, and therefore don't do workings, there's no point.
So yes, thankyou in advance :)
many goats
I've been trying to do these basically all evening, so any help would be appreciated muchos.
Right, I have 4 right-angled triangles here, and need to find the perimeter and area of each. I'll list the sides given after each question. I think you have to use pythagoras to work out the last side before finding the area and perimeter.
Any help with putting the square root sign in instead of writing it would be great also :D
1: 6 (a), squareroot24(o)
2: squareroot20 (h), 4 (a)
3: sqaureroot6 - 1 (a), squareroot6 + 1 (o)
4: 4squareroot5, 10, 10 (Isocoles, so two of the longest length?)
Hope that sort of makes sense. I've answers in front of me, but obviously if I don't understand it, and therefore don't do workings, there's no point.
So yes, thankyou in advance :)
many goats