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maths help pleASE

ill post the question

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Reply 1
here it is
Can you post what you've done so far?
Reply 3
Original post by RogerOxon
Can you post what you've done so far?


sorry i have no idea what to do... im really stuck, andd there is no markscheme so i cant check the answers
We have AC and BC, so need AB to be able to calculate the perimeter. Place a point D on BC, such that ADB is a right angle. We know AC and ACB, so can calculate AD and CD. Take CD from BC to get DB and then use Pythagoras to get AB. Now add the sides to get the perimeter.
(edited 6 years ago)
For 6, we have AB and the ratio of AE to BE, so can calculate AE and BE (Pythagoras again).

We know the ratio of AD to AE, so can calculate CD from BE (similar triangles).

Finally, we know AE and AD, so can calculate ED, and get the area from ED*(BE+CD)/2.
(edited 6 years ago)
Reply 6
Original post by usernamenew
here it is

Remember, this is a show that question. This is arguably the nicest question, because if you found got the result that is stipulated in the question -in this case it is: paremeter=(4+surd7)x-, then you know that your working out was correct.

Essentially, you have a triangle but it doesn't say whether it is right angled or not. You'll need to assume that it isn't a right angle triangle so you will need to use the formulas.

You can either the sin, cosine and area formulas.

You will need to find the paremeter so due to this you will need to use the cosine formula to find the side CB. This formula is: c2 = a2+b2-2ab*cos(C). All you need to do is to sub in the known angle (which is 60) and the lengths. Then you obvously need to then simplify the the result.

Then, with the side which you have just found out, find the perimeter and modify it to the format which it wants.

All I am trying to do is to 'push' you in doing the question yourself.
(edited 6 years ago)
Reply 7
Original post by stoyfan
Remember, this is a show that question. This is arguably the nicest question, because if you found got the result that is stipulated in the question -in this case it is: paremeter=(4+surd7)x-, then you know that your working out was correct.

Essentially, you have a triangle but it doesn't say whether it is right angled or not. You'll need to assume that it isn't a right angle triangle so you will need to use the formulas.

You can either the sin, cosine and area formulas.

You will need to find the paremeter so due to this you will need to use the cosine formula to find the side CB. This formula is: c2 = a2+b2-2ab*cos(C). All you need to do is to sub in the known angle (which is 60) and the lengths. Then you obvously need to then simplify the the result.

Then, with the side which you have just found out, find the perimeter and modify it to the format which it wants.

All I am trying to do is to 'push' you in actually doing the question.


thank you i really appreciate it, when i get an answer, do you mind checking if its right?
Reply 8
Original post by usernamenew
thank you i really appreciate it, when i get an answer, do you mind checking if its right?


Sure, in these type of questions, it is your working out that is really going to get you the marks. You know if you got the correct answer if the paremeter you found outis the same as the one given to you in the question.
Reply 9
Original post by stoyfan
Sure, in these type of questions, it is your working out that is really going to get you the marks. You know if you got the correct answer if the paremeter you found outis the same as the one given to you in the question.

yeh but i doubt i will get it right since i cant even substitute the numbers as its x and 3x? am i not supposed to be finding out what ab is, not cb?
if i substituted in would it be
x2 = x2 + 3x2- 2(x)(3x)cos60?
thank you so much! just one question, where did the 1 and 3 come from at the end, to make 4?
thank you, i get it now:smile:
For Q6, I got 16. Let me know if you disagree.
I'm not the OP - I was posting the answer for them to check theirs.
I keep getting 26.6 for question 6:frown:(((
From the diagram we know AE=2BE
Therefore BE=x and AE=2x
We are also told that AB is 3 root5
Using pythagoras theorem (a^2 + b^2 = c^2) on triangle ABE we can find the value of x
2x^2+x^2=(3root5)^2
3x^2=3root5 x 3root5
3x^2=9 x 5
3x^2= 45
x^2=45/3=15
x= root15
therefore, BE= root15

we know that AE must be 2root15
and that
3AD=5AE
therefore
3AD=5 x 2root15
3AD= 10root15
AD= 10root15/3
ED = AD- AE = 10root15/3 - 2 root 15 = 4root15/3
therefore, ED= 4root15/3

Triangles ABE and ACD are similar
To find the scale factor we do the length of AD divided by the corresponding length of AE
AD/AE = 10root15 over 3 / 2root15 = 10/6
the length of CD is the corresponding side length multiplied by the scale factor
so CD= 10/6 x root15 = 5root15/3
Therefore CD = 5root15/3

The shaded area forms a trapezium
the area of a trapezium is 1/2(a+b)xh
a=the length of the top side=BE= root15
b= the length of the bottom side=CD= 5root15/3
c= the height of the trapezium= ED= 4root15/3
Therefore the area = 1/2(root15+5root15/3)x 4root15 over3 =26.6
Which means the area is 26.6??:s-smilie:

Any help appreciated thanks

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