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helppppppp, difficult maths question

Can any one help me with this maths problem? Any help would be useful, thanks

AC=16cm
Angle ABC=90 degrees
Angle CAB= 30 degrees

BC=BD
CD=12cm

Calculate the area of triangle BCD. Give your answer correct to 3 significant figures.

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Reply 1

Kallisto

Entertainment Forum Helper

Original post by AceOfSpade

Maybe a sketch helps you to understand? so you (and we) can see the coherences of that construction.

From the clues you have given, I would advice you to begin with the triangle ABC and to calculate BC and AB.

(edited 8 years ago)

Reply 2

AceOfSpade

Original post by Kallisto

I hope this helps

Scan 2015-10-16 0002.jpg51.7KB

Reply 3

10tillyh

Draw a pic then use pithagourous theory ( a^2+b^2=c^2) :smile:

Reply 4

PlayerBB

Is it 32 ?

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Reply 5

AceOfSpade

Original post by PlayerBB

Is it 32 ?

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I'm not sure. How did you get 32?

Reply 6

PlayerBB

h= 16Sin30/Sin90
h=8 then area of triangle BCD= 0.5*8*12*Sin(45) = 33.9
P.s sorry wrong calculation, not 32 but 33.9

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Reply 7

AceOfSpade

Original post by PlayerBB

h= 16Sin30/Sin90
h=8 then area of triangle BCD= 0.5*8*12*Sin(45) = 33.9
P.s sorry wrong calculation, not 32 but 33.9

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Thank you so much, i'm not sure if that's the answer because I don't have the answers but thanks anyway

Reply 8

username2176591

I got 6 root 38
(rushed)

Reply 9

AceOfSpade

Original post by Scott_dp

I got 6 root 38
(rushed)

How, please?

Reply 10

username2176591

CB=16sin(30)=8
BC=BD=8
So triangle CBD is isoceles with side lengths 12, 8 and 8 cm.
Use pythagoras to find height:
Root(8^2-6^2)=Root(38)
Area = half base * height = (12/2)*root(38)=6 root 38

Reply 11

PlayerBB

Original post by AceOfSpade

Thank you so much, i'm not sure if that's the answer because I don't have the answers but thanks anyway

You're welcome and when you have the answer, post it please

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Reply 12

AceOfSpade

Original post by PlayerBB

You're welcome and when you have the answer, post it please

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Okay

Reply 13

PlayerBB

But from the pic you have submitted, it seems that ACD is 90 degrees, if it is the BCD is 30 and therefore since BC=BD the BC= 8 from Pythagoras theorm and sin30*8*12*0.5 = 24 so the area is 24 if the angle is as the pic which is 90

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Reply 14

AceOfSpade

Original post by PlayerBB

They haven't given me the angle for ACD so it does look like 90degrees but I'm not 100% sure

Reply 15

PlayerBB

Original post by AceOfSpade

They haven't given me the angle for ACD so it does look like 90degrees but I'm not 100% sure

Hope it is!

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Reply 16

rede121

Original post by AceOfSpade

I did this really quickly, based on your diagram, check through to make sure the maths is ok

Original post by rede121

I did this really quickly, based on your diagram, check through to make sure the maths is ok

Your maths looks fine to me but I still don't what the answer is because everyone seems to be getting different ones. But thank you so much for your solution and I will post the answer on Wednesday so you can see if you were correct.

Reply 18

Absent Agent

Original post by rede121

I did this really quickly, based on your diagram, check through to make sure the maths is ok

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

Reply 19

rede121

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