maths challenge question.
i am stuck on this question, not really sure how to approach it. i tried to find the height of the triangle, but couldn't work it out because i had too many unknowns. i can't figure out a way to use the info to find the answer.
the answer, for those who are wondering is D
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Who is this aimed at? 15-16 year olds or 16-18? You know all the angles in the centre are 90 degrees (or you should). And the only values you have are 8 and 12. I'd work with them and attempt to solve them using the info available.I'm giving it an attempt.
(edited 9 years ago)
Original post by The Big Stone
i am stuck on this question, not really sure how to approach it. i tried to find the height of the triangle, but couldn't work it out because i had too many unknowns. i can't figure out a way to use the info to find the answer.
the answer, for those who are wondering is D
i am stuck on this question, not really sure how to approach it. i tried to find the height of the triangle, but couldn't work it out because i had too many unknowns. i can't figure out a way to use the info to find the answer.
the answer, for those who are wondering is D
You can find the area of the quadrilateral TSRQ with the information given
That eliminates answers A, B, and C
It also suggests that E is impossible
Original post by MathMeister
It is no coincidence 12*8= 96 and 96/2= the answer
96/2 = 48
Original post by The Big Stone
i am stuck on this question, not really sure how to approach it. i tried to find the height of the triangle, but couldn't work it out because i had too many unknowns. i can't figure out a way to use the info to find the answer.
the answer, for those who are wondering is D
i am stuck on this question, not really sure how to approach it. i tried to find the height of the triangle, but couldn't work it out because i had too many unknowns. i can't figure out a way to use the info to find the answer.
the answer, for those who are wondering is D
1. Note that triangles PQS and QSR have the same base and height. (So what?)
2. Note that the medians of a triangle intersect each other in the ratio 2:1
3. Note that you have a bunch of right triangles to work with.
any tips? hints?
Original post by MathMeister
any tips? hints?
Between post 3 and 7 there are all the hints needed
Original post by TenOfThem
You can find the area of the quadrilateral TSRQ with the information given
That eliminates answers A, B, and C
It also suggests that E is impossible
That eliminates answers A, B, and C
It also suggests that E is impossible
Okay, but imagine i wasn't given possible answers, i couldn't eliminate any answers, so how would i get to 64?
Original post by The Big Stone
Okay, but imagine i wasn't given possible answers, i couldn't eliminate any answers, so how would i get to 64?
Well I would use the similar triangles
Thats exactly what i thought of!!!
Original post by The Big Stone
Okay, but imagine i wasn't given possible answers, i couldn't eliminate any answers, so how would i get to 64?
Did you not understand the hints I gave earlier? I've done most of the work for you there.
Original post by atsruser
Did you not understand the hints I gave earlier? I've done most of the work for you there.
Interestingly you went in a completely different direction to me
I thought so.
Original post by Super199
Is this a senior challenge question?
I would have thought intermediate the knowledge required is not a high level
Original post by atsruser
Did you not understand the hints I gave earlier? I've done most of the work for you there.
i understand now. the 2 triangles have the equal area, and so working out the area of one x2 would give you the answer, also the ratios you work out the lengths, but wouldn't it be easier to say that PTR = QTR instead of saying QSR=PQS ?
Original post by Super199
Is this a senior challenge question?
Yh, but it was from 2007
Original post by TenOfThem
Interestingly you went in a completely different direction to me
What's your solution?
Original post by The Big Stone
i understand now. the 2 triangles have the equal area, and so working out the area of one x2 would give you the answer, also the ratios you work out the lengths,
Right.
but wouldn't it be easier to say that PTR = QTR instead of saying QSR=PQS ?
Yes, a bit. Once I found a solution, I couldn't be bothered to see if there was a quicker one though.
Original post by atsruser
Yes, a bit. Once I found a solution, I couldn't be bothered to see if there was a quicker one though.
Yes, a bit. Once I found a solution, I couldn't be bothered to see if there was a quicker one though.
kay, thanks.
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