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secondary 4 | A Maths
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How to work out the steps part ii ? Tks
Date Posted: 2 years ago
Views: 458
So for part ii, we reflect Q along line AB to get an "image" R.
The shortest distance between P and R is the straight line joining them (i.e. the straight line PR).
But PR = PC + CR and then it's pretty obvious that CR = CQ. so this leads to PR = PC + CQ being the shortest.
From here, triangles PAC and RBC are similar
- angles PAC and RBC are right angles
- angles PCA and RCB are vertically opposite each other
The length ratio PA : RB is 10 : 15, which simplifies to 2 : 3.
Correspondingly, AC : BC = 2 : 3 as well due to the similarity of both triangles.
So, AC : AB = 2 : 5, and AC = 2/5 x 20 = 8 km.
The shortest distance between P and R is the straight line joining them (i.e. the straight line PR).
But PR = PC + CR and then it's pretty obvious that CR = CQ. so this leads to PR = PC + CQ being the shortest.
From here, triangles PAC and RBC are similar
- angles PAC and RBC are right angles
- angles PCA and RCB are vertically opposite each other
The length ratio PA : RB is 10 : 15, which simplifies to 2 : 3.
Correspondingly, AC : BC = 2 : 3 as well due to the similarity of both triangles.
So, AC : AB = 2 : 5, and AC = 2/5 x 20 = 8 km.
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lemme know if its right~
Part i is correct , how about part ii? Tks
The shortest distance between two points is basically a line joining two points.
You need an idea like this for part ii.
You need an idea like this for part ii.
Kindly present the workings part ii, tks
Will be letting someone else present this as I don’t have time to write for the moment. If I can remember at night (~ 2 am) or over the next few days, I will write down my workings.
Part ii’s workings uses similar triangles.
Part ii’s workings uses similar triangles.
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