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secondary 3 | A Maths
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need help with this qn, pls explain too
Date Posted: 3 years ago
Views: 350
For part d there does not seem to be any convincing answer to me at all. I considered all other triangles there and none of them fits my criteria for congruency. There are close ones, but none of them seem congruent at all.
If I draw my own one, I would draw a line from G perpendicular to BC to put AD at a point H. Then triangle HGD would be congruent.
If I draw my own one, I would draw a line from G perpendicular to BC to put AD at a point H. Then triangle HGD would be congruent.
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I have thought long and hard for the fourth part. I do not see any triangle which is congruent to ADE and CDG. The closest is AGF, but I dismissed it.
I update you if I can find the triangle.
I update you if I can find the triangle.
Date Posted:
3 years ago
For the fourth part, I need the triangle to have one length equal to the large square DEFG, one length equal to the small square ABCD and one angle which is a right angle.
Triangle DAG has DA and DG as the lengths, but there are no 90 degrees in there.
Triangle AGF has the 90 degree and the length FG, but the problem is this. In triangle CDG, length CD is longer than length CG, so the longer of the two non-hypotenuse sides is CD. Yet for triangle AGF, the side AG is already longer than CD, so AGF cannot be congruent to triangle
Triangle DAG has DA and DG as the lengths, but there are no 90 degrees in there.
Triangle AGF has the 90 degree and the length FG, but the problem is this. In triangle CDG, length CD is longer than length CG, so the longer of the two non-hypotenuse sides is CD. Yet for triangle AGF, the side AG is already longer than CD, so AGF cannot be congruent to triangle
vased on the ans key its triangle FEB, but FEB does not look like a triangle thi as the line isnt connected to B
can we form our own triangles ourselves by joining random lines or...
can we form our own triangles ourselves by joining random lines or...
i dont really understand part c tho
FEB is definitely out. Length EB is way too Long already.
If it had been asking for similar triangles, many triangles would qualify.
For part c, DEFG is a rectangle. To prove that it is a square, we must prove that the corresponding non-opposite sides are equal.
This can be obtained from the congruency statement which we have proved. Because we have proved that the two triangles ADE and CDG are congruent, we can conclude that
- AD = CD
- AE = CG
- DE = DG
- angle ADE = angle CDG
- angle DAE = angle DCG
- angle AED = angle CGD
We extract the DE = DG statement to say that in rectangle DEFG, the adjacent lengths are also equal in length. This completes our proof that DEFG is also a square. Because if this happens, the fourth length will automatically be the same as well.
If it had been asking for similar triangles, many triangles would qualify.
For part c, DEFG is a rectangle. To prove that it is a square, we must prove that the corresponding non-opposite sides are equal.
This can be obtained from the congruency statement which we have proved. Because we have proved that the two triangles ADE and CDG are congruent, we can conclude that
- AD = CD
- AE = CG
- DE = DG
- angle ADE = angle CDG
- angle DAE = angle DCG
- angle AED = angle CGD
We extract the DE = DG statement to say that in rectangle DEFG, the adjacent lengths are also equal in length. This completes our proof that DEFG is also a square. Because if this happens, the fourth length will automatically be the same as well.
thx for helping me with this qn :)
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