Sec 3 Math Problem Help
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Sec 3 Math Problem Help
by raristy » Sun May 31, 2015 9:24 pm
Please let me know if you can please help me solve this problem. Thank you!

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Re: Sec 3 Math Problem Help
by jxue1015@gmail.com » Wed Jun 10, 2015 1:24 am
angle CXM = angle AXB (directly opp angle)
angle CMX = angle ABX (alternate angle)
angle MCX = angle BAX (alternate angle)
Therefore triangle CMX and triangle AXB is similar based on AAA property
angle CMX = angle ABX (alternate angle)
angle MCX = angle BAX (alternate angle)
Therefore triangle CMX and triangle AXB is similar based on AAA property

jxue1015@gmail.com  YellowBelt
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Re: Sec 3 Math Problem Help
by raristy » Wed Jun 17, 2015 1:39 am
jxue1015@gmail.com wrote:angle CXM = angle AXB (directly opp angle)
angle CMX = angle ABX (alternate angle)
angle MCX = angle BAX (alternate angle)
Therefore triangle CMX and triangle AXB is similar based on AAA property
Thanks so much for your response. We got that. Any thoughts on part (b) (ii) of this problem? Greatly appreciate your help on this!

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Re: Sec 3 Math Problem Help
by jxue1015@gmail.com » Thu Jun 18, 2015 1:00 am
Given that CM = 3MD, Therefor the ratio of CM : CD is equal to 3 : 4.
and since AB = CD, therefore CM : AB is also equal to 3 : 4
Since Triangle CMX and AXB are similar,
Their 1 dimensional ratio (which is length ratio) is 3:4
Their 2 dimensional ratio (which is area ratio) will be the square of the 1 dimensional ratios which is 9:16 [answer for part b(i)]
For part b(ii),
The 1 dimensional ratio of the two similar triangles, is not only applicable to the base of the triangle (ie CM : AB is equal to 3 : 4), the ratio of the (perpendicular height of CM to X): (Perpendicular height of AB to X) is also 1D ratio 3:4
[ie base ratio and perpendicular height ratio of similar triangles are the same]
Therefore CB = (perpendicular height of CM to X) + (Perpendicular height of AB to X)= 3+4=7 units
area ratio of Triangle AXB : Triangle ACB
is equal to 0.5*AB * 4units (perpendicular ht) : 0.5 * AB * 7 units (perpendicular ht)
Cancel common terms and we get the ratio to be 4:7
so area of triangle BXC is 3units squared, area of rectangle is 2*area of ACB = 14 units squared.
Therefore the ratio of area of BXC : ABCD is equal to 3:14 [answer for part b(ii)]
and since AB = CD, therefore CM : AB is also equal to 3 : 4
Since Triangle CMX and AXB are similar,
Their 1 dimensional ratio (which is length ratio) is 3:4
Their 2 dimensional ratio (which is area ratio) will be the square of the 1 dimensional ratios which is 9:16 [answer for part b(i)]
For part b(ii),
The 1 dimensional ratio of the two similar triangles, is not only applicable to the base of the triangle (ie CM : AB is equal to 3 : 4), the ratio of the (perpendicular height of CM to X): (Perpendicular height of AB to X) is also 1D ratio 3:4
[ie base ratio and perpendicular height ratio of similar triangles are the same]
Therefore CB = (perpendicular height of CM to X) + (Perpendicular height of AB to X)= 3+4=7 units
area ratio of Triangle AXB : Triangle ACB
is equal to 0.5*AB * 4units (perpendicular ht) : 0.5 * AB * 7 units (perpendicular ht)
Cancel common terms and we get the ratio to be 4:7
so area of triangle BXC is 3units squared, area of rectangle is 2*area of ACB = 14 units squared.
Therefore the ratio of area of BXC : ABCD is equal to 3:14 [answer for part b(ii)]

jxue1015@gmail.com  YellowBelt
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Re: Sec 3 Math Problem Help
by sg.angel » Mon Jul 27, 2015 11:08 am
Any one can help on this question? Thanks in advance.

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