Sec 3 Math Problem Help

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Sec 3 Math Problem Help

raristy
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Re: Sec 3 Math Problem Help

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

jxue1015@gmail.com
YellowBelt

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Re: Sec 3 Math Problem Help

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!

raristy
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Re: Sec 3 Math Problem Help

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)]

jxue1015@gmail.com
YellowBelt

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Re: Sec 3 Math Problem Help

Any one can help on this question? Thanks in advance.

sg.angel
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