Thread: Triangle Problem

Could someone please help me with this question from a MAOL math competition Matemaattisten Aineiden Opettajien Liitto MAOL ry : Lukion matematiikka ?





We have a triangle. The height of the triangle divides the base of the triangle into two parts, which are 14 cm and 36 cm. Another line parallel to the height is drawn. This line divides the triangle into two equal ( in terms of area) parts. How long are the segments into which this line divides the base of the triangle?



The ratios of the segment of bases and the areas are related so











Then the sample answer says something weird. Is there any reason we would raise the ratio of two segments to the power of two?






Thank you!

Anyone?

So in the figure, you mean AD = 36 cm, and DB = 14cm?

Okay.

Then The base of the triangle, AB = 36 +14 = 50cm.
Call the altitude, CD, as h.
Then, area of triangle ABC = (1/2)(50)(h) = 25h sq.cm.

Hence, area of triangle AEF = (1/2)(25h) = 12.5h sq.cm.

Triangles AEF and ADC are similar, so they are proportional.

Call AE = x
And EF = y

By proportion,
y/x = h/36
Cross multiply,
36y = hx
y = hx /36 ------------**

Area of triangle AEF = (1/2)(x)(y) = 12.5h
(1/2)(x)(hx/36) = 12.5h
(x)(hx/36) = 25h
(x^2)/36 = 25
Take the square roots of both sides,
x/6 = 5
So,
x = 30 cm.

Therefore, the other vertical line divides the base of triangle ABC into 30cm and 20cm segments. ------------answer.

Ok, I understand your method of doing this problem, but I still don't know why the sample answer I posted in my first post, squares the fraction? Why does on have to square AE and AD when doing this problem through another method?




(Sorry, I know this is a stupid question).