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Solution of the Week #524 - Four Areas

June 30, 2025 Elliott Line

We’ll start by drawing a line dividing the 19 region into two. We’ll call one ‘a’ and by definition the other will be 19-a.

Now, we don’t know where the line from the apex hits the base, but we can say that the ratio of the left half to the right half is the same whether we are talking about the full height triangle or just the 6 and 19-a regions. In other words:

(19-a)/6 = 27/(x+6)

Let’s manipulate to find an expressions for a:

19-a = 162/(x+6)

a = 19-162/(x+6)

 

If we were to rotate the figure we can do exactly the same but with the 8 and a regions along the base:

a/8 = 25/(x+8)

a = 200/(x+8)

 

Now we have an equation involving both expressions for a:

19-162/(x+6) = 200/(x+8)

19(x+8)-162*(x+8)/(x+6) = 200

19x+152-200 = 162*(x+8)/(x+6)

(19x-48)*(x+6) = 162*(x+8)

19x^2-48x+114x-288 = 162x+1296

19x^2-96x-1584=0

Using the quadratic formula the positive value of the missing area x is 12.

 

 

Solution of the Week #523 - Equal Sequences →

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