Listing Details
| ID: | 1812 | |||||
| Title: | Homeschool Math Blog | |||||
| URL: | http://homeschoolmath.blogspot.com/ | |||||
| Category: | Science: Maths | |||||
| Description: | Blog of a teacher and homeschooling mother who loves teaching maths. | |||||
| Free ebook "All Children Can Be Great Listeners" - 2012-01-31 23:15:00 | ||||||
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| Triangle problem with equal areas - solution - 2012-01-30 08:50:00 | ||||||
This is the solution for thetriangle problemwith equal areas that I posted earlier. We are given that the areas of the three right triangles are equal, that is the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD. We will make an equation based on that fact. For that, I like to use x as my variable, so I denote the longer side of the rectangle with a, the other side with b, the distance AE with x, and the distance BF with y. We are asked the ratio AE:EB, which is the same as x : (a − x) using my notation, and the ratio BF:FC, which is the same as y : (b − y) using my notation.  The area of triangle ADE is its base times altitude divided by 2, or bx/2. The area of triangle EBF is its base times altitude divided by 2, or y(a − x)/2. The area of triangle CDF is its base times altitude divided by 2, or a(b − y)/2. And these three are equal. Basically you just make two equations from the above information, and manipulate your equations until you get the ratios expressedwithoutx and y. Since the area of triangle ADE = area of triangle EBF, we get:
bx = y(a − x) bx = ay − yx Since the area of triangle EBF = area of triangle CDF, we get:
y(a − x) = a(b − y) ay − yx = ab − ay Now I have two equations with two unknowns (x and y) to solve, or a system of equations: bx = ay − yx ay − yx = ab − ay We'll solve them, and then write the ratios x : (a − x) and y : (b − y). ======================================================== Well, to save time... I did solve these on paper the other day but lost it, and writing it in html is not my favorite hobby... so I just putthese equations into Wolframalpha, separated by a semicolon, and got x = (3a - √5a)/2, y = (√5 − 1)b / 2 OR x = (3a + √5a)/2, y = (√5 − 1)b / 2 Then the two ratios. We need to substitute the value of x from above into the ratio x : (a − x). I'll use the first solution first. x : (a − x) = x / (a − x) = ((3a - √5 a)/2 ) / a − (3a - √5 a)/2) Hey, maybe Wolframalpha willsimplify this expressionfor me, too. And yes, it did. The simplified form is 1/2 (√5 - 1). What happens if I use the second solution for x? In the ratio x / (a − x), we get (3a + √5a)/2 /(a − ((3a + √5a)/2)) The a's will cancel out from this expression, so we get (3a + √5a)/2 /(a − ((3a + √5a)/2)) Simplifying with Wolframalphaagain, I get (−1 − √5)/2 which is a negative number. So I discard that solution. Then on to the second ratio, y : (b − y), using the (only) solution y = (√5 − 1)b / 2 first. I substitute that and get((√5 − 1)b / 2) / (b − ((√5 − 1)b / 2)). Again, b's cancel out so we have ((√5 − 1)/2) / (1 − ((√5 − 1)/ 2)) Simplifying that, we get (1 + √5)/2, which is the GOLDEN RATIO (or Phi). So, the two ratios are (√5 + 1)/2 and (√5 − 1)/2, or the golden ratio and its conjugate. Notice these don't depend onaorb(the sides of the rectangle) — meaning that these ratios are the same in any rectangle.  | ||||||
| - 2012-01-28 17:13:00 | ||||||
Bon fromMath is not a four-letter wordmade this little counting song, sung to the tune of "Twinkle Twinkle Little Star". Hope your little ones enjoy it! | ||||||