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Find the area of the colore square.

 

 Jul 10, 2020
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A bunch of these triangles are similar.

 

Let's take the bottom square. Name the bottom-left point B, the bottom-right point C, the point 5 units above C as D. Then BCD is right and BD = 5sqrt(10). Now, name the points on BD from left to right as E and F. Name the point under E as G. Call BF = x, then FD = 5sqrt(10)-x. Also call CF = y. Then we have two equations using Pythagoras

 

x^2+y^2=15^2, (5sqrt(10)-x)^2 +y^2=25. (We get these from triangles BFC and FDC.) Solving gets x = (9sqrt(10))/2 and y = (3sqrt(10))/2. That means that the area of BFC is xy/2, or in other words, 135/4. There are four of these triangles surrounding the square, so then 135/4 * 4 = 135. Subtracting from the total square area of 15^2 = 225 gets us the final answer of 225 - 135 = 90.

 

You are very welcome!

:P

 Jul 10, 2020

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