Brain Teasers
Square Flag
A square flag has a blue field with two white diagonal bars that form a cross. Each bar extends across the entire diagonal, from corner to corner. The cross occupies half of the flag's area. The flag measures 60 inches on each side. To the nearest 0.01, what is the width of the bars?
Answer
12.43 inchesa) The square flag measures 60 inches on a side, so its area is 60*60 = 3600.
b) The white cross occupies half of the area, or 3600/2 = 1800.
c) Since the flag is square, the cross divides the flag's blue field into four equal-sized isosceles right triangles. Each of those triangles has an area of 1800/4 = 450.
d) The area (A) of an isosceles right triangle is calculated by the formula 0.5(X^2) = A, where X is the length of one of the two perpendicular legs. 0.5(X^2) = 450, or X^2 = 900, or X = 30. These perpendicular legs form the border between the field and the cross.
e) The hypotenuse (H) of an isosceles right triangle is calculated by the formula H = (2(X^2))^(1/2), where X is the length of one of the two perpendicular legs. H = (2(30^2))^(1/2) = (2(900)^(1/2) = 1800^(1/2) = 42.43.
f) The hypotenuse of each triangle lies along the perimeter in the center of each side of the flag. The remaining length of each side is split equally between the two bars, leaving (60-42.43)/2 , or 8.79, for each bar.
g) Where the bars extend into the corners of the flag, isosceles right triangles with legs of 8.79 are formed. Again, use the formula H = (2(X^2))^(1/2). H = (2(8.79^2))^(1/2) = (2(77.26))^(1/2) = 154.52^(1/2) = 12.43 = H. The hypotenuse of 12.43 is the width of the bars.
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Fun!
You made getting the answer harder than necessary.
Initially, picture two isosceles right triangles. The first has legs of one, a hypothenuse the square root of two, aka (2^(1/2)), and area one-half square, and the second has legs (2^(1/2)), a hypothenuse of two, and area one square. Next, picture an isosceles right triangle representing one-eighth of the flag. (one-half the width as the base, one-half the height as the altitude and one-half a diagonal as the hypothenuse)
Using the laws of similar traingles, the flag section has legs/sides of 30 inches, so its hypothenuse is 30 * (2^(1/2)) inches.
Comparing the second imagined triangle to the first shows that dividing the area by 2 requires dividing the dimensions by (2^(1/2)). The blue field section's hypothenuse = 30 * (2^(1/2)) / (2^(1/2)) = 30 inches.
The trapezoid formed between the two hypothenuses has an altitude equal half the width of the white bar. A perpendicular projection of the minor hypothenuse on the major forms two new 90-45-45 triangles, which show us that the the difference between major and minor equals the full width of a white bar. (30 * (2^(1/2))) - 30 = 30 * ((2^(1/2)) - 1) = 30 * (1.4142... - 1) = 30 * 0.4142... = 12.43 (rounded).
Initially, picture two isosceles right triangles. The first has legs of one, a hypothenuse the square root of two, aka (2^(1/2)), and area one-half square, and the second has legs (2^(1/2)), a hypothenuse of two, and area one square. Next, picture an isosceles right triangle representing one-eighth of the flag. (one-half the width as the base, one-half the height as the altitude and one-half a diagonal as the hypothenuse)
Using the laws of similar traingles, the flag section has legs/sides of 30 inches, so its hypothenuse is 30 * (2^(1/2)) inches.
Comparing the second imagined triangle to the first shows that dividing the area by 2 requires dividing the dimensions by (2^(1/2)). The blue field section's hypothenuse = 30 * (2^(1/2)) / (2^(1/2)) = 30 inches.
The trapezoid formed between the two hypothenuses has an altitude equal half the width of the white bar. A perpendicular projection of the minor hypothenuse on the major forms two new 90-45-45 triangles, which show us that the the difference between major and minor equals the full width of a white bar. (30 * (2^(1/2))) - 30 = 30 * ((2^(1/2)) - 1) = 30 * (1.4142... - 1) = 30 * 0.4142... = 12.43 (rounded).
liitle hard
but challenging
but challenging
Wow, lots more math than you need. Stil also, but less so.
I was going to use the similar triangles, but I figured that was more work than needed. I took
sqrt(((60-sqrt(60^2/2))/2)^2*2) = 12.426...
Working from the inside out, 60^2/2 is half the size of the square, so the square root of this is the length of the side of the square formed by the four triangles. Subtract this from the length of the side and divide by 2 to get one side of the isosceles right triangle in the corner of the flag whose hypothenuse is the width of the strip. Square this value, multiply by two and take the square root to get the hypothenuse. Done.
I was going to use the similar triangles, but I figured that was more work than needed. I took
sqrt(((60-sqrt(60^2/2))/2)^2*2) = 12.426...
Working from the inside out, 60^2/2 is half the size of the square, so the square root of this is the length of the side of the square formed by the four triangles. Subtract this from the length of the side and divide by 2 to get one side of the isosceles right triangle in the corner of the flag whose hypothenuse is the width of the strip. Square this value, multiply by two and take the square root to get the hypothenuse. Done.
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