Hello. Similar method than you : (with your notations) Quite similar to you till 7:20 to find the obvious values : 3*y=6 then y=2 z=3*r (with z=3*r=6-2*QR=6-2*c then c=3-3/2*r) We can express "cos(x)" in the right-angled triangle with hypotenuse = TU : cos(x)=r/b=r/sqrt(r^2+y^2)=r/sqrt(r^2+4) We can express "cos(x)" in the right-angled triangle RSV : cos(x)=RS/RV=b/RV=b/(QV-QR)=b/(6-c)=b/(6-(3-3/2*r))=b/(3+3/2*r)=sqrt(r^2+4)/(3+3/2*r) cos(x)=r/sqrt(r^2+4)=sqrt(r^2+4)/(3+3/2*r) r*(3+3/2*r)=(sqrt(r^2+4))^2 3/2*r^2+3*r=r^2+4 3*r^2+6*r=2*r^2+8 r^2+6*r-8=0 r=sqrt(17)-3 Blue area=6/2*(6-(r+c)) =3*(6-(3-1/2*r)) =3*(3+r/2) =3*(3+sqrt(17)/2-3/2) =3/2*(3+sqrt(17))
Side of square: s = 6 = 2c + 3r --> c = 3 - 3r/2 Similarity of right triangles: r / b = b / (s-c) r = b² / [6 - (3-3r/2)] r = (2² + r²) / (3 + 3r/2) 3r + 3r²/2 = 4 + r² ½r² + 3r - 4 = 0 r² + 6r - 8 = 0 ---> r = 1,1231 mts Similarity of right triangles: a/2 = 2/r ---> a = 4/r = 3,5615 mts Area of blue triangle: A = ½b.h = ½.s.a = ½*6*3,5615 A = 10,685 m² ( Solved √ )
My solution Is (9+3sqrt17)/2 solved in this way: The height of triangle PTU is 2 and it is easy to see if you extend the height above perpendiculary to QV side. You get two similar right triangle whose hypotenuses are b and 2b. Tracing the perpendicular from R to PW side we get two similar triangle having angle in U in common, whose hypotenuse are 3b and b, and the two major legs 6 and 2, so we can write by similarity: 6 : 2 = (6 - 2c) : r r = (6 - 2c)/3 a = 6 - c - r = 6 - c - (6 - 2c)/3 a = (12 - c)/3 = 4 - c/3 now we can find c applying Euclid's theorem on PTU 2² = (6 - 2c)/3 * (12 - c)/3 c² - 15c + 18 = 0 c = (15 - 3√ 17)/2 a = 4 - 1/3*(15 - 3√ 17)/2 = (3 - √ 17)/2 area = 1/2*6* (3 - √ 17)/2 = (9+3sqrt17)/2
Hello. Similar method than you : (with your notations)
Quite similar to you till 7:20 to find the obvious values :
3*y=6 then y=2
z=3*r
(with z=3*r=6-2*QR=6-2*c then c=3-3/2*r)
We can express "cos(x)" in the right-angled triangle with hypotenuse = TU :
cos(x)=r/b=r/sqrt(r^2+y^2)=r/sqrt(r^2+4)
We can express "cos(x)" in the right-angled triangle RSV :
cos(x)=RS/RV=b/RV=b/(QV-QR)=b/(6-c)=b/(6-(3-3/2*r))=b/(3+3/2*r)=sqrt(r^2+4)/(3+3/2*r)
cos(x)=r/sqrt(r^2+4)=sqrt(r^2+4)/(3+3/2*r)
r*(3+3/2*r)=(sqrt(r^2+4))^2
3/2*r^2+3*r=r^2+4
3*r^2+6*r=2*r^2+8
r^2+6*r-8=0
r=sqrt(17)-3
Blue area=6/2*(6-(r+c))
=3*(6-(3-1/2*r))
=3*(3+r/2)
=3*(3+sqrt(17)/2-3/2)
=3/2*(3+sqrt(17))
Similarity of right triangles:
a/2 = 2/r --> a = 4/r
Side of square:
s= 2c + 3r = 6 --> c= ½(6 - 3r)
s= a+r+c= 4/r+r+c > c= 6- 4/r -r
Equalling:
½(6 - 3r) = 6 - 4/r - r
6 - 3r = 12 - 8/r - 2r
r - 8/r + 6 = 0
r² + 6r - 8 = 0 ---> r = 1,1231 mts
Height of blue triangle:
a = 4/r = 3,5615 mts
Area of blue triangle:
A = ½b.h = ½.s.a = ½*6*3,5615
A = 10,685 m² ( Solved √ )
Side of square:
s = 6 = 2c + 3r --> c = 3 - 3r/2
Similarity of right triangles:
r / b = b / (s-c)
r = b² / [6 - (3-3r/2)]
r = (2² + r²) / (3 + 3r/2)
3r + 3r²/2 = 4 + r²
½r² + 3r - 4 = 0
r² + 6r - 8 = 0 ---> r = 1,1231 mts
Similarity of right triangles:
a/2 = 2/r ---> a = 4/r = 3,5615 mts
Area of blue triangle:
A = ½b.h = ½.s.a = ½*6*3,5615
A = 10,685 m² ( Solved √ )
My solution Is (9+3sqrt17)/2
solved in this way:
The height of triangle PTU is 2 and it is easy to see if you extend the height above perpendiculary to QV side. You get two similar right triangle whose hypotenuses are b and 2b.
Tracing the perpendicular from R to PW side we get two similar triangle having angle in U in common, whose hypotenuse are 3b and b, and the two major legs 6 and 2, so we can write by similarity:
6 : 2 = (6 - 2c) : r
r = (6 - 2c)/3
a = 6 - c - r = 6 - c - (6 - 2c)/3
a = (12 - c)/3 = 4 - c/3
now we can find c applying Euclid's theorem on PTU
2² = (6 - 2c)/3 * (12 - c)/3
c² - 15c + 18 = 0
c = (15 - 3√ 17)/2
a = 4 - 1/3*(15 - 3√ 17)/2 = (3 - √ 17)/2
area = 1/2*6* (3 - √ 17)/2 = (9+3sqrt17)/2
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