A sailing boat has three sails whose dimensions are given in the figure below at
ID: 2997952 • Letter: A
Question
A sailing boat has three sails whose dimensions are given in the figure below at full mast. All coordinates shown are in meter. a- Demonstrate how each sail can be divided into several areas and: b- Determine the x and y coordinates for centroid of sails 1, 2, and 3.
A sailing boat has three sails whose dimensions are given in the figure below at full mast. All coordinates shown are in meter. a- Demonstrate how each sail can be divided into several areas and: b- Determine the x and y coordinates for centroid of sails 1, 2, and 3.Explanation / Answer
1) centroid of a triangle
x = (x1+x2+x3)/3 = (0+20+16)/3 = 12
y = (y1+y2+y3)/3 = (0+0+21)/3 = 7
centroind of figure one is at (12,7)
2)
Any quadrilateral can be divided into two triangles by drawing one of the diagonals. Find the centroids of these two triangles and then connect the line segment between them. Create two new triangles in the quadrilateral by drawing the other diagonal. Find the centroids of these two triangles and then connect the line segment between them. The two line segments of the four centroids intersect at G.
coordinates
A(0,0),B(3,20),C(12.5,23),D(10,0)
centroind of ABD
x1 = (0+3+ 10)/3
y1 = (0+20+0) /3
G1(x1,y1) = ( 13/3,20/3)
centroid of BCD
G2(x1,y1) = (22.5/3 , 43/3)
centroid of ABC
G3(x1,y1) = (15.5/3,43/3)
centroid of ACD
G4(X1,y1) = (22.5/3,23/3)
the point of intersection of line G1G2 and line G3G4 gives the centroind of entire figure
G1(x1,y1) = ( 13/3,20/3) , G2(x1,y1) = (22.5/3 , 43/3)
eqaution of line G1G2 = (x- 13/3) = 2.421(y-20/3) = x-2.421y+ 11.8 =0
G3(x1,y1) = (15.5/3,43/3), G4(X1,y1) = (22.5/3,23/3)
equation of line G3G4 = (x - 22.5/3) = -2.857( y- 23/3)= x+2.875y=29.40
therefore point of intersection is
G (7.304,7.779)
this is the centroid of figure 2
3)same method for this
A(0,0) B(3.5,21) C(14,29) D(23,0)
i am solving this problem directly
G1( 17.5/3,50/3) from traingle ABC
G2(37/3,29/3) FROM traingle ACD
G3(40.5/3 , 50/3) from traingle BCD
G4( 16.5/3,7) from traingle ABD
line G1G2 = y = -1.076x+22.94
line G3G4 = y =1.208x +0.354
therefore point of intersection is
G (9.889,12.3)
therefore centroinf of figure 3 is G (9.889,12.3)
Any quadrilateral can be divided into two triangles by drawing one of the diagonals. Find the centroids of these two triangles and then connect the line segment between them. Create two new triangles in the quadrilateral by drawing the other diagonal. Find the centroids of these two triangles and then connect the line segment between them. The two line segments of the four centroids intersect at G.
coordinates
A(0,0),B(3,20),C(12.5,23),D(10,0)
centroind of ABD
x1 = (0+3+ 10)/3
y1 = (0+20+0) /3
G1(x1,y1) = ( 13/3,20/3)
centroid of BCD
G2(x1,y1) = (22.5/3 , 43/3)
centroid of ABC
G3(x1,y1) = (15.5/3,43/3)
centroid of ACD
G4(X1,y1) = (22.5/3,23/3)
the point of intersection of line G1G2 and line G3G4 gives the centroind of entire figure
G1(x1,y1) = ( 13/3,20/3) , G2(x1,y1) = (22.5/3 , 43/3)
eqaution of line G1G2 = (x- 13/3) = 2.421(y-20/3) = x-2.421y+ 11.8 =0
G3(x1,y1) = (15.5/3,43/3), G4(X1,y1) = (22.5/3,23/3)
equation of line G3G4 = (x - 22.5/3) = -2.857( y- 23/3)= x+2.875y=29.40
therefore point of intersection is
G (7.304,7.779)
this is the centroid of figure 2
3)same method for this
A(0,0) B(3.5,21) C(14,29) D(23,0)
i am solving this problem directly
G1( 17.5/3,50/3) from traingle ABC
G2(37/3,29/3) FROM traingle ACD
G3(40.5/3 , 50/3) from traingle BCD
G4( 16.5/3,7) from traingle ABD
line G1G2 = y = -1.076x+22.94
line G3G4 = y =1.208x +0.354
therefore point of intersection is
G (9.889,12.3)
therefore centroinf of figure 3 is G (9.889,12.3)
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