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| | ==Problem== | | ==Problem== |
| − | Sides <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively of convex quadrilateral <math>ABCD</math> are extended past <math>B</math>, <math>C</math>, <math>D</math>, and <math>A</math> to points <math>B'</math>, <math>C'</math>, <math>D'</math>, and <math>A'</math>. If <math>AB = BB' = 6</math>, <math>BC = CC' = 7</math>, <math>CD = DD' = 8</math>, and <math>DA = AA' = 9</math>, and the area of <math>ABCD</math> is 10, determine the area of quadrilateral <math>A'B'C'D'</math>. | + | Sides <math>AB,~ BC, ~CD</math> and <math>DA</math>, respectively, of convex quadrilateral <math>ABCD</math> are extended past <math>B,~ C ,~ D</math> and <math>A</math> to points <math>B',~C',~ D'</math> and <math>A'</math>. Also, <math>AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8</math> and <math>DA = AA' = 9</math>; and the area of <math>ABCD</math> is <math>10</math>. The area of <math>A 'B 'C'D'</math> is |
| − | | |
| − | <math></math>
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| − | [asy]
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| − | unitsize(1 cm);
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| − | | |
| − | pair[] A, B, C, D;
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| − | | |
| − | A[0] = (0,0);
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| − | B[0] = (0.6,1.2);
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| − | C[0] = (-0.3,2.5);
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| − | D[0] = (-1.5,0.7);
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| − | B[1] = interp(A[0],B[0],2);
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| − | C[1] = interp(B[0],C[0],2);
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| − | D[1] = interp(C[0],D[0],2);
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| − | A[1] = interp(D[0],A[0],2);
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| − | | |
| − | draw(A[1]--B[1]--C[1]--D[1]--cycle);
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| − | draw(A[0]--B[1]);
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| − | draw(B[0]--C[1]);
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| − | draw(C[0]--D[1]);
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| − | draw(D[0]--A[1]);
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| − | | |
| − | label("<math>A</math>", A[0], SW);
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| − | label("<math>B</math>", B[0], SE);
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| − | label("<math>C</math>", C[0], NE);
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| − | label("<math>D</math>", D[0], NW);
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| − | label("<math>A'</math>", A[1], SE);
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| − | label("<math>B'</math>", B[1], NE);
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| − | label("<math>C'</math>", C[1], N);
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| − | label("<math>D'</math>", D[1], SW);
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| − | | |
| − | [\asy]
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| − | <cmath>
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| − | | |
| − | </cmath>
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| − | [asy]
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| − | unitsize(1 cm);
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| − | | |
| − | pair[] A, B, C, D;
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| − | | |
| − | A[0] = (0,0);
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| − | B[0] = (0.6,1.2);
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| − | C[0] = (-0.3,2.5);
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| − | D[0] = (-1.5,0.7);
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| − | B[1] = interp(A[0],B[0],2);
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| − | C[1] = interp(B[0],C[0],2);
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| − | D[1] = interp(C[0],D[0],2);
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| − | A[1] = interp(D[0],A[0],2);
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| − | | |
| − | draw(A[1]--B[1]--C[1]--D[1]--cycle);
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| − | draw(A[0]--B[1]);
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| − | draw(B[0]--C[1]);
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| − | draw(C[0]--D[1]);
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| − | draw(D[0]--A[1]);
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| − | | |
| − | label("<math>A</math>", A[0], SW);
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| − | label("<math>B</math>", B[0], SE);
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| − | label("<math>C</math>", C[0], NE);
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| − | label("<math>D</math>", D[0], NW);
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| − | label("<math>A'</math>", A[1], SE);
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| − | label("<math>B'</math>", B[1], NE);
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| − | label("<math>C'</math>", C[1], N);
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| − | label("<math>D'</math>", D[1], SW);
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| − | | |
| − | [\asy]
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| − | <math></math>
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| | | | |
| | ==Solution== | | ==Solution== |
Problem
Sides 
and 
, respectively, of convex quadrilateral 
are extended past 
and 
to points 
and 
. Also, 
and 
; and the area of is 
. The area of 
is
Solution
Notice that the area of 
is the same as that of 
(same base, same height). Thus, the area of is twice that (same height, twice the base). Similarly, [ 
] = 2 
[ 
], and so on.
Adding all of these, we see that the area the four triangles around is twice [ ] + [ ] + [ 
] + [ 
], which is itself twice the area of the quadrilateral . Finally, [
] = [] + 4 [] = 5 [] = 
.
~ Mathavi
Note: Anyone with a diagram would be of great help (still new to LaTex).
and 
, respectively, of convex quadrilateral 
are extended past 
and 
to points 
and 
. Also, 
and 
; and the area of 
. The area of 
is
is the same as that of 
(same base, same height). Thus, the area of 
] = 2 
[
], and so on.
] + [
], which is itself twice the area of the quadrilateral 
] = [
.
