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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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| − | <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;
| |
| − | | |
| − | 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 
] = [
.
