Difference between revisions of "1978 AHSME Problems/Problem 29"
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==Problem== | ==Problem== | ||
− | Sides <math>AB, | + | 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>. Also, <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. Find the area of <math>A'B'C'D'</math>. |
+ | |||
+ | [asy] | ||
+ | unitsize(1 cm); | ||
+ | |||
+ | pair[] A, B, C, D; | ||
+ | |||
+ | A[0] = (0,0); | ||
+ | B[0] = (0.6,1.2); | ||
+ | C[0] = (-0.3,2.5); | ||
+ | D[0] = (-1.5,0.7); | ||
+ | B[1] = interp(A[0],B[0],2); | ||
+ | C[1] = interp(B[0],C[0],2); | ||
+ | D[1] = interp(C[0],D[0],2); | ||
+ | A[1] = interp(D[0],A[0],2); | ||
+ | |||
+ | draw(A[1]--B[1]--C[1]--D[1]--cycle); | ||
+ | draw(A[0]--B[1]); | ||
+ | draw(B[0]--C[1]); | ||
+ | draw(C[0]--D[1]); | ||
+ | draw(D[0]--A[1]); | ||
+ | |||
+ | label("<math>A</math>", A[0], SW); | ||
+ | label("<math>B</math>", B[0], SE); | ||
+ | label("<math>C</math>", C[0], NE); | ||
+ | label("<math>D</math>", D[0], NW); | ||
+ | label("<math>A'</math>", A[1], SE); | ||
+ | label("<math>B'</math>", B[1], NE); | ||
+ | label("<math>C'</math>", C[1], N); | ||
+ | label("<math>D'</math>", D[1], SW); | ||
+ | [/asy] | ||
− | |||
==Solution== | ==Solution== | ||
Revision as of 14:15, 23 August 2022
Problem
Sides ,
,
, and
, respectively of convex quadrilateral
are extended past
,
,
, and
to points
,
,
, and
. Also,
,
,
, and
, and the area of
is 10. Find the area of
.
[asy] unitsize(1 cm);
pair[] A, B, C, D;
A[0] = (0,0); B[0] = (0.6,1.2); C[0] = (-0.3,2.5); D[0] = (-1.5,0.7); B[1] = interp(A[0],B[0],2); C[1] = interp(B[0],C[0],2); D[1] = interp(C[0],D[0],2); A[1] = interp(D[0],A[0],2);
draw(A[1]--B[1]--C[1]--D[1]--cycle); draw(A[0]--B[1]); draw(B[0]--C[1]); draw(C[0]--D[1]); draw(D[0]--A[1]);
label("", A[0], SW);
label("
", B[0], SE);
label("
", C[0], NE);
label("
", D[0], NW);
label("
", A[1], SE);
label("
", B[1], NE);
label("
", C[1], N);
label("
", D[1], SW);
[/asy]
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).