Difference between revisions of "1978 AHSME Problems/Problem 29"

Latest revision as of 12:44, 26 September 2024

Problem

Sides $AB,~ BC, ~CD$ and $DA$ , respectively, of convex quadrilateral $ABCD$ are extended past $B,~ C ,~ D$ and $A$ to points $B',~C',~ D'$ and $A'$ . Also, $AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8$ and $DA = AA' = 9$ ; and the area of $ABCD$ is $10$ . The area of $A 'B 'C'D'$ is

$\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad \textbf{(E) }60$

Solution

Notice that the area of $\triangle$ $DAB$ is the same as that of $\triangle$ $A'AB$ (same base, same height). Thus, the area of $\triangle$ $A'AB$ is twice that (same height, twice the base). Similarly, [ $\triangle$ $BB'C$ ] = 2 $\cdot$ [ $\triangle$ $ABC$ ], and so on.

Adding all of these, we see that the area the four triangles around $ABCD$ is twice [ $\triangle$ $DAB$ ] + [ $\triangle$ $ABC$ ] + [ $\triangle$ $BCD$ ] + [ $\triangle$ $CDA$ ], which is itself twice the area of the quadrilateral $ABCD$ . Finally, [ $A'B'C'D'$ ] = [ $ABCD$ ] + 4 $\cdot$ [ $ABCD$ ] = 5 $\cdot$ [ $ABCD$ ] = $\fbox{50}$ .

~ Mathavi

Note: Anyone with a diagram would be of great help (still new to LaTex).

@@ Line 1: / Line 1: @@
 ==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>
+<math>\textbf{(A) }20\qquad
-[asy]
+\textbf{(B) }40\qquad
-unitsize(1 cm);
+\textbf{(C) }45\qquad
+\textbf{(D) }50\qquad
-pair[] A, B, C, D;
+\textbf{(E) }60     </math>
-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]
-<cmath>
-</cmath>
-[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]
-<math></math>
 ==Solution==

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Difference between revisions of "1978 AHSME Problems/Problem 29"

Latest revision as of 12:44, 26 September 2024

Problem

Solution