Difference between revisions of "1993 AJHSME Problems/Problem 18"
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draw(B--D--F); | draw(B--D--F); | ||
dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); | dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); | ||
− | label("$A$",A, | + | label("$A$",A,NW); |
− | label("$ | + | label("$B$",B,N); |
label("$C$",C,NE); | label("$C$",C,NE); | ||
label("$D$",D,SE); | label("$D$",D,SE); |
Latest revision as of 00:05, 11 November 2019
Problem
The rectangle shown has length , width , and and are midpoints of and , respectively. The area of quadrilateral is
Solution
The area of the quadrilateral is equal to the areas of the two right triangles and subtracted from the area of the rectangle . Because and are midpoints, we know the dimensions of the two right triangles.
See Also
1993 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.