Difference between revisions of "2000 AMC 8 Problems/Problem 25"
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<math>[\triangle AMN] = 72 - 18 - 9 - 18</math> | <math>[\triangle AMN] = 72 - 18 - 9 - 18</math> | ||
− | <math>[\triangle AMN] = 27</math>, and the answer is <math>\boxed{ | + | <math>[\triangle AMN] = 27</math>, and the answer is <math>\boxed{B}</math> |
==Solution 2== | ==Solution 2== | ||
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<math>[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}</math> | <math>[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}</math> | ||
− | <math>[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27</math>, and the answer is <math>\boxed{ | + | <math>[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27</math>, and the answer is <math>\boxed{B}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2000|num-b=24|after=Last<br />Question}} | {{AMC8 box|year=2000|num-b=24|after=Last<br />Question}} |
Revision as of 13:18, 11 September 2011
Contents
Problem
The area of rectangle is . If point and the midpoints of and are joined to form a triangle, the area of that triangle is
Solution 1
To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that can have any dimension. Give the rectangle dimensions of and , which is the easiest way to avoid fractions. Labelling the right midpoint as , and the bottom midpoint as , we know that , and .
, and the answer is
Solution 2
The above answer is fast, but not satisfying, and assumes that the area of is independent of the dimensions of the rectangle. Label and
Labelling and as the right and lower midpoints respectively, and redoing all the work above, we get:
, and the answer is
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 |