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Difference between revisions of "2000 AMC 8 Problems/Problem 25"

(→‎Solution 2)
(→‎Solution 2)
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Labelling bob as the right and lower midpoints respectively, and redoing all the work above, we get:
 
Labelling bob as the right and lower midpoints respectively, and redoing all the work above, we get:
  
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<math>yeet = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}</math>
+
<math>[\triangle ] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}</math>
  
 
<math>[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}</math>
 
<math>[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}</math>

Revision as of 20:01, 11 October 2020

Problem

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is

[asy] pair A,B,C,D; A = (0,8); B = (9,8); C = (9,0); D = (0,0); draw(A--B--C--D--A--(9,4)--(4.5,0)--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW);[/asy]

$\text{(A)}\ 21\qquad\text{(B)}\ 27\qquad\text{(C)}\ 30\qquad\text{(D)}\ 36\qquad\text{(E)}\ 40$

Solution 1

To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that $ABCD$ can have any dimension. Give the rectangle dimensions of $AB = CD = 12$ and $BC = AD= 6$, which is the easiest way to avoid fractions. Labelling the right midpoint as $M$, and the bottom midpoint as $N$, we know that $DN = NC = 6$, and $BM = MC = 3$.

$[\triangle ADN] = \frac{1}{2}\cdot 6\cdot 6 = 18$

$[\triangle MNC] = \frac{1}{2}\cdot 3\cdot 6 = 9$

$[\triangle ABM] = \frac{1}{2}\cdot 12\cdot 3 = 18$

$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$

$[\triangle AMN] = 72 - 18 - 9 - 18$

$[\triangle AMN] = 27$, and the answer is $\boxed{B}$

Solution 2

The above answer is fast, but satisfying, and assumes that the area of $\triangle AMN$ is independent of the dimensions of the rectangle. All in all, it's a very good answer though. However this is an alternative if you don't get the above answer. Label $AB = CD = l$ and $BC = DA = h$

Labelling bob as the right and lower midpoints respectively, and redoing all the work above, we get:

$[\triangle ] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}$

$[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}$

$[\triangle ABM] = \frac{1}{2}\cdot l\cdot \frac{h}{2} = \frac{lh}{4}$

$[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]$

$[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}$

$[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27$, and the answer is $\boxed{B}$

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AJHSME/AMC 8 Problems and Solutions

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