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

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<math>[\triangle AMN] = 27</math>, and the answer is <math>\boxed{B}</math>  
 
<math>[\triangle AMN] = 27</math>, and the answer is <math>\boxed{B}</math>  
 +
==Solution 2==
 +
 +
The above answer is fast, but satisfying, and assumes that the area of <math>\triangle AMN</math> 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 <math>AB = CD = l</math> and <math>BC = DA = h</math>
 +
 +
Labelling <math>m</math> and <math>n</math> as the right and lower midpoints respectively, and redoing all the work above, we get:
 +
 +
<math>[\triangle ABN] = \frac{1}{2}\cdot h\cdot \frac{l}{2} = \frac{lh}{4}</math>
  
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==Solution 2==
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<math>[\triangle MNC] = \frac{1}{2}\cdot \frac{l}{2}\cdot \frac{w}{2} = \frac{lh}{8}</math>
−
Let's assume, for simplicity, that the sides of the rectangle are <math>9</math> and <math>8.</math> The area of the 3 triangles would then be <math>8\cdot\frac{9}{2}\cdot\frac{1}{2} = 18,</math>
+
 
 +
<math>[\triangle ABM] = \frac{1}{2}\cdot l\cdot \frac{h}{2} = \frac{lh}{4}</math>
 +
 
 +
<math>[\triangle AMN] = [\square ABCD] - [\triangle ADN] - [\triangle MNC] - [\triangle ABM]</math>
 +
 
 +
<math>[\triangle AMN] = lh - \frac{lh}{4} - \frac{lh}{8} - \frac{lh}{4}</math>
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 +
<math>[\triangle AMN] = \frac{3}{8}lh = \frac{3}{8}\cdot 72 = 27</math>, and the answer is <math>\boxed{B}</math>
 +
==Solution 3==
 +
Let's assume WLOG that the sides of the rectangle are <math>9</math> and <math>8.</math> The area of the 3 triangles would then be <math>8\cdot\frac{9}{2}\cdot\frac{1}{2} = 18,</math>
 
<math>4\cdot\frac{9}{2}\cdot\frac{1}{2} = 9,</math>
 
<math>4\cdot\frac{9}{2}\cdot\frac{1}{2} = 9,</math>
 
<math>4\cdot 9\cdot\frac{1}{2} = 18.</math>
 
<math>4\cdot 9\cdot\frac{1}{2} = 18.</math>
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==See Also==
 
==See Also==
 
M4st3r0fm4th
 
M4st3r0fm4th
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{{AMC8 box|year=2000|num-b=24|after=Last<br />Question}}
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 +
SlimeKnight
 +
 
 +
{{AMC8 box|year=2000|num-b=24|after=Last<br>Question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:26, 27 October 2024

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 $m$ and $n$ as the right and lower midpoints respectively, and redoing all the work above, we get:

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

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

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

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

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

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

Solution 3

Let's assume WLOG that the sides of the rectangle are $9$ and $8.$ The area of the 3 triangles would then be $8\cdot\frac{9}{2}\cdot\frac{1}{2} = 18,$ $4\cdot\frac{9}{2}\cdot\frac{1}{2} = 9,$ $4\cdot 9\cdot\frac{1}{2} = 18.$ Adding these up, we get $45$, and subtracting that from $72$, we get $27$, so the answer is $\boxed{B}$

~ilee0820

Video Solution

https://youtu.be/yoIO9q_GTig. Soo, DRMS, NM

https://www.youtube.com/watch?v=XxQwfirFn4M ~David

See Also

M4st3r0fm4th

SlimeKnight

2000 AMC 8 (Problems • Answer Key • Resources)
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