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2000 AMC 8 Problems/Problem 25

Revision as of 00:07, 6 August 2024 by Agrimupadhyay (talk | contribs) (→‎Solution 3)

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

Let's assume, for simplicity, 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

2000 AMC 8 (Problems • Answer Key • Resources)
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