Difference between revisions of "1985 AJHSME Problem 4"
Coolmath34 (talk | contribs) (Created page with "== Problem == The area of polygon <math>ABCDEF</math>, in square units, is <math>\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \tex...") |
Aadhya2012 (talk | contribs) (→Solution) |
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== Solution == | == Solution == | ||
Notice that the polygon is made up of two smaller rectangles with dimensions <math>5 \times 6</math> and <math>4 \times 4.</math> That makes the area <math>30 + 16 = 46,</math> so the answer is <math>\text{(C) 46.}</math> | Notice that the polygon is made up of two smaller rectangles with dimensions <math>5 \times 6</math> and <math>4 \times 4.</math> That makes the area <math>30 + 16 = 46,</math> so the answer is <math>\text{(C) 46.}</math> | ||
| + | |||
| + | == Easier Method == | ||
| + | Notice that the polygon's total area including the missing piece is 54, by multiplying <math>9 x </math>6. Figuring out that the missing sides are <math>4 and </math>2, we conclude that the missing area is 8. <math>54 - </math>8 = <math>46, so the answer is </math>\text{(C) 46.}$ | ||
Revision as of 14:36, 10 January 2025
Problem
The area of polygon 
, in square units, is
![[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]](http://latex.artofproblemsolving.com/a/4/1/a41b89f4f50edbafedb2363591428f6c617d5ebb.png)
Solution
Notice that the polygon is made up of two smaller rectangles with dimensions 
and 
That makes the area 
so the answer is 
Easier Method
Notice that the polygon's total area including the missing piece is 54, by multiplying 
6. Figuring out that the missing sides are 
2, we conclude that the missing area is 8. 
8 = 
\text{(C) 46.}$
