Revision as of 19:50, 19 November 2022

Solution 1 (List)

Let's label some points.

$[asy] fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,mediumgray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("A",(0,0),S); label("B",(3,0),S); label("C",(7,0),S); label("D",(7.5,3),S); label("E",(7.5,7.8),S); label("F",(5.5,7.8),S); label("G",(-.5,7.8),S); label("H",(-.5,5),S); [/asy]$

By finding the dimensions of the middle rectangle, we need to find the dimensions of the other 4 rectangles. By doing this, there is going to be a rule.

Rule: $AB + BC = CD + DE = EF + FG = GH + AH$

Let's make a list of all the dimensions of the rectangles from the diagram. We have to fill in the dimensions from up above, but still apply to the rule.

$AB\times AH$

$CD\times BC$

$EF\times DE$

$GH\times FG$

By applying the rule, we get $AB=5, BC=7, CD=2, DE=6, EF=1, FG=3, GH=2$ , and $AH=6$ .

By substitution, we get this list

$2\times 7$

$5\times 6$

$2\times 3$

$1\times 6$

Notice how the only dimension not used in the list was $2\times 4$ and that corresponds with B so the answer is, $\textbf{(B) }B.$

~ghfhgvghj10 & Education, the study of everything.

@@ Line 38: / Line 38: @@
 By substitution, we get this list
+<math>2\times 7</math>
 <math>5\times 6</math>
-<math>2\times 7</math>
+<math>2\times 3</math>
 <math>1\times 6</math>
-<math>2\times 3</math>
 Notice how the only dimension not used in the list was <math>2\times 4</math> and that corresponds with B so the answer is, <math>\textbf{(B) }B.</math>
 ~ghfhgvghj10 & Education, the study of everything.

Page

Toolbox

Search

Difference between revisions of "2022 AMC 12A Problems/Problem 3"

Revision as of 19:50, 19 November 2022

Solution 1 (List)