Difference between revisions of "2022 AMC 12A Problems/Problem 3"
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==Solution 3 (Area, Perimeter of Square)== | ==Solution 3 (Area, Perimeter of Square)== | ||
− | The area of this square is equal to <math>6 + 8 + 30 + 14 + 6 = 64</math>, and thus its side lengths are <math>8</math>. The sum of the dimensions of the rectangles are <math>2 + 7 + 5 + 6 + 2 + 3 + 1 + 6 + 2 + 4 = 38</math>. Thus, because the perimeter of the rectangle is <math>32</math>, the rectangle on the inside must have a perimeter of <math>6</math>. The only rectangle that works is <math>\boxed{\textbf{(B) }B}</math>. | + | The area of this square is equal to <math>6 + 8 + 30 + 14 + 6 = 64</math>, and thus its side lengths are <math>8</math>. The sum of the dimensions of the rectangles are <math>2 + 7 + 5 + 6 + 2 + 3 + 1 + 6 + 2 + 4 = 38</math>. Thus, because the perimeter of the rectangle is <math>32</math>, the rectangle on the inside must have a perimeter of <math>6 \cdot 2 = 12</math>. The only rectangle that works is <math>\boxed{\textbf{(B) }B}</math>. |
~mathboy100 | ~mathboy100 |
Revision as of 23:55, 30 November 2022
Contents
Problem 3
Five rectangles, ,
,
,
, and
, are arranged in a square as shown below. These rectangles have dimensions
,
,
,
, and
, respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle?
Solution 1 (List)
Let's label some points.
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:
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.
By applying the rule, we get , and
.
By substitution, we get this list
This also tells us that the diagram is not drawn to scale, lol.
Notice how the only dimension not used in the list was and that corresponds with B so the answer is,
~ghfhgvghj10 & Education, the study of everything.
Solution 2 (Perimeter of Square)
Note that the perimeter of the square is the sum of 4 pairs of dimensions (8 values are added). Adding up all of the dimensions () gives us 38. We know that as the square's side length is an integer, the perimeter must be divisible by 4. Testing out by subtracting all 5 pairs of dimensions from 38, only
works (
), which corresponds with
.
~iluvme
Solution 3 (Area, Perimeter of Square)
The area of this square is equal to , and thus its side lengths are
. The sum of the dimensions of the rectangles are
. Thus, because the perimeter of the rectangle is
, the rectangle on the inside must have a perimeter of
. The only rectangle that works is
.
~mathboy100