Difference between revisions of "2019 AMC 8 Problems/Problem 2"
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− | + | =Problem 2= | |
+ | Three identical rectangles are put together to form rectangle <math>ABCD</math>, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle <math>ABCD</math>? | ||
− | We | + | <asy> |
+ | draw((0,0)--(3,0)); | ||
+ | draw((0,0)--(0,2)); | ||
+ | draw((0,2)--(3,2)); | ||
+ | draw((3,2)--(3,0)); | ||
+ | dot((0,0)); | ||
+ | dot((0,2)); | ||
+ | dot((3,0)); | ||
+ | dot((3,2)); | ||
+ | draw((2,0)--(2,2)); | ||
+ | draw((0,1)--(2,1)); | ||
+ | label("A",(0,0),S); | ||
+ | label("B",(3,0),S); | ||
+ | label("C",(3,2),N); | ||
+ | label("D",(0,2),N); | ||
+ | </asy> | ||
+ | |||
+ | <math>\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150</math> | ||
+ | |||
+ | |||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | We can see that there are 2 rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is 5 (Because the problem says so) and the bigger side is 10 (Double 5). Now we get the sides of the big rectangles being 15 and 10 so the area is <math>\boxed{\textbf{(E)}\ 150}</math>. ~avamarora | ||
+ | |||
+ | ==Solution 2== | ||
+ | Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is <math>5 \cdot 2 = 10</math>. So the area of the identical rectangles is <math>5 \cdot 10 = 50</math>. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is <math>50 \cdot 3 = \boxed{\textbf{(E)}\ 150}</math>. ~~fath2012 | ||
+ | |||
+ | ==Solution 3== | ||
+ | We see that if the short sides are 5, the long side has to be <math>5\cdot2=10</math> because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle <math>ABCD</math>) | ||
+ | is <math>10+5=15</math> because long side + short side of the small rectangle is <math>15</math>. The short side of rectangle <math>ABCD</math> is <math>10</math> because it is the long side of the short rectangle. Multiplying <math>15</math> and <math>10</math> together gets us <math>15\cdot10</math> which is <math>\boxed{\textbf{(E)}\ 150}</math>. | ||
+ | ~~mathboy282 | ||
+ | |||
+ | (also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC8 box|year=2019|num-b=1|num-a=3}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 20:42, 27 November 2020
Problem 2
Three identical rectangles are put together to form rectangle , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle ?
Solution 1
We can see that there are 2 rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is 5 (Because the problem says so) and the bigger side is 10 (Double 5). Now we get the sides of the big rectangles being 15 and 10 so the area is . ~avamarora
Solution 2
Using the diagram we find that the larger side of the small rectangle is 2 times the length of the smaller side. Therefore the longer side is . So the area of the identical rectangles is . We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is . ~~fath2012
Solution 3
We see that if the short sides are 5, the long side has to be because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle ) is because long side + short side of the small rectangle is . The short side of rectangle is because it is the long side of the short rectangle. Multiplying and together gets us which is . ~~mathboy282
(also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s
See also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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