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Difference between revisions of "2019 AMC 8 Problems/Problem 2"

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==Solution 1==
 
==Solution 1==
  
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We know that the length of the shorter side of the 3 identical rectangles are all 5 so we can use that by seeing that the longer side of the right rectangle is the same as 2 of the shorter sides of the other 2 left rectangles. This means that <math>2\cdot{5}\ = 10</math> which is the longer side of the right rectangle, and because all the rectangles are congruent, we see that each of the rectangles have a longer side of 10 and a shorter side of 5. Now the bigger rectangle has a shorter length of 10(because the shorter side of the bigger rectangle is the bigger side of the shorter rectangle, which is 10) and so the bigger side of the bigger rectangle is the bigger side of the smaller rectangle + the smaller side of the smaller rectangle, which is <math>10 + 5 = 15</math> . Thus, the area is <math>15\cdot{10}\ = 150</math>  for choice <math>\boxed{\textbf{(E)}\ 150}</math>                 ~~Saksham27
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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==
 
==Solution 2==

Revision as of 15:12, 8 November 2020

Problem 2

Three identical rectangles are put together to form rectangle $ABCD$, 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 $ABCD$?

[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]

$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$


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 $\boxed{\textbf{(E)}\ 150}$. ~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 $5 \cdot 2 = 10$. So the area of the identical rectangles is $5 \cdot 10 = 50$. We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is $50 \cdot 3 = \boxed{\textbf{(E)}\ 150}$. ~~fath2012

Solution 3

We see that if the short sides are 5, the long side has to be $5\cdot2=10$ 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 $ABCD$) is $10+5=15$ because long side + short side of the small rectangle is $15$. The short side of rectangle $ABCD$ is $10$ because it is the long side of the short rectangle. Multiplying $15$ and $10$ together gets us $15\cdot10$ which is $\boxed{\textbf{(E)}\ 150}$. ~~mathboy282 

(also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s
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