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

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=Problem 2=
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==Problem==
 
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>?
 
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>?
  
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<math>\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150</math>
 
<math>\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150</math>
  
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==Solutions==
  
 
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===Solution 1===
==Solution 1==
 
  
 
We can see that there are <math>2</math> 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 <math>5</math>, if we take the information of the problem, and the bigger side is <math>10</math>, if we do <math>5 \cdot 2 = 10</math>. Now we get the sides of the big rectangles being <math>15</math> and <math>10</math>, so the area is <math>\boxed{\textbf{(E)}\ 150}</math>. ~avamarora
 
We can see that there are <math>2</math> 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 <math>5</math>, if we take the information of the problem, and the bigger side is <math>10</math>, if we do <math>5 \cdot 2 = 10</math>. Now we get the sides of the big rectangles being <math>15</math> and <math>10</math>, so the area is <math>\boxed{\textbf{(E)}\ 150}</math>. ~avamarora
  
==Solution 2==
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===Solution 2===
 
Using the diagram we find that the larger side of the small rectangle is <math>2</math> 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
 
Using the diagram we find that the larger side of the small rectangle is <math>2</math> 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==
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===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>)
 
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>.
 
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
 
~~mathboy282
  
(also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s
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== Video Solution ==
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Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eEqtoI8BQKE&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=3
  
 
==See also==
 
==See also==

Revision as of 21:00, 12 May 2021

Problem

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]

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

Solutions

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 $5$, if we take the information of the problem, and the bigger side is $10$, if we do $5 \cdot 2 = 10$. 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

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eEqtoI8BQKE&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=3

See also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AJHSME/AMC 8 Problems and Solutions

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