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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==
  
 +
===Solution 1===
  
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==Solution 1==
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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>, so the bigger side is <math>10</math>, if we do <math>5 \cdot 2 = 10</math>. Now we get the sides of the big rectangle being <math>15</math> and <math>10</math>, so the area is <math>\boxed{\textbf{(E)}\ 150}</math>. ~avamarora
  
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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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===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
  
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==Solution 2==
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===Solution 3===
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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
 
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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
  
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==Also See==
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===Solution 4===
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We see that the <math>2</math> rectangles lying on top of each other give us the height of the rectangle. Using what we know, we can find out that the <math>2</math> rectangles put together is a square. So, we can infer that the length of the rectangles is <math>10</math>. Adding that to the width of the third rectangle which is <math>5</math>, we get that the length of the rectangle is <math>15</math>. Multiplying <math>10</math> and <math>15</math> gives us <math>15\cdot10</math> which is <math>\boxed{\textbf{(E)}\ 150}</math>.
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~~awesomepony566
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===Solution 5===
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There are two rectangles lying on the side and one standing up. Given that one small side is 5, we can determine that two of the small sides make up a big side which means that the long side is equal to 10. The top side of the rectangle is made up of one small side and one long side, therefore the dimensions for the rectangle is 10x15. 10 multiplied by 15 is 150, hence the answer <math>\boxed{\textbf{(E)}\ 150}</math>.
 +
~elenafan
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 +
=== Video Solution ===
 +
 
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/IgpayYB48C4?si=N-pH5zJQY7O5lVWk&t=162
 +
 
 +
~Math-X
 +
 
 +
The Learning Royal: https://youtu.be/IiFFDDITE6Q
 +
 
 +
=== Video Solution 2 ===
 +
 
 +
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=eEqtoI8BQKE&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=3
 +
 
 +
==Video Solution 3==
 +
https://youtu.be/IAfEqEGRcF0
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/51K3uCzntWs?t=1209
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/wpZXXu3Vedg
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 
 +
~Hayabusa1
 +
 
 +
==See also==
 
{{AMC8 box|year=2019|num-b=1|num-a=3}}
 
{{AMC8 box|year=2019|num-b=1|num-a=3}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:36, 24 April 2024

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 $5$, so the bigger side is $10$, if we do $5 \cdot 2 = 10$. Now we get the sides of the big rectangle 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

Solution 4

We see that the $2$ rectangles lying on top of each other give us the height of the rectangle. Using what we know, we can find out that the $2$ rectangles put together is a square. So, we can infer that the length of the rectangles is $10$. Adding that to the width of the third rectangle which is $5$, we get that the length of the rectangle is $15$. Multiplying $10$ and $15$ gives us $15\cdot10$ which is $\boxed{\textbf{(E)}\ 150}$. ~~awesomepony566

Solution 5

There are two rectangles lying on the side and one standing up. Given that one small side is 5, we can determine that two of the small sides make up a big side which means that the long side is equal to 10. The top side of the rectangle is made up of one small side and one long side, therefore the dimensions for the rectangle is 10x15. 10 multiplied by 15 is 150, hence the answer $\boxed{\textbf{(E)}\ 150}$. ~elenafan

Video Solution

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/IgpayYB48C4?si=N-pH5zJQY7O5lVWk&t=162

~Math-X

The Learning Royal: https://youtu.be/IiFFDDITE6Q

Video Solution 2

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

Video Solution 3

https://youtu.be/IAfEqEGRcF0

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/51K3uCzntWs?t=1209

~ pi_is_3.14

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/wpZXXu3Vedg

~Education, the Study of Everything

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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