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2019 AMC 8 Problems/Problem 2

Revision as of 15:20, 16 January 2021 by Etvat (talk | contribs)

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

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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