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Difference between revisions of "2000 CEMC Gauss (Grade 8) Problems/Problem 15"

(Created page with "==Problem== <math>ABCD</math> is a square that is made up of two identical rectangles and two squares of area <math>4 </math> <math>\text{cm}^{2}</math> and <math>16 </math>...")
 
 
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{{Duplicate|[[2000 CEMC Gauss (Grade 8) Problems|2000 CEMC Gauss (Grade 8) #15]] and [[2000 CEMC Gauss (Grade 7) Problems|2000 CEMC Gauss (Grade 7) #17]]}}
 
==Problem==   
 
==Problem==   
 
<math>ABCD</math> is a square that is made up of two identical rectangles and two squares of area <math>4 </math> <math>\text{cm}^{2}</math> and <math>16 </math> <math>\text{cm}^{2}</math>. What is the area, in <math>\text{cm}^{2}</math>, of the square <math>ABCD</math>?
 
<math>ABCD</math> is a square that is made up of two identical rectangles and two squares of area <math>4 </math> <math>\text{cm}^{2}</math> and <math>16 </math> <math>\text{cm}^{2}</math>. What is the area, in <math>\text{cm}^{2}</math>, of the square <math>ABCD</math>?
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~anabel.disher
 
~anabel.disher
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{{CEMC box|year=2000|competition=Gauss (Grade 8)|num-b=14|num-a=16}}
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{{CEMC box|year=2000|competition=Gauss (Grade 7)|num-b=16|num-a=18}}

Latest revision as of 12:57, 20 October 2025

The following problem is from both the 2000 CEMC Gauss (Grade 8) #15 and 2000 CEMC Gauss (Grade 7) #17, so both problems redirect to this page.

Problem

$ABCD$ is a square that is made up of two identical rectangles and two squares of area $4$ $\text{cm}^{2}$ and $16$ $\text{cm}^{2}$. What is the area, in $\text{cm}^{2}$, of the square $ABCD$?

$\text{ (A) }\ 64 \qquad\text{ (B) }\ 49 \qquad\text{ (C) }\ 25 \qquad\text{ (D) }\ 36 \qquad\text{ (E) }\ 20$

Solution

First, we can draw a diagram to make the problem easier to visualize.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

We can find the side lengths of the squares because the side lengths are just the square roots of their areas. We then have:

$\sqrt{4} = 2$ $\text{cm}$

$\sqrt{16} = 4$ $\text{cm}$

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

We can now see that the side length of $ABCD$ is just the sum of the side lengths of the smaller squares. Thus, the side length is $2 + 4 = 6$ $\text{cm}$.

From the side length, $[ABCD] = 6^2 = \boxed {\textbf {(D) } 36}$ $\text{cm}^{2}$.

~anabel.disher

2000 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
CEMC Gauss (Grade 8)
2000 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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
CEMC Gauss (Grade 7)
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