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Difference between revisions of "2000 AMC 8 Problems/Problem 6"

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==Solution 4==
 
==Solution 4==
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In the bottpom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of <math>4\cdot4=16</math>. In the top right of that 4 by 4 square is a 3 by 3 square with an area of <math>3\cdot3=9</math>. When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is <math>16-9=\boxed{\text{(A)}\ 7}</math>
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In the bottom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of <math>4\cdot4=16</math>. In the top right of that 4 by 4 square is a 3 by 3 square with an area of <math>3\cdot3=9</math>. When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is <math>16-9=\boxed{\text{(A)}\ 7}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 06:43, 20 December 2024

Problem

Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$-shaped region is

[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy]

$\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution 1

The side of the large square is $1 + 3 + 1 = 5$, so the area of the large square is $5^2 = 25$.

The area of the middle square is $3^2$, and the sum of the areas of the two smaller squares is $2 * 1^2 = 2$.

Thus, the big square minus the three smaller squares is $25 - 9 - 2 = 14$. This is the area of the two congruent L-shaped regions.

So the area of one L-shaped region is $\frac{14}{2} = 7$, and the answer is $\boxed{A}$

Solution 2

The shaded area can be divided into two regions: one rectangle that is 1 by 3, and one rectangle that is 4 by 1. (Or the reverse, depending on which rectangle the 1 by 1 square is "joined" to.) Either way, the total area of these two regions is $3 + 4 = 7$, and the answer is $\boxed{A}$.

Solution 3

Chop the entire 5 by 5 region into $25$ squares like a piece of graph paper. When you draw all the lines, you can count that only $7$ of the small 1 by 1 squares will be shaded, giving $\boxed{A}$ as the answer.

Solution 4

In the bottom left corner of the 5 by 5 square there is a 4 by 4 square which has an area of $4\cdot4=16$. In the top right of that 4 by 4 square is a 3 by 3 square with an area of $3\cdot3=9$. When we remove the 3 by 3 square from the 4 by 4 square we get the L-shaped figure so our answer is $16-9=\boxed{\text{(A)}\ 7}$

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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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