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

Latest revision as of 20:02, 28 July 2023

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 bottpom 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 $16-9=\boxed{\text{(A)}\ 7}$

@@ Line 25: / Line 25: @@
 <math>\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15</math>
-==Solution==
+==Solution 1==
-Square FECG¡ square FHIJ = <math>4 \times 4 - 3 \times 3 = 16 - 9 = \boxed{\text{(B) 7}}</math>.
+The side of the large square is <math>1 + 3 + 1 = 5</math>, so the area of the large square is <math>5^2 = 25</math>.
+The area of the middle square is <math>3^2</math>, and the sum of the areas of the two smaller squares is <math>2 * 1^2 = 2</math>.
+Thus, the big square minus the three smaller squares is <math>25 - 9 - 2 = 14</math>.  This is the area of the two congruent L-shaped regions.
+So the area of one L-shaped region is <math>\frac{14}{2} = 7</math>, and the answer is <math>\boxed{A}</math>
+==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 <math>3 + 4 = 7</math>, and the answer is <math>\boxed{A}</math>.
+==Solution 3==
+Chop the entire 5 by 5 region into <math>25</math> squares like a piece of graph paper.  When you draw all the lines, you can count that only <math>7</math> of the small 1 by 1 squares will be shaded, giving <math>\boxed{A}</math> as the answer.
+==Solution 4==
+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>
 ==See Also==
 {{AMC8 box|year=2000|num-b=5|num-a=7}}
+{{MAA Notice}}

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

Page

Toolbox

Search