Difference between revisions of "2016 AMC 10A Problems/Problem 11"
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<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math> | <math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math> | ||
− | == Solution == | + | == Solution 1 == |
First, split the rectangle into <math>4</math> triangles: | First, split the rectangle into <math>4</math> triangles: | ||
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The bases of these triangles are all <math>1</math>, and their heights are <math>4</math>, <math>\frac{5}{2}</math>, <math>4</math>, and <math>\frac{5}{2}</math>. Thus, their areas are <math>2</math>, <math>\frac{5}{4}</math>, <math>2</math>, and <math>\frac{5}{4}</math>, which add to the area of the shaded region, which is <math>\boxed{6\frac{1}{2}}</math>. | The bases of these triangles are all <math>1</math>, and their heights are <math>4</math>, <math>\frac{5}{2}</math>, <math>4</math>, and <math>\frac{5}{2}</math>. Thus, their areas are <math>2</math>, <math>\frac{5}{4}</math>, <math>2</math>, and <math>\frac{5}{4}</math>, which add to the area of the shaded region, which is <math>\boxed{6\frac{1}{2}}</math>. | ||
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== Solution 2 == | == Solution 2 == |
Revision as of 16:39, 4 February 2016
Contents
Problem
What is the area of the shaded region of the given rectangle?
Solution 1
First, split the rectangle into triangles:
The bases of these triangles are all , and their heights are , , , and . Thus, their areas are , , , and , which add to the area of the shaded region, which is .
Solution 2
Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. This will result in - = $\boxed{6$ (Error compiling LaTeX. ! File ended while scanning use of \boxed.)\frac{1}{2}}$
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AMC 10 Problems and Solutions |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.