Difference between revisions of "1985 AHSME Problems/Problem 13"
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==Problem== | ==Problem== | ||
− | Pegs are put in a board <math> 1 </math> unit apart both horizontally and vertically. A rubber band is stretched over <math> 4 </math> pegs as shown in the figure, forming a | + | Pegs are put in a board <math>1</math> unit apart both horizontally and vertically. A rubber band is stretched over <math>4</math> pegs as shown in the figure, forming a quadrilateral. Its area in square units is |
<asy> | <asy> | ||
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<math> \mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 5.5 \qquad \mathrm{(E) \ }6 </math> | <math> \mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 5.5 \qquad \mathrm{(E) \ }6 </math> | ||
− | + | ==Solution 1== | |
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<asy> | <asy> | ||
int i,j; | int i,j; | ||
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label("$H$",(0,1),W); | label("$H$",(0,1),W); | ||
</asy> | </asy> | ||
− | + | We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle <math>ABCD</math> is <math>(3)(4) = 12</math>, while the areas of the triangles <math>AEH</math>, <math>EBF</math>, <math>FCG</math>, and <math>GDH</math> are, respectively, <cmath>\begin{align*}&\text{area of } \triangle AEH = \frac{1}{2}(1)(2) = 1, \ &\text{area of } \triangle EBF = \frac{1}{2}(3)(2) = 3, \ &\text{area of } \triangle FCG = \frac{1}{2}(1)(1) = \frac{1}{2}, \text{ and} \ &\text{area of } \triangle GDH = \frac{1}{2}(3)(1) = \frac{3}{2}.\end{align*}</cmath> | |
+ | Hence the area of the given quadrilateral <math>EFGH</math> is <math>12-\left(1+3+\frac{1}{2}+\frac{3}{2}\right) = \boxed{\text{(E)} \ 6}</math>. | ||
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+ | ==Solution 2== | ||
+ | The number of lattice points (i.e. pegs on the board) strictly inside the quadrilateral is <math>5</math> and the number of lattice points on its boundary is <math>4</math>. Therefore, by [[Pick's theorem]], its area is <math>5+\frac{4}{2}-1 = \boxed{\text{(E)} \ 6}</math>. | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=12|num-a=14}} | {{AHSME box|year=1985|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:05, 19 March 2024
Contents
[hide]Problem
Pegs are put in a board unit apart both horizontally and vertically. A rubber band is stretched over pegs as shown in the figure, forming a quadrilateral. Its area in square units is
Solution 1
We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle is , while the areas of the triangles , , , and are, respectively, Hence the area of the given quadrilateral is .
Solution 2
The number of lattice points (i.e. pegs on the board) strictly inside the quadrilateral is and the number of lattice points on its boundary is . Therefore, by Pick's theorem, its area is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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