Difference between revisions of "1985 AHSME Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | + | Diagonal <math>DB</math> of rectangle <math>ABCD</math> is divided into three segments of length <math>1</math> by parallel lines <math>L</math> and <math>L'</math> that pass through <math>A</math> and <math>C</math> and are perpendicular to <math>DB</math>. The area of <math>ABCD</math>, rounded to the one decimal place, is | |
<asy> | <asy> | ||
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==Solution== | ==Solution== | ||
− | Let <math> E </math> be the intersection of | + | Let <math>E</math> be the point of intersection of <math>L</math> and <math>\overline{BD}</math>. Then, because <math>AE</math> is the altitude to the hypotenuse of right triangle <math>ABD</math>, triangles <math>ADE</math> and <math>BAE</math> are similar, giving <cmath>\frac{AE}{BE} = \frac{ED}{EA},</cmath> and so <cmath>\begin{align*}AE &= \sqrt{BE \cdot ED} \ &= \sqrt{(1+1)(1)} \\ &= \sqrt{2}.\end{align*}</cmath> Thus, taking <math>BD</math> and <math>AE</math> as the base and perpendicular height, respectively, of triangle <math>ABD</math>, we may compute its area as <math>\frac{1}{2}(3)\left(\sqrt{2}\right) = \frac{3\sqrt{2}}{2}</math>. By symmetry, the area of the entire rectangle <math>ABCD</math> is <cmath>2\left(\frac{3\sqrt{2}}{2}\right) = 3\sqrt{2} \approx (3)(1.4) = \boxed{\text{(B)} \ 4.2}.</cmath> |
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=16|num-a=18}} | {{AHSME box|year=1985|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:39, 19 March 2024
Problem
Diagonal of rectangle is divided into three segments of length by parallel lines and that pass through and and are perpendicular to . The area of , rounded to the one decimal place, is
Solution
Let be the point of intersection of and . Then, because is the altitude to the hypotenuse of right triangle , triangles and are similar, giving and so Thus, taking and as the base and perpendicular height, respectively, of triangle , we may compute its area as . By symmetry, the area of the entire rectangle is
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AHSME Problems and Solutions |
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