Difference between revisions of "2019 AMC 8 Problems/Problem 4"
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<math>\overline{AB}</math> = <math>13</math>, and <math>\overline{AE}</math> = <math>12</math>. Using Pythagorean theorem, we find that <math>\overline{BE}</math> = <math>5</math>. | <math>\overline{AB}</math> = <math>13</math>, and <math>\overline{AE}</math> = <math>12</math>. Using Pythagorean theorem, we find that <math>\overline{BE}</math> = <math>5</math>. | ||
− | + | You may recall the famous Pythagorean triple, (5, 12, 13). | |
Thus the values of the two diagonals are <math>\overline{AC}</math> = <math>24</math> and <math>\overline{BD}</math> = <math>10</math>. | Thus the values of the two diagonals are <math>\overline{AC}</math> = <math>24</math> and <math>\overline{BD}</math> = <math>10</math>. | ||
The area of a rhombus is = <math>\frac{d_1\cdot{d_2}}{2}</math> = <math>\frac{24\cdot{10}}{2}</math> = <math>120</math> | The area of a rhombus is = <math>\frac{d_1\cdot{d_2}}{2}</math> = <math>\frac{24\cdot{10}}{2}</math> = <math>120</math> | ||
− | <math>\boxed{\textbf{(D)}\ 120}</math> | + | <math>\boxed{\textbf{(D)}\ 120}</math> |
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==See also== | ==See also== |
Latest revision as of 21:47, 26 December 2020
Problem 4
Quadrilateral is a rhombus with perimeter meters. The length of diagonal is meters. What is the area in square meters of rhombus ?
Solution 1
A rhombus has sides of equal length. Because the perimeter of the rhombus is , each side is . In a rhombus, diagonals are perpendicular and bisect each other, which means = = .
Consider one of the right triangles:
= , and = . Using Pythagorean theorem, we find that = . You may recall the famous Pythagorean triple, (5, 12, 13).
Thus the values of the two diagonals are = and = . The area of a rhombus is = = =
See also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.