Difference between revisions of "2019 AMC 8 Problems/Problem 4"
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− | == Solution == | + | == Problem 4 == |
+ | |||
+ | Quadrilateral <math>ABCD</math> is a rhombus with perimeter <math>52</math> meters. The length of diagonal <math>\overline{AC}</math> is <math>24</math> meters. What is the area in square meters of rhombus <math>ABCD</math>? | ||
+ | |||
+ | <asy> | ||
+ | draw((-13,0)--(0,5)); | ||
+ | draw((0,5)--(13,0)); | ||
+ | draw((13,0)--(0,-5)); | ||
+ | draw((0,-5)--(-13,0)); | ||
+ | dot((-13,0)); | ||
+ | dot((0,5)); | ||
+ | dot((13,0)); | ||
+ | dot((0,-5)); | ||
+ | label("A",(-13,0),W); | ||
+ | label("B",(0,5),N); | ||
+ | label("C",(13,0),E); | ||
+ | label("D",(0,-5),S); | ||
+ | </asy> | ||
+ | |||
+ | <math>\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144</math> | ||
+ | |||
+ | |||
+ | == Solution 1 == | ||
<asy> | <asy> | ||
draw((-12,0)--(0,5)); | draw((-12,0)--(0,5)); | ||
Line 20: | Line 42: | ||
</asy> | </asy> | ||
− | + | A rhombus has sides of equal length. Because the perimeter of the rhombus is <math>52</math>, each side is <math>\frac{52}{4}=13</math>. In a rhombus, diagonals are perpendicular and bisect each other, which means <math>\overline{AE}</math> = <math>12</math> = <math>\overline{EC}</math>. | |
− | |||
− | Consider one of the right triangles | + | Consider one of the right triangles: |
<asy> | <asy> | ||
Line 36: | Line 57: | ||
</asy> | </asy> | ||
− | <math>\overline{AB}</math> = <math>13</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>. |
− | <math>\overline{AE}</math> = <math>12</math>. | + | "You may recall the famous Pythagorean triple, (5, 12, 13), that's how I did it" - Zack2008 |
− | |||
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> | |
+ | |||
+ | <math>\boxed{\textbf{(D)}\ 120}</math> ~phoenixfire | ||
− | + | Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s | |
− | ==See | + | ==See also== |
− | {{AMC8 box|year=2019|num-b= | + | {{AMC8 box|year=2019|num-b=3|num-a=5}} |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:41, 27 November 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), that's how I did it" - Zack2008
Thus the values of the two diagonals are = and = . The area of a rhombus is = = =
~phoenixfire
Video Solution (Also includes problems 1-10)= https://www.youtube.com/watch?v=5i69xiEF-pk&t=2s
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.