Difference between revisions of "2019 AMC 8 Problems/Problem 4"
Phoenixfire (talk | contribs) (Created page with "== Solution == <asy> draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); draw((0,0)--(13,0)); draw((0,0)--(0,5)); draw((0,0)--(-13,0)); dra...") |
|||
(32 intermediate revisions by 22 users not shown) | |||
Line 1: | Line 1: | ||
− | == | + | == 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> | <asy> | ||
draw((-13,0)--(0,5)); | draw((-13,0)--(0,5)); | ||
Line 5: | Line 8: | ||
draw((13,0)--(0,-5)); | draw((13,0)--(0,-5)); | ||
draw((0,-5)--(-13,0)); | draw((0,-5)--(-13,0)); | ||
− | |||
− | |||
− | |||
− | |||
dot((-13,0)); | dot((-13,0)); | ||
dot((0,5)); | dot((0,5)); | ||
Line 16: | Line 15: | ||
label("B",(0,5),N); | label("B",(0,5),N); | ||
label("C",(13,0),E); | 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> | ||
+ | draw((-12,0)--(0,5)); | ||
+ | draw((0,5)--(12,0)); | ||
+ | draw((12,0)--(0,-5)); | ||
+ | draw((0,-5)--(-12,0)); | ||
+ | draw((0,0)--(12,0)); | ||
+ | draw((0,0)--(0,5)); | ||
+ | draw((0,0)--(-12,0)); | ||
+ | draw((0,0)--(0,-5)); | ||
+ | dot((-12,0)); | ||
+ | dot((0,5)); | ||
+ | dot((12,0)); | ||
+ | dot((0,-5)); | ||
+ | label("A",(-12,0),W); | ||
+ | label("B",(0,5),N); | ||
+ | label("C",(12,0),E); | ||
label("D",(0,-5),S); | label("D",(0,-5),S); | ||
label("E",(0,0),SW); | label("E",(0,0),SW); | ||
</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> | ||
− | draw((- | + | draw((-12,0)--(0,5)); |
− | draw((0,0)--(- | + | draw((0,0)--(-12,0)); |
draw((0,0)--(0,5)); | draw((0,0)--(0,5)); | ||
− | dot((- | + | dot((-12,0)); |
dot((0,5)); | dot((0,5)); | ||
− | label("A",(- | + | label("A",(-12,0),W); |
label("B",(0,5),N); | label("B",(0,5),N); | ||
label("E",(0,0),SE); | label("E",(0,0),SE); | ||
</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 the Pythagorean theorem, we find that <math>\overline{BE}</math> = <math>5</math>. |
− | <math>\overline{AE}</math> = <math>12</math>. | + | You know the 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> | |
+ | |||
+ | <math>\boxed{\textbf{(D)}\ 120}</math> | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | The Learning Royal: https://youtu.be/IiFFDDITE6Q | ||
+ | |||
+ | == Video Solution 2 == | ||
+ | |||
+ | Solution detailing how to solve the problem:https://www.youtube.com/watch?v=-yHfOUapg7I&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=5 | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://youtu.be/mL6gIb5y3B0 | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC8 box|year=2019|num-b=3|num-a=5}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 12:26, 7 February 2022
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 the Pythagorean theorem, we find that = . You know the Pythagorean triple, (5, 12, 13).
Thus the values of the two diagonals are = and = . The area of a rhombus is = = =
Video Solution
The Learning Royal: https://youtu.be/IiFFDDITE6Q
Video Solution 2
Solution detailing how to solve the problem:https://www.youtube.com/watch?v=-yHfOUapg7I&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=5
Video Solution 3
~savannahsolver
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.