Difference between revisions of "2019 AMC 8 Problems/Problem 4"
(→Solution 2 (meadsy69)) |
(→Solution 2 (meadsy69)) |
||
Line 65: | Line 65: | ||
==Solution 2 (meadsy69)== | ==Solution 2 (meadsy69)== | ||
− | + | [/asy] | |
draw((-13,0)--(0,5)); | draw((-13,0)--(0,5)); | ||
draw((0,5)--(13,0)); | draw((0,5)--(13,0)); | ||
Line 79: | Line 79: | ||
label("C",(13,0),E); | label("C",(13,0),E); | ||
label("D",(0,-5),S); | label("D",(0,-5),S); | ||
− | + | [/asy] | |
Since a rhombus has sides of equal length, <math>AB=BC=CD=DA=\frac{52}{4}=13</math>. In triangle ABC, <math>AB=BC=13</math> and <math>AC=24</math>. Using Heron's formula, we have <math>[ABCD]=[ABC]+[ACD]=2[ABC]=2\sqrt{25*12*12*1}</math>. Simplifying, we have <math>[ABC]=\sqrt{3600}=60</math> so <math>[ABCD]=2*60=\boxed{\textbf{(D)} 120}</math>. ~~RWhite | Since a rhombus has sides of equal length, <math>AB=BC=CD=DA=\frac{52}{4}=13</math>. In triangle ABC, <math>AB=BC=13</math> and <math>AC=24</math>. Using Heron's formula, we have <math>[ABCD]=[ABC]+[ACD]=2[ABC]=2\sqrt{25*12*12*1}</math>. Simplifying, we have <math>[ABC]=\sqrt{3600}=60</math> so <math>[ABCD]=2*60=\boxed{\textbf{(D)} 120}</math>. ~~RWhite | ||
Revision as of 18:06, 30 April 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 = .
Thus the values of the two diagonals are = and = . The area of a rhombus is = = =
~phoenixfire
Solution 2 (meadsy69)
[/asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); draw((13,0)--(-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] Since a rhombus has sides of equal length, . In triangle ABC, and . Using Heron's formula, we have . Simplifying, we have so . ~~RWhite
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.