Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 8 Problems/Problem 6"
(→‎Solution 2)
Line 7: Line 7:
  
 
===Solution 2===
 
===Solution 2===
−
Splitting the isosceles triangle in half, we get a right triangle with hypotenuse 29 and leg 21. Using the Pythagorean Theorem , we get <math>841-441</math>. Now that we know the height, the area is
+
Splitting the isosceles triangle in half, we get a right triangle with hypotenuse <math>29</math> and leg <math>21</math>. Using the Pythagorean Theorem , we get <math>841-441</math>. Now that we know the height, the area is
 
<math>\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}</math>.
 
<math>\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}</math>.
 +
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=5|num-a=7}}
 
{{AMC8 box|year=2015|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:49, 26 November 2015

In $\bigtriangleup ABC$, $AB=BC=29$, and $AC=42$. What is the area of $\bigtriangleup ABC$?

$\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701$

Solution 1

We know the semi-perimeter of $\triangle ABC$ is $\frac{29+29+42}{2}=50$. Next, we use heron's formula to find that the area of the triangle is just $\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\boxed{\textbf{(B) }420}$.

Solution 2

Splitting the isosceles triangle in half, we get a right triangle with hypotenuse $29$ and leg $21$. Using the Pythagorean Theorem , we get $841-441$. Now that we know the height, the area is $\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}$.

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_6&oldid=73209"