Difference between revisions of "2015 AMC 8 Problems/Problem 6"

Latest revision as of 19:37, 26 June 2025

Problem

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

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

Solutions

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 (easier)

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

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/ddif3hlBWTk

~Education, the Study of Everything

Video Solution 1

https://www.youtube.com/watch?v=Bl3_W2i5zwc ~David

Video Solution 2

https://youtu.be/HNj4U3S827E

~savannahsolver

Note

20-21-29 is a Pythagorean Triple (only for right triangles!)

~SaxStreak

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 37: / Line 37: @@
 {{AMC8 box|year=2015|num-b=5|num-a=7}}
 {{MAA Notice}}
+[[Category:Introductory Geometry Problems]]

Page

Toolbox

Search