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

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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 know the height is <math>\sqrt{29^2-21^2}=20</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 know the height is <math>\sqrt{29^2-21^2}=20</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>.
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 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/ddif3hlBWTk
 +
 +
~Education, the Study of Everything
 +
  
 
==Video Solution 1==
 
==Video Solution 1==
  
−
https://www.youtube.com/watch?v=Bl3_W2i5zwc
+
https://www.youtube.com/watch?v=Bl3_W2i5zwc   ~David
  
 
==Video Solution 2==
 
==Video Solution 2==
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~savannahsolver
 
~savannahsolver
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 +
==Note==
 +
20-21-29 is a Pythagorean Triple (only for right triangles!)
 +
 +
~SaxStreak
  
 
==See Also==
 
==See Also==
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{{AMC8 box|year=2015|num-b=5|num-a=7}}
 
{{AMC8 box|year=2015|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

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

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

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

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