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

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===Solution 1===
 
===Solution 1===
 
We know the semi-perimeter of <math>\triangle ABC</math> is <math>\frac{29+29+42}{2}=50</math>. Next, we use heron's formula to find that the area of the triangle is just <math>\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\textbf{(B) }420.</math>
 
We know the semi-perimeter of <math>\triangle ABC</math> is <math>\frac{29+29+42}{2}=50</math>. Next, we use heron's formula to find that the area of the triangle is just <math>\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\textbf{(B) }420.</math>
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===Solution 2===
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Splitting the isosceles triangle in half, we get a right triangle with hypotenuse 29 and leg 21. By Pythagorean Theorem the height is 20 so the area is
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<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 17:35, 25 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 $\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\textbf{(B) }420.$

Solution 2

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

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
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

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