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

 
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==Problem==
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In <math>\bigtriangleup ABC</math>, <math>AB=BC=29</math>, and <math>AC=42</math>. What is the area of <math>\bigtriangleup ABC</math>?
 
In <math>\bigtriangleup ABC</math>, <math>AB=BC=29</math>, and <math>AC=42</math>. What is the area of <math>\bigtriangleup ABC</math>?
  
 
<math>\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701</math>
 
<math>\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701</math>
  
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==Solutions==
 
===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>
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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}=\boxed{\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 <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
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<math>\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}</math>.
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==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}}

Latest revision as of 16:33, 16 January 2021

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

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 $\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}$.

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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