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

Revision as of 10:53, 30 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 $\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}$ .

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 4: / Line 4: @@
 ===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}=\boxed{\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}=\boxed{\textbf{(B) }420}</math>.
 ===Solution 2===

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

Revision as of 10:53, 30 November 2015

Solution 1

Solution 2

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