2015 AMC 8

Complete problem set with solutions and individual problem pages

Problem 6 Easy

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

  • A.

    100

  • B.

    420

  • C.

    500

  • D.

    609

  • E.

    701

Answer:B

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}.

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