2015 AMC 8

Complete problem set with solutions and individual problem pages

Problem 19 Hard

A triangle with vertices as A=(1,3), B=(5,1), and C=(4,4) is plotted on a 6\times5 grid. What fraction of the grid is covered by the triangle?

  • A.

    \frac 16

  • B.

    \frac 15

  • C.

    \frac 14

  • D.

    \frac13

  • E.

    \frac12

Answer:A

Solution 1

The area of \triangle ABC is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is \sqrt{1^2+2^2}=\sqrt{5}, and its base is \sqrt{2^2+4^2}=\sqrt{20}. We multiply these and divide by 2 to find the area of the triangle is \frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5. Since the grid has an area of 30, the fraction of the grid covered by the triangle is \frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}.

 

Solution 2

Note angle \angle ACB is right; thus, the area is \sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=10 \times \dfrac{1}{2}=5; thus, the fraction of the total is \dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}.

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