2001 AMC 8 Problems/Problem 7


To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid. [asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } }  draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]

What is the number of square inches in the area of the small kite?

$\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25$

Solution 1

The area of a kite is half the product of its diagonals. The diagonals have lengths of $6$ and $7$, so the area is $\frac{(6)(7)}{2}=21, \boxed{\text{A}}$.

Solution 2

Drawing in the diagonals of the kite will form four right triangles on the "inside" part of the grid. Drawing in the border of the 7 by 6 grid will form four right triangles on the "outside" part of the grid. Since each right triangle on the inside can be paired with a congruent right triangle that is on the outside, the area of the kite is half the total area of the grid, or $\frac{(6)(7)}{2}=21, \boxed{\text{A}}$.

Solution 3

Pick's Theorem states: \[\frac{\text{number of boundary points}}{2}+\text{number of interior points}-1\] as the area of a figure on a grid. Counting, we see there are $4$ boundary points and $20$ interior points. Therefore, we have \[\frac{4}{2}+20-1\implies 20+1\implies 21.\] Hence, the answer is $\boxed{\text{(A)}}$

See Also

2001 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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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