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Difference between revisions of "2001 AMC 8 Problems/Problem 7"

(Second solution / explanation of first solution)
(→‎Solution 1)
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<math>\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25</math>
 
<math>\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 23 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 25</math>
 
==Solution 1==
 
==Solution 1==
−
The area of a kite is twice the product of its diagonals. The diagonals have lengths of <math> 6 </math> and <math> 7 </math>, so the area is <math> \frac{(6)(7)}{2}=21, \boxed{\text{A}} </math>.
+
The area of a kite is half the product of its diagonals. The diagonals have lengths of <math> 6 </math> and <math> 7 </math>, so the area is <math> \frac{(6)(7)}{2}=21, \boxed{\text{A}} </math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 01:17, 23 September 2011

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]

Problem

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 $\frac{(6)(7)}{2}=21, \boxed{\text{A}}$.

See Also

2001 AMC 8 (Problems • Answer Key • Resources)
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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