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Difference between revisions of "1996 AJHSME Problems/Problem 22"

(Created page with "==Problem== The horizontal and vertical distances between adjacent points equal 1 unit. The area of triangle <math>ABC</math> is <asy> for (int a = 0; a < 5; ++a) { for (i...")
 
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}
 
}
 
draw((0,0)--(3,2)--(4,3)--cycle);
 
draw((0,0)--(3,2)--(4,3)--cycle);
−
label("$A$",(0,0),SW);
+
label("$A$",(0,0),SW);wwqaerhqe
 
label("$B$",(3,2),SE);
 
label("$B$",(3,2),SE);
 
label("$C$",(4,3),NE);
 
label("$C$",(4,3),NE);

Revision as of 22:57, 10 October 2012

Problem

The horizontal and vertical distances between adjacent points equal 1 unit. The area of triangle $ABC$ is 

for (int a = 0; a < 5; ++a)
{
    for (int b = 0; b < 4; ++b)
    {
        dot((a,b));
    }
}
draw((0,0)--(3,2)--(4,3)--cycle);
label("$A$",(0,0),SW);wwqaerhqe
label("$B$",(3,2),SE);
label("$C$",(4,3),NE);

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$\text{(A)}\ 1/4 \qquad \text{(B)}\ 1/2 \qquad \text{(C)}\ 3/4 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 5/4$

Solution 1

[asy] for (int a = 0; a < 5; ++a) { for (int b = 0; b < 4; ++b) { dot((a,b)); } } draw((0,0)--(3,2)--(4,3)--cycle); draw((0,0)--(3,2)--(4,0)--cycle); draw((4,2)--(3,2)--(4,3)--cycle); draw((0,0)--(4,0)--(4,3)--cycle); draw((3,2)--(3,0)--cycle); label("$A$",(0,0),SW); label("$B$",(3,2),SE); label("$C$",(4,3),NE); label("$D$",(4,0),SE); label("$E$",(4,2),SE); label("$F$",(3,0),SE); [/asy]

$\triangle ADC$ takes up half of the 4x3 grid, so it has area of $6$.

$\triangle ABD$ has height of $BF = 2$ and a base of $AD = 4$, for an area of $\frac{1}{2}\cdot 2 \cdot 4 = 4$.

$\triangle CBD$ has height of $BE = 1$ and a base of $CD = 3$, for an area of $\frac{1}{2}\cdot 1 \cdot 3 = \frac{3}{2}$

Note that $\triangle ABC$ can be found by taking $\triangle ADC$, and subtracting off $\triangle ABD$ and $\triangle CBD$.

Thus, the area of $\triangle ABC = 6 - 4 - \frac{3}{2} = \frac{1}{2}$, and the answer is $\boxed{B}$.

There are other equivalent ways of dissecting the figure; right triangles $\triangle ABF, \triangle BCE$ and rectangle $\square BEDF$ can also be used.

Solution 2

Using the Shoelace Theorem, and labelling the points $A(0,0), B(3,2), C(4,3)$, we find the area is:

$(0,0)$

$(3,2)$

$(4,3)$

$(0,0)$

Area = $\frac{1}{2} = |(3\cdot 0 + 4\cdot 2 + 0\cdot 3) - (0\cdot 2 + 3\cdot 3 + 4\cdot 0)| = \frac{1}{2}$, which is option $\boxed{B}$.

See Also

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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