Difference between revisions of "1992 AJHSME Problems/Problem 10"

Latest revision as of 23:09, 4 July 2013

Problem

An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown. The shaded area is

$[asy] for (int a=0; a <= 3; ++a) { for (int b=0; b <= 3-a; ++b) { fill((a,b)--(a,b+1)--(a+1,b)--cycle,grey); } } for (int c=0; c <= 3; ++c) { draw((c,0)--(c,4-c),linewidth(1)); draw((0,c)--(4-c,c),linewidth(1)); draw((c+1,0)--(0,c+1),linewidth(1)); } label("$8$",(2,0),S); label("$8$",(0,2),W); [/asy]$

$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$

Solution

Solution 1

Because the smaller triangles are congruent, the shaded area take $\frac{10}{16}$ of the largest triangles area, which is $\frac{8 \times 8}{2}=32$ , so the shaded area is $\frac{10}{16} \times 32= \boxed{\text{(B)}\ 20}$ .

Solution 2

Each of the triangle has side length of $\frac{1}{4} \times 8=2$ , so the area is $\frac{1}{2} \times 2 \times 2=2$ . Because there are $10$ triangles is the shaded area, its area is $2 \times 10 =\boxed{\text{(B)}\ 20}$ .

1992 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 31: / Line 31: @@
 === Solution 2 ===
 Each of the triangle has side length of <math> \frac{1}{4} \times 8=2 </math>, so the area is <math> \frac{1}{2} \times 2 \times 2=2 </math>. Because there are <math> 10 </math> triangles is the shaded area, its area is <math> 2 \times 10 =\boxed{\text{(B)}\ 20} </math>.
+==See Also==
+{{AJHSME box|year=1992|num-b=9|num-a=11}}
+{{MAA Notice}}

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