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Difference between revisions of "1969 AHSME Problems/Problem 22"

(Solution to Problem 22)
m (→‎Solution)
 
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<asy>
 
<asy>
  
−
import graph; size(10.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black;
+
import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black;
 
real xmin=-5.2,xmax=9.2,ymin=-5.2,ymax=13.2;  
 
real xmin=-5.2,xmax=9.2,ymin=-5.2,ymax=13.2;  
 
pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);  
 
pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);  

Latest revision as of 03:12, 10 June 2018

Problem

Let $K$ be the measure of the area bounded by the $x$-axis, the line $x=8$, and the curve defined by

\[f={(x,y)\quad |\quad y=x \text{ when } 0 \le x \le 5, y=2x-5 \text{ when } 5 \le x \le 8}.\]

Then $K$ is:

$\text{(A) } 21.5\quad \text{(B) } 36.4\quad \text{(C) } 36.5\quad \text{(D) } 44\quad \text{(E) less than 44 but arbitrarily close to it}$

Solution

[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=9.2,ymin=-5.2,ymax=13.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw((0,0)--(5,5)--(8,11)--(8,0)--(0,0)); dot((0,0)); dot((5,5)); dot((8,11)); dot((8,0)); [/asy]

The shape can be divided into a triangle and a trapezoid. For the triangle, the base is $5$ and the height is $5$, so the area is $\frac{5 \cdot 5}{2} = \frac{25}{2}$. For the trapezoid, the two bases are $5$ and $11$ and the height is $3$, so the area is $\frac{3(5+11)}{2} = 24$. Thus, the total area is $\frac{25}{2} + 24 = \frac{73}{2} = \boxed{\textbf{(C) } 36.5}$.

See also

1969 AHSC (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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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