1969 AHSME Problems/Problem 22

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 $\frac{25}{2} + 24 = \frac{73}{2} = \boxed{\textbf{(C) } 36.5}$ .

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AHSME Problems and Solutions

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

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1969 AHSME Problems/Problem 22

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