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Difference between revisions of "2008 iTest Problems/Problem 2"

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size(200);
 
pair A=(0,0),B=(15,0),C=(135,0),D=(150,0);
 
pair A=(0,0),B=(15,0),C=(135,0),D=(150,0);
 
draw(A--D);
 
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==See Also==
 
==See Also==
 
{{2008 iTest box|num-b=1|num-a=3}}
 
{{2008 iTest box|num-b=1|num-a=3}}
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[[Category:Introductory Geometry Problems]]

Latest revision as of 23:15, 21 June 2018

Problem

One day while Tony plays in the back yard of the Kubik's home, he wonders about the width of the back yard, which is in the shape of a rectangle. A row of trees spans the width of the back of the yard by the fence, and Tony realizes that all the trees have almost exactly the same diameter, and the trees look equally spaced. Tony fetches a tape measure from the garage and measures a distance of almost exactly $12$ feet between a consecutive pair of trees. Tony realizes the need to include the width of the trees in his measurements. Unsure as to how to do this, he measures the distance between the centers of the trees, which comes out to be around $15$ feet. He then measures $2$ feet to either side of the first and last trees in the row before the ends of the yard. Tony uses these measurements to estimate the width of the yard. If there are six trees in the row of trees, what is Tony's estimate in feet?

[asy] size(350); defaultpen(linewidth(0.8)); draw((0,-3)--(0,3)); int d=8; for(int i=0;i<=5;i=i+1) { draw(circle(7/2+d*i,3/2)); } draw((5*d+7,-3)--(5*d+7,3)); draw((0,0)--(2,0),Arrows(size=7)); draw((5,0)--(2+d,0),Arrows(size=7)); draw((7/2+d,0)--(7/2+2*d,0),Arrows(size=7)); label("$2$",(1,0),S); label("$12$",((7+d)/2,0),S); label("$15$",((7+3*d)/2,0),S);[/asy]

Solution

[asy] size(200); pair A=(0,0),B=(15,0),C=(135,0),D=(150,0); draw(A--D); draw(circle(A,15)); draw(circle(D,15)); dot(A); dot(B); dot(C); dot(D); label("$x$",(7.5,0),N); label("$12$",(75,0),N); label("$x$",(142.5,0),N); [/asy]

Let $x$ be the distance from the center to the edge of each tree (since all the trees have the same distance). Because the distance between the two trees is $15$ feet, we can write an equation to solve for $x$. \[2x+12=15\] \[2x=3\] \[x=1.5\] Thus, the distance from the center to the edge of each tree is $1.5$ feet, so the distance between opposite edges is $3$ feet. Therefore, Tony’s estimated length of his yard is $2(2)+6(3)+12(5)=\boxed{82}$ feet.

See Also

2008 iTest (Problems)
Preceded by:
Problem 1 		Followed by:
Problem 3
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