Length of a ramp
The post office in the town of Eagleville has a front door which is 4 feet off
the ground. There is a wheelchair ramp running up to it. In order to be the right steepness,
the ramp has to extend 48 feet horizontally from the door. The red triangle
$▵ABC$ drawn to the left shows a cross-section of this ramp.
The people in Eagleville have decided to build a ramp to the back door
of the post office, which is 6 feet off the ground. They want it to be as steep as the ramp
to the front door. Will this ramp need to be longer or
shorter than the ramp going up to the front door?
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Click . Now a cross-section of the back door ramp is
drawn as the blue triangle $▵A'B'C'$. We would like to find the
length of the side $\cl"blue"\ov{A'B'}$ of this triangle.
If the two ramps are equally steep, that means that the Angles
$\cl"red"{∠A}$ and $\cl"blue"{∠A'}$ have equal measure. Also, the Angles $\cl"red"{∠B}$
and $\cl"blue"{∠B'}$ have equal measure, because the sides $\cl"red"\ov{BC}$ and
$\cl"blue"\ov{B'C'}$ are both measurements of height off the ground, so they must both be at
right angles to the ground. What similarity rule do the triangles
$\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$ obey?
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Are $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$ similar triangles?
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Because $\cl"red"{▵ABC}$ and $\cl"blue"{▵A'B'C'}$ are similar triangles,
there is some number $r$ which you can multiply the length of each side of $\cl"red"{▵ABC}$ by
to get the length of the corresponding side of $\cl"blue"{▵A'B'C'}$.
Notice that $\cl"red"\ov{BC}$ and $\cl"blue"\ov{B'C'}$ correspond to
each other. Since the length of $\cl"red"\ov{BC}$ is 4 and the length of $\cl"blue"\ov{B'C'}$ is
6, $r$ must be a solution to the equation $4r=6$. Solve this equation, and check your
solution. Write your solution as a fraction in lowest terms.
The side $\cl"red"\ov{AB}$ has length 48. What is the length of the
side $\cl"blue"\ov{A'B'}$?
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How far from the back door should its ramp extend horizontally?
| feet |
Height of a flagpole
The flagpole outside Eagleville High School is very tall. Some of
the students at the high school would like to figure out exactly how tall it is without having
to get to the top of it.
At 7:00 in the morning on a certain day of the year, the students decide to measure the
length of the flagpole’s shadow. This is drawn to the left. The orange
arrow in the top left of the picture shows the direction the sun is shining from. The
points $A$ and $B$ are the bottom and top of the flagpole, and the point $\cl"orange"C$ is the
end of its shadow. This means that a ray from the sun which passes through $B$ (the top of the
flagpole) will hit the ground at $\cl"orange"C$. The students measure the flagpole’s shadow (the
line segment between $A$ and $\cl"orange"C$) and find it is approximately 117.1 feet long.
By sliding the slider, you can see the length of the shadow at
different times of the morning. Approximately how long is the shadow at 8:00 in the morning?
| feet |
Approximately how long is the shadow at noon (12:00)?
| feet |
In order to connect the length of the flagpole to the length of its shadow, the students
put a yardstick (which is 3 feet tall) next to the flagpole, sticking straight up, and measure
the length of its shadow. Click to see a picture of
the yardstick below the picture of the flagpole. (The two pictures are drawn at different
scales.)
Because the sun is shining from the same direction on both the
flagpole and the yardstick, the Angles $\cl"orange"{∠C}$ and $\cl"orange"{∠C'}$ have
equal measure. Because the flagpole and the yardstick are both sticking straight up
(that is, perpendicular to the ground), the Angles $∠A$ and $∠A'$ have equal measure.
What similarity rule do the triangles $▵ABC$ and $▵A'B'C'$ obey?
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Are $▵ABC$ and $▵A'B'C'$ similar triangles?
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Because $▵ABC$ and $▵A'B'C'$ are similar triangles, there is some number $r$
which you can multiply the length of each side of $▵A'B'C'$ by to get the length of the
corresponding side of $▵ABC$. We can find this $r$ by looking at the corresponding sides
$\cl"gray"\ov{A'C'}$ and $\cl"gray"\ov{AC}$ (the shadows of the yardstick and the
flagpole).
At 8:00 in the morning, approximately how long is the yardstick’s
shadow ($\cl"gray"\ov{A'C'}$)?
| feet |
At 8:00 in the morning, the flagpole’s shadow is approximately 79.8
feet long. Using this and your previous answer, what is an equation involving $r$ that must be
true?
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Using scratch paper and a calculator, solve that equation for $r$,
rounded to one decimal place.
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Using the value of $r$ you just found, find the height of the
flagpole. (Round to the nearest foot.)
| feet |
At 11:00 in the morning, approximately how long is the yardstick’s
shadow ($\cl"gray"\ov{A'C'}$)?
| feet |
Using the pictures and the slider, find an equation for $r$ at 11:00
in the morning.
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Using scratch paper and a calculator, solve that equation for $r$,
rounded to one decimal place.
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Using the value of $r$ you just found, find the height of the
flagpole. (Round to the nearest foot.)
| feet |
Did this process give you the same answer for the height of the
flagpole (rounded to the nearest foot) at both 8:00 and 11:00 in the morning?
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Circumference of the earth
The circumference of a spherical
object (like the earth) is the largest possible distance around it. More than two thousand years
ago, the mathematician and astronomer Eratosthenes measured the circumference of the earth by
making a few simple observations.
First, he observed that at noon on a certain day, the sun was directly overhead in the city
of Syene (in what is now southern Egypt). At the same time, the sun was roughly 7˚ away from
directly overhead in Alexandria (in northern Egypt). This is pictured to the left, with the
sun’s direction in Alexandria ($A$) on top and the sun’s direction in Syene ($S$) on the bottom.
The sun’s rays are drawn in orange, and the gray
ray drawn in the Alexandria picture is vertical (that is, perpendicular to
the ground).
Click to see a picture which explains this
discrepancy. The orange rays represent the sun’s rays. Because the
sun is so far away from the earth, these rays are all approximately
parallel. Even so, the sun’s rays hit different points on the earth at different angles to the
ground because the earth is curved. The gray rays meet at the center
of the earth, $C$.
The ray from the sun to Alexandria has been extended as a dashed line through the earth.
Because the sun’s rays are parallel, this orange line (through the
point $A$) is parallel to the gray segment and
orange ray through the point $S$ (at Syene). So the two marked
angles are corresponding angles.
Eratosthenes measured the angle at $A$
(Alexandria) to be approximately 7˚. What is the approximate measure of the marked angle at
$C$ (the center of the earth)?
| ˚ |
Approximately what fraction of a complete circle (360˚) is the marked
angle at $C$? Round your answer to four decimal places. (We write $≈$ — meaning “approximately
equal to” — as a reminder that your answer is rounded.)
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Eratosthenes also knew that Alexandria was 5000 stades (an ancient Mediterranean
unit of length) north of Syene. In modern units, 5000 stades is roughly 925 kilometers.
From this measurement and circumference-qn-1, we
know that 925 kilometers represents $0.0194$ of the circumference of the earth. That is, if
$E$ is the circumference of the earth in kilometers, we have an equation for $E$:
$$
0.0194\,E = 925
$$
Solve this equation using scratch paper and a calculator. What is
Eratosthenes’ estimate for the circumference of the earth? Round to the nearest kilometer.
| kilometers |
The modern generally accepted value for the circumference of the earth
is 40075 kilometers. Compare this to Eratosthenes’ estimate. Did Eratosthenes come close to the
modern value?
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An article with more details about Eratosthenes’ calculation can be found
here.