A student is standing next to a vertical flagpole. The top of the student's shadow

coincides with the top of the flagpole's shadow as shown.

not to scale

The student is 62 inches tall. The student estimates that the distance from the flagpole

to the point where the student is standing is between 21 and 22 feet. The student also

estimates that the length of the student's shadow is between 7 and 7 feet.

Based on the given information, what are the least and greatest possible heights, in

feet, of the flagpole? Explain how you arrived at your answers.

Answer:

The least possible height of the flagpole is 227.33 inches and the greatest possible height is 248 inches

Step-by-step explanation:

Given

See attachment for the system

Required

Determine the least and greatest heights of the flagpole

From the question, we have:

[tex]AB = 7\ to\ 7\frac{1}{2} ft[/tex]

[tex]BE =62in[/tex]

[tex]BC = 20\ to\ 21ft[/tex]

Convert AB and BC to feet (multiply by 12)

[tex]AB = 84in\ to\ 90in[/tex]

[tex]BC = 240in\ to\ 252in[/tex]

The height of the flagpole is represented as: CD

To calculate CD, we make use of the following equivalent ratios:

[tex]CD : AC = BE : AB[/tex]

Express as fraction

[tex]\frac{CD }{ AC } = \frac{BE }{ AB}[/tex]

Make CD the subject

[tex]CD = \frac{BE }{ AB} * AC[/tex]

From the attachment:

[tex]AC = AB + BC[/tex]

Where:

[tex]AB = 84in\ to\ 90in[/tex]

[tex]BC = 240in\ to\ 252in[/tex]

The possible values of AC are:

[tex]AC = 84 + 240 =324in[/tex]

[tex]AC = 84 + 252 =336in[/tex]

[tex]BC = 90 + 240 = 330in[/tex]

[tex]AC = 90 + 252 =342in[/tex]

So:

[tex]CD = \frac{BE }{ AB} * AC[/tex]

[tex]BE =62in[/tex]

When AC = 324 ; AB = 84

[tex]CD = \frac{62}{84} * 324[/tex]

[tex]CD = \frac{62* 324}{84}[/tex]

[tex]CD = \frac{20088}{84}[/tex]

[tex]CD = 239.14in[/tex]

When AC = 336 ; AB = 84

[tex]CD = \frac{62}{84} * 336[/tex]

[tex]CD = \frac{62* 336}{84}[/tex]

[tex]CD = \frac{20832}{84}[/tex]

[tex]CD = 248in[/tex]

When AC = 330 ; AB = 90

[tex]CD = \frac{62}{90} * 330[/tex]

[tex]CD = \frac{62 * 330}{90}[/tex]

[tex]CD = \frac{20460}{90}[/tex]

[tex]CD = 227.33in[/tex]

When AC = 342 ; AB = 90

[tex]CD = \frac{62}{90} * 342[/tex]

[tex]CD = \frac{62* 342}{90}[/tex]

[tex]CD = \frac{21204}{90}[/tex]

[tex]CD = 235.6in[/tex]

So, we have:

[tex]CD = 239.14in[/tex]

[tex]CD = 248in[/tex]

[tex]CD = 227.33in[/tex]

[tex]CD = 235.6in[/tex]

The least possible height of the flagpole is 227.33 inches and the greatest possible height is 248 inches

The given range of the length of the shadow and distance of the student

from the tree of between 7 and [tex]7\frac{1}{2}[/tex] feet and 21 and 22 feet gives;

The greatest possible height of the flagpole in feet ≈ 21.8 feet

The least possible height of the flagpole in feet = 19.3 feet

Height of the student = 62 inches = [tex]5\frac{1}{6}[/tex] feet

Distance of the flagpole to the student = 21 to 22 feet = 252 to 264 inches

Length of the student's shadow = Between 7 and [tex]\mathbf{7\frac{1}{2}}[/tex] feet = Between 84 and 90 inches

Required:

The least and greatest height of the flagpole.

Solution;

By using similar triangles formed by the student and the flagpole, we

have;

[tex]\dfrac{Height \ of \ the \ student}{Length\ of \ the \ student's \ shadow} = \mathbf{\dfrac{Height \ of \ the \ flagpole}{Length\ of \ the \ flagpole's \ shadow}}[/tex]

Which gives;

[tex]\dfrac{62}{84 \ or \ 90} =\mathbf{ \dfrac{h}{252+(84 \ or \ 90) \ or \ 264 + (84 \ or \ 90)}}[/tex]

Based on the method of cross multiplication, h will be the greatest

possible value, when we have, the largest possible value for the

denominator on the right hand side and the least possible value for the

denominator on the left hand side of the equation, and vice versa.

Which gives;

[tex]\dfrac{62}{90} = \mathbf{\dfrac{h_{min}}{252 + 84}}[/tex]

[tex]h_{min} \ in \ feet = \mathbf{\dfrac{\dfrac{62}{90} \times (252 + 84)}{12}} \approx 19.3[/tex]

The least possible height of the flagpole in feet, [tex]h_{min}[/tex] ≈ 19.3 feet

Similarly. we have;

[tex]\dfrac{62}{84} = \mathbf{\dfrac{h_{max}}{264 + 90}}[/tex]

Which gives;

[tex]h_{max} = \dfrac{62}{84} \times (264 + 90) \approx 261.3[/tex]

[tex]h_{max} \ in \ feet \approx \dfrac{261.3}{12} \approx\mathbf{ 21.8}[/tex]

The greatest possible height of the flagpole in feet, [tex]h_{max}[/tex] ≈ 21.8 feet

Learn more about the characteristics of similar triangles here:

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