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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

So I was completing this lesson about triangles than the last 4 questions say

OK for this one I divided 3 by 1 and got 3 then I divided 57 by 3 and got 19 as an answer Is that a correct way to do it or not.

17. We are trying to measure the height of a building. We have a 1-meter long stick. When we set it on the ground, its shadow is 3 meters long. When we measure the shadow of the building, it is 57 meters long. How high is the building?

A 27 m

B40 m

C22 m

D26 m

E 19 m

F 30 m

for this one I am really confused I mean is their a special formula to follow?

18. I can look out my window and see the top of a television transmitter tower. On the map, I see that it is 2 miles away. I read somewhere that the tower is 500 feet tall. As I look at the tower, I see that the very top leaves of a tree sometimes get in the way of the top of the tower. The tree is 50 yards from where I sit. How tall is the tree?

A 7.1 ft

B9.2 ft

C5.1 ft

D3.8 ft

E 2.6 ft

F 1.5 ft

I think the answer is B

20. The sun is shining and I am on a hill. I want to measure the height of a tree downhill from me, using my one-meter stick, and a tape measure for measuring shadows. What do I do to take the slope of the hill into account?

A Bend the meter stick.

B Place the meter stick on the hill.

C Call your mother.

D Using the meter stick and the tape measure, recreate the slope

E It's not possible.

F You don't need to do anything different so long as your meter stick is on the same slope as the tree.

One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,227

Hi;

OK for this one I divided 3 by 1 and got 3 then I divided 57 by 3 and got 19 as an answer Is that a correct way to do it or not.

Where is the question?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

Hey,

This is the question

17. We are trying to measure the height of a building. We have a 1-meter long stick. When we set it on the ground, its shadow is 3 meters long. When we measure the shadow of the building, it is 57 meters long. How high is the building?

One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,227

Hi;

19 m is the correct answer. You would divide by 3.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

That is what I did, but I am confused on the rest do you know what this lesson is called so I can search up a video to explain how to solve the rest?

One, who adopts patience, will never be deprived of success though it may take a long time to reach him. Imam ali (as)<3

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,227

Hi;

Do you have access to some drawing programs?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

Hi,

Like what ? No I don't.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,227

I will make a drawing of a simpler one for you to learn with. Take a look at this simpler drawing.

The tower is 12 ft away the tree is 4 feet away. The ratio of the tree's distance to the tower's distance is

so the tree's height must be 1 / 3 of the towers. See the diagram it is 4 ft high.

**In mathematics, you don't understand things. You just get used to them.**

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

Thank you so much the example was really helpful here is what I came up with:

150/10560 = 70.4

and 500 divided by 70.4= 7.1022727

So my answer would be A 7.1 ft

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

for #20 I think the answer is E

for #19 am really stuck

19. Here are two triangles. I am trying to measure the area of triangle ABC. The formula for area of a triangle is

base*height/2.

I know the base, but I need to find the height. I know the top of triangle ABC is directly above a point 4.5 units from point A. I also know that <CAB is the same as <FDE, and that ls_DF has a length of 1, and ls_FE has a length of 1.43. What is the area of triangle ABC?

(ls stands for line segment) and (< stands for angle)

A 15.26 square units

B 19.31 square units

C 25.37 square units

D 27.16 square units

E 27.19 square units

F 13.02 square units

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,227

Hi;

I sure wished I had a drawing to look at.

**In mathematics, you don't understand things. You just get used to them.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

hi zee-f and bobbym

This topic is called 'proportion' and particularly 'similar triangles'.

When two shapes have the same angles or one is an exact enlargement of the other, they are said to be **similar**.

I've had a go at a diagram based on what you have written but I've hit two problems with trying to offer a solution.

Firstly, you need all the angles in the pair of triangles to be equal for the triangles to be similar. I've taken that to be true in my diagram, but you only say CAB = FDE. The base lines (BC and EF) could still be going in different directions.

Secondly, to work out the scale factor for the enlargement (once you know there is an enlargement) you need to know a measurement on one shape and the corresponding measurement on the other. From your description we haven't got that.

You could do it if the '1' is the height of the smaller triangle. Then the enlargement is 4.5/1 so the base BC = 1.43 x 4.5/1

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

I answered this question and I choose (B) and it was correct thanks for all the help!:D

But I am stuck on this question bellow(#14) well first I answered B and it was incorrect then I said:

My new answer would be (F) a triangle with an angle of 90o, then a side of SQRT(2), then an angle of 45o.

I have two corresponding angles that are congruent which are angles (45°,45°) and (90°,90°) and the corresponding sides (SQRT(2)),(SQRT(2)) between the angles are congruent.

So,according to the angle-side-angle theorem (ASA)the triangles are congruent.

And that was incorrect my teacher said #14 is incorrect. The right angle would not be adjacent to the hypotenuse of the triangle. I am really confused help plzzzz

14. I have a triangle with sides of 1 and a side of SQRT(2), with an angle of 45o and an angle of 90o. Which of the following would be congruent to it? (You will need to use what you've learned about triangles and angle / side relations, as well as your knowledge of the rules of congruence to fill in the gaps and answer the question. Sketches may be helpful.)

A a triangle with a side of 1, then an angle of 90o, and a side of 1

B a triangle with a side of 1, then an angle of 90o, then a side of SQRT(2)

C a triangle with the angles 45o, 45o, 90o

D a triangle with sides of 1 and 1

E a triangle with a side of 1, then an angle of 45o, then a side of 1

F a triangle with an angle of 90o, then a side of SQRT(2), then an angle of 45o

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

hi zee-f

The longest side of any triangle is always opposite the largest angle; and shortest is opposite smallest.

As root 2 is larger than 1, it must be opposite the 90 degrees, not adjacent to it.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

So (A) can be the answer because it has one as a side and 90 degree so side and side and angle.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

Stay on-line. I'll draw a diagram.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

ok

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

Below one diagram for this problem.

The triangle is isosceles because it has two sides equal and the third is the longest.

Oh. Just looked again at your latest answer. A is correct!

Bob

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

Thank you so much for all the help and time you are giving me ! )

But the thing I don't get is now I have an angle and a side that are congruent because (A) doesn't say it has a side of SQRT(2) and the question says the triangle has sides of 1 and a side of SQRT(2), with an angle of 45° and an angle of 90°. So in what theorem are they congruent side angle side SAS ?

Zee

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

When you do pythagoras theorem you will learn the the right angled triangle with two sides of 1 has a longest side of root 2

But you don't have to know this to do the question.

I have a triangle with sides of 1 and a side of SQRT(2), with an angle of 45o and an angle of 90o.

You are told all three sides and all three angles. Because the longest side must be opposite the largest angle the 90 must be between the two sides of 1.

If you do the following construction, only one triangle is possible (if you allow that all rotations and reflections are really the same)

(i) Draw a side with length 1.

(ii) At one end make a 90 degree angle.

(iii) Make this second side also have length 1.

(iv) You now have all three points for the vertices of the triangle so you can complete by drawing the third side. It's length will be root 2 but you don't need to know this. You have the congruency by SAS ( S = S = 1 and A = 90)

Bob

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

Ok I will show you what I drew in conclusion to what my mind came up with

Zee

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

ok that should be fun

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**zee-f****Member**- Registered: 2011-05-12
- Posts: 1,220

3aaaa here is what I came up with

*Last edited by zee-f (2012-10-08 07:46:22)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,227

Hi;

I see the image. Put the lines that make up the sides on a diet.

**In mathematics, you don't understand things. You just get used to them.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,731

Those are Euclidean lines ... so they are infinitesimally thin.

I'm going to pretend that your drawing shows the first triangle **and** the one that is congruent to it.

Bob

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