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#1 2021-06-23 08:21:51

zemole
Member
Registered: 2021-06-23
Posts: 1

Help Me! geometry review

Due to current family matters and certain personal issues, I'm in a steep time crunch and need help with answers( i just need answers, no this is not homework but simply a review). if it is no trouble I would like to know how you got your answers.

I have left out questions I can do on my own easily.
file:///C:/Users/skull/Downloads/geometry%20pictures.pdf
This link is for the pictures for the problems that have them( they are out of order but they are labeled for which question they belong to)

1. Two angles are complementary. The smaller angle is  35 degrees less than the larger angle. What is the measure of the larger angle?

A. 27.5 degrees
B. 55 degrees
C. 62.5 degrees
D. 125 degrees



3. In the figure shown, AC is congruent to CE .  D is the midpoint of CE(with a line over it).  DE = 5 and BC = 2 .
What is the measure of AB(with a line over it)?


A. 5
B. 7
C. 8
D. 10




6. Calculate the area of the regular octagon shown.


A. 21.9 square meters
B. 48.0 square meters
C. 91.2 square meters
D. 350.4 square meters

7. The apothem of a regular 9-sided figure is 6 feet and the area is 117.67 square feet. What is the perimeter of the regular 9-sided figure? 

A. 4.36 feet
B. 19.61 feet
C. 39.22 feet
D. 54.00 feet

8. Select which statement is true about the relationship between the measures of  (all letters have small triangles with an open end facing the letter) a and b in the diagram shown.



A. The measure of  a    is greater than the measure of  b  .
B. The measure of  b  is less than the measure of a .
C. The measure of  b    is equal to the measure of  a  .
D. Cannot be determined without any other calculations.



10. In the diagram shown below,  PN   is parallel to JQ  , KN=16 centimeters, PN=8 centimeters, and JQ=14 centimeters.



What is the length of  NQ?

A. 28 cm
B. 12 cm
C. 9.1 cm
D. 7 cm

11. Which is the contrapositive of the following statement? If quadrilateral ABCD is a rectangle, then quadrilateral ABCD is a square.

A. If quadrilateral ABCD is a square, then quadrilateral ABCD is a rectangle.
B. If quadrilateral ABCD is not a square, then quadrilateral ABCD is not a rectangle.
C. If quadrilateral ABCD is not a rectangle, then quadrilateral ABCD is not a square.
D. If quadrilateral ABCD is not a square, then quadrilateral ABCD is a rectangle.

12. How does the sum of the exterior angles of a square compare to the sum of the exterior angles of a pentagon?

A. The sum of the exterior angles of a square is greater.
B. The sum of the exterior angles of a pentagon is greater.
C. The two sums are equal.
D. There is not enough information to compare the two sums.



14. What is the value of x in this diagram?



A. 60 degrees
B. 72 degrees
C. 90 degrees
D. 120 degrees

15. What is the length of the line segment with endpoints  (-8,6)and (-10, -1)?

A. 53
B. square root of 373
C. square root of 53
D. 373

16. A right circular cylinder has a base diameter of 12 meters and a height of 16 meters. What is the approximate total lateral surface area of the cylinder? 

A. 602.9 meters squared(this goes for all)
B. 829.4
C. 1507.2
D. 2034.7

17. The volume of a cone is 218 cubic centimeters and the height of the cone is 13 centimeters. What is the radius of the cone to the nearest whole number? 

A. 4 cm
B. 5 cm
C. 8 cm
D. 16 cm

18. In the diagram shown, a 12-foot slide is attached to a swing set. The slide makes a 65 degree angle with the swing set. Which answer most closely represents the height of the top of the slide?

SIN 65degrees= 0.91(squiggly = this goes for all functions)
COS 65 degrees =0.42
TAN 65 degreesv=2.14

A. 5.0 feet
B. 5.6 feet
C. 10.9 feet
D. 25.7 feet

19. A regular hexagon is inscribed in a circle with a radius of 10 meters (as shown). What is the length of each side of the hexagon?




A. 5 meters
B.5 square root of three meters
C. 10 meters
D. Cannot be determined

20. Calculate the area of the shaded region in the figure shown.



A. 144-54 square root of three square meters
B. 144-18 square root of three square meters
C. 144-6 square root of three square meters
D. 54 square root of three square meters

Last edited by zemole (2021-06-23 08:30:16)

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#2 2021-06-23 20:02:28

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Help Me! geometry review

hi zemole

Welcome to the forum.

Review or homework? what's the difference?  This looks like a CompuHigh worksheet and you shouldn't be posting the whole set of questions like this.

I will try to help you to work out the answers yourself.  I'm happy to check any answers you post.

Most of the sheet, I cannot help with anyway as the link you have provided is to your own computer's hard drive.  Hence no pictures.  You could try describing the diagram in words.

Here's the ones I can help with:

1. Two angles are complementary. The smaller angle is  35 degrees less than the larger angle. What is the measure of the larger angle?

Complementary means they add up to 90.  If I call the larger one x, then the smaller is x - 35.  So solve the equation:

x + x - 35 = 90

7. The apothem of a regular 9-sided figure is 6 feet and the area is 117.67 square feet. What is the perimeter of the regular 9-sided figure?

The angle at the centre of a regular nonagon is 360/9 = θ .  If you join two adjacent vertices (A and B) to the centre ( O ) then AOB =  θ. Split this isosceles triangle in half to make a right angled triangle with 6 feet as the height.  From this and some trig. you can work out a side and hence the perimeter.

11. Which is the contrapositive of the following statement? If quadrilateral ABCD is a rectangle, then quadrilateral ABCD is a square.

If A => B, then the contrapositive statement is "not B => not A"

12. How does the sum of the exterior angles of a square compare to the sum of the exterior angles of a pentagon?

Imagine an ant crawling around the perimeter of a polygon.  At each vertex it turns through an exterior angle before continuing its route.  After completing the journey it has turned around 360, so the total for any polygon is the same.

15. What is the length of the line segment with endpoints  (-8,6)and (-10, -1)?

To calculate this you need to use Pythagoras Theorem.  Calculate the difference in x coordinates and square this number.  Do the same with the difference in y coordinates.  Add the two answers and square root to get the hypotenuse.  If you plot the points and draw in a right angled triangle this might help.

16. A right circular cylinder has a base diameter of 12 meters and a height of 16 meters. What is the approximate total lateral surface area of the cylinder?

Imagine the cylinder is split down its length and unfolded to make a rectangle.  The height is the same as the height of the cylinder and the width is the circumference of the circular end.

17. The volume of a cone is 218 cubic centimeters and the height of the cone is 13 centimeters. What is the radius of the cone to the nearest whole number?

The formula is

So use this to find the base area, then the radius squared and finally the radius.

19. A regular hexagon is inscribed in a circle with a radius of 10 meters (as shown). What is the length of each side of the hexagon?

If you divide a regular hexagon into 6 triangles made by joining each vertex to the centre, these are all equilateral triangles so their sides are equal.  So this question is easy!

For help with any others I need the diagrams.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2021-06-26 09:49:17

Rana
Member
Registered: 2021-06-26
Posts: 1

Re: Help Me! geometry review

A cube ABCDEFGH is given. determine what is the intersection of planes ACH and BFH. what is the relatin b/w these two planes

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#4 2021-06-26 21:28:35

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Help Me! geometry review

Hi Rana,

Welcome to the forum.  This question is quite difficult.  I'll try to explain as I go.

Firstly, here's an experiment for you to try.  I'm assuming you are in a room with walls at right angles.  Get some card and cut out a scalene triangle ABC.  That is a triangle with three different angles. Make one of them obtuse.

Now try to fit your triangle against a pair of walls so that A is in the corner, AB lying against one wall and AC lying against the other.  Whatever shape your triangle, this is possible. Now repeat with B in the corner, then C.

Clearly none of these angles is the true angle between the walls.  To 'see' the true angle you need to be looking along the line of intersection.  Above the walls looking down on the corner. Then the angle looks like 90 degrees.

So now to your question. I hope I've got the correct diagram. I'm assuming a square base ABCD, with E above A, F above B and so on.

The plane BFH is the plane BFHD.  Join BD and also AC and call their intersection J.  The line HJ lies in both planes so it must be the line of intersection.  We now need to find a triangle that has one point on the line of intersection, let's choose J.  And a line in the plane AJH (= ACH) that is at right angles to HJ.  call it JK. And a line in the plane BHJ ( = BFH) that is at right angles to HJ call it JL. The angle KJL is the angle you want.

So draw the triangle AJH and use trig to calculate JK. And the triangle JHB and calculate JL.

Argh. Just realised I cannot calculate KL.  Have to use vectors.  Do you know the scalar product?

I'll have a longer think to see if I can simplify this.  Back later.

Later edit:. I've got a sketch that makes it look like the answer is 90.  But I haven't got access to my geometry software at the moment.  When I have I'll post a diagram that maybe will demonstrate this.

Bob

Last edited by Bob (2021-06-26 23:03:05)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#5 2021-06-28 01:13:54

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Help Me! geometry review

hi again,

I now have a decent diagram for this.

I have assumed that ABCD is the base, and EFGH are above these points in that order.

J is the point where AC crosses BD.

As J lies on AC, the line HJ lies in the plane ACH.

Plane BFH includes point D.

As BD is in that plane, and J is on BD, HJ lies in the plane.

Thus HJ is the line of intersection of the planes.

In order to 'see' the true angle between the planes it is necessary to look along the line of intersection.  In the diagram that follows I have shown H 'over' J so that we are looking along that line.

a3YaLNa.gif

Notice that ACH and FBH appear as straight lines.  That is as it should be as we are looking along the line of intersection.

The angle between the planes appears to be 90 degrees.  Can we be certain of this, rather than 89 degrees say?

The plane BFHD is a plane of symmetry for the cube so an 'object' such as point A and it's 'image' C will be such that the angle betwen AC and the plane is 90. This follows from the laws of reflection.  So the angle is 90.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#6 2021-07-23 04:52:45

Mark Dater
Member
Registered: 2021-07-23
Posts: 12

Re: Help Me! geometry review

Who did assignment 18?

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#7 2021-07-23 06:53:17

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Help Me! geometry review

This poster just copy/pasted the whole worksheet but without the diagrams.  Q18 is not 'do-able' if you don't have a diagram.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#8 2022-08-09 18:34:56

Hicies87
Member
From: Portsmouth
Registered: 2022-01-22
Posts: 15

Re: Help Me! geometry review

I try to understand the last image but can't; can you please remind me of the reflection laws? It is summer, and I forgot a little bit about some of the math theories. During the study year, I used https://plainmath.net/secondary/geometry to get successful results and A grades. My worst difficulties were with geometry exercises. Fortunately, the Plainmath service saves me every time. I copy my problems and look through the high school geometry questions and answers database.

Last edited by Hicies87 (2022-08-16 07:01:53)

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#9 2022-08-09 21:23:52

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Help Me! geometry review

hi Hicies87

I'll have a go at explaining but it is tricky.  If you are able to make a model using straws that may help.

The cube is ABCD as a base with EFGH above it.  I've added one more point, J, which is the middle of the base ABCD, where the diagonals AC and BD cross.

Trying to work out the angle between two planes can be difficult.  In a house you'd expect the wall to be at right angles to each other but even with that sinple task you could get it wrong.  It'll be worth trying the following experiment.  With a large piece of card make a triangle with one obtuse angle.  It doesn't matter what size. Let's say it is XYZ with the obtuse angle at Y.  You can fit XYZ into the corner of a room so that both ZY and XY are touching the wall, with Y in the corner.  You'll have to lean the trangle a bit but it is possible.  So you have three points against the two walls, ZY against one wall and XY against the other.  Does that mean the angle between the walls is XYZ ?  As this works for any obtuse angle the answer must be no.

Now try to fit the trinagle so that X is in the corner, ZX against one wall and YX against the other.  Now it might seem that the angle between the walls is ZXY, which is acute.

To get the true angle between the walls you need to be on the ceiling looking down the line where the walls meet. Then the angle looks like 90.  It's only when you are lookig along the lone of intersection that you se the true angle between the surfaces.

So my first job was to identify the line that is in both planes; the line of intersection. J is on AC and H is given as a point in the plane ACH so JH is a line in the plane ACH.  It's not so obvious that J is also in the plane BFH.  HF is a horizontal line and FB is a vertiucal line, so the plane BFH is vertical.  So if we were looking down from above J would appear to be where FH and EG cross.  So J is in the plane.  Thus JH lies in the plane BFH.

That means it is the line where the planes intersect; so we need to look along this line to 'see' the true angle between the planes.  So I tried to make a picture of the cube looking directly at corner H, with J exactly behind it.  As a result of getting into this position ACH appears to be a line.  We're looking at ACH edge on.  Similarly BFH; again we're looking at BFH edge on.  So we can 'see' the true angle betwen the planes; it's the angle beween AC and BD.

Laws of reflection:  Stand in front of a mirror.  Imagine a point on the floor and it's image in the mirror.  Imagine a line joining them.  Such a line is always at right angles to the mirror  Let's prove that with some lettering.  Call the point P and its reflection Q.  Let R be the point on the mirror where PQ crosses.  Let S be another point on the mirror.  Consider the triangles PRS and QRS. Thye must be congruent as the image is an exact copy of the object. So angle PRS = QRS. But PRQ is a straight line so angle PRQ = 180.  So PRS =QRS = 180/2 = 90.



Back to the problem: AC is a line in one plane, BD a line in the other.  So the angle between the planes must be 90.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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