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April 20, 2014

April 20, 2014

**Recent Homework Questions About Geometry**

Post a New Question | Current Questions

**Geometry**

correct. and volume is cubed.
*Friday, January 17, 2014 at 12:09pm*

**Geometry**

Use the figure of a square pyramid to answer the following questions. a.) Find the total surface area of the pyramid. b.) Find the volume of the pyramid. Do not answer them. I know How to find the answer. Surface Area is squared right?
*Friday, January 17, 2014 at 12:08pm*

**Geometry**

5 feet, 6 inches = 5.5 ft not 5.6 ft I don't have your diagram, so have no clue where your N, S etc are, but I can guess that 5.5/12 = height/30 height = 30(5.5/12) = 13.75 ft = 13 ft, 9 inches
*Friday, January 17, 2014 at 10:46am*

**Geometry**

A person who is 5 feet, 6 inches (5.6 feet) tall casts a shadow (from N to S) that is 12 feet long. The distance along the ground from the person (N) to the flagpole (G) is 18 feet. Find the height of the flagpole (FG) showing all calculations.
*Friday, January 17, 2014 at 10:34am*

**Geometry**

A=10.6cm H^2=84.27 H=9.1 cm
*Friday, January 17, 2014 at 12:40am*

**Geometry - 8th grade**

Well, I am trying to get geometry polygons and angles, but it's really hard to understand what to do. I'm not one to exaggerate and say I'm dying, but I'm pretty desperate for some help. If someone could explain the basics in simple terms or even post a helpful...
*Thursday, January 16, 2014 at 8:51pm*

**Geometry**

What is the reason called when there are two legs on an isosceles triangle, are bisected using a mid segment to form a trapezoid, and you are trying to prove the legs of the trapezoid are congruent?
*Tuesday, January 14, 2014 at 8:50pm*

**Geometry/Math**

According to my diagram, h/445 = 10/45 h = 10(445)/45 = ...
*Tuesday, January 14, 2014 at 5:24pm*

**Geometry/Math**

Judy lies on the ground 45 feet from her tent. Both the top of the tent and the top of a tall cliff are in her line of sight Her tent is 10 feet tall. About how is the cliff? Assume the two triangles are similar. Postscript: The distance from the tent to the cliff is 400 feet...
*Tuesday, January 14, 2014 at 5:15pm*

**geometry**

ratio of height to shadow length is the same for both 5.5/12 = FG/length of flagpole shadow I do not have your diagram so I do not know if the flagpole shadow is 18 feet or 18+12 or whatever
*Tuesday, January 14, 2014 at 11:22am*

**geometry**

A person who is 5 feet, 6 inches (5.5 feet) tall casts a shadow (from N to S) that is 12 feet long. The distance along the ground from the person (N) to the flagpole (G) is 18 feet. Find the height of the flagpole (FG) showing all calculations. What is the length of SP and PF?
*Tuesday, January 14, 2014 at 10:52am*

**Geometry**

You forgot to change 29m to 2900cm.
*Tuesday, January 14, 2014 at 8:26am*

**GEOMETRY**

50 degrees
*Monday, January 13, 2014 at 2:22am*

**Geometry**

Use a proportion to solve this problem. 1.2 / 1.8 = 10/x Cross multiply and solve for x.
*Saturday, January 11, 2014 at 8:45pm*

**Geometry**

Stephanie casts a shadow of 1.2m and she is 1.8m tall. A wind turbine casts a shadow of 10m at the same time that Stephanie measured her shadow. Draw a diagram of this situation and then calculate how tall the wind turbine is.
*Saturday, January 11, 2014 at 8:37pm*

**geometry**

efgh has half the area of abcd, so its side is 6/√2.
*Friday, January 10, 2014 at 5:34pm*

**Geometry**

If we are in one dimension here (on a number line) then -2 + 4 = 2 and -2 - 4 = -6
*Friday, January 10, 2014 at 2:58pm*

**Geometry**

Suppose point J has coordinate of -2 and JK=4. Then what is the possible coordinate(s) for K?
*Friday, January 10, 2014 at 2:42pm*

**geometry**

abcd is a square picture frame. efgh is a square inscribed within abcd as a space for a picture. the area of efgh (for the picture) is equal to the area of the picture frame (the area of abcd minus the area of efgh). if AB=6, what is the length of EF
*Friday, January 10, 2014 at 1:44pm*

**geometry**

Consider the right triangles with AD and BC as hypotenuses. Their height is the same, so their base legs are also the same. So, with all three sides congruent, the angles are also congruent.
*Friday, January 10, 2014 at 12:56pm*

**geometry**

Suppose ABCD is a trapezium in which AB is parallel to CD and AD=BC.prove that angle A=angle B and angle C=angle D.
*Friday, January 10, 2014 at 11:34am*

**geometry**

I don't know
*Wednesday, January 8, 2014 at 8:34pm*

**geometry**

y = [x/4] where [n] is the greatest integer less than or equal to n.
*Wednesday, January 8, 2014 at 5:45am*

**geometry**

what is the algebraic rule for x=4,y=1;x=6,y=1;x=8,y=2;x=10,y=2;x=12,y=3;x=14,y=3;x=16,y=4;x=18,y=4...
*Wednesday, January 8, 2014 at 5:39am*

**geometry**

just use a simple ratio .... 5/2 = h/8 h = 8(5/2) = 20
*Tuesday, January 7, 2014 at 8:30pm*

**geometry**

if a 5-foot tall bush casts a 2-foot shadow how tall is a tree standing next to the bush that casts an 8-foot shadow at the same time?
*Tuesday, January 7, 2014 at 8:27pm*

**Geometry**

its 15
*Tuesday, January 7, 2014 at 7:05pm*

**geometry**

so we are doing: (x,y) --->(x-1, y+3) ----> (x+4, y-1) so for R(-2,-2) (-2,-2) --->(-3, 1) ----> (1, 0) do the same for the other points (I am not sure why you have quotations around S and T in the final result. Are you saying that only S and T are transformed? If ...
*Monday, January 6, 2014 at 9:09pm*

**geometry**

RST with vertices R (-2, -2) , S (-3, 1) , and T (1, 1) is translated by (x, y) → (x - 1, y + 3) . Then the image, RST , is translated by (x, y) → (x + 4, y - 1) , resulting in R "S"T ". a. Find the coordinates for the vertices of R &...
*Monday, January 6, 2014 at 9:01pm*

**Geometry**

Open the book and learn the material. The examples in the textbook you have are pretty easy to follow and correspond to what you are asked on the assessments. HS math is not so difficult that you cannot get at least 60% of the material by your own work with a little effort. ...
*Monday, January 6, 2014 at 2:06pm*

**help hexagon geometry**

drwls you were close it's 4/3
*Friday, January 3, 2014 at 8:53pm*

**geometry**

You have 6 identical equilateral triangles. the area of one of them = (1/2)(11)(11)sin60° = (121/2)(√3/2) so the are of the hexagon = 6(121/2)(√3/2) =appr 314 to the nearest whole number
*Friday, January 3, 2014 at 5:03pm*

**geometry**

what is the area of a hexagon with a base of 11m to the nearest whole number?
*Friday, January 3, 2014 at 12:25pm*

**Geometry**

Since the diagonals of a rectangle are equal, 2x+30 = 36 x = 3
*Thursday, January 2, 2014 at 6:56pm*

**Geometry**

In rectamhle ABCD, diagonals AC and BD intersect at E. if AC=36 and BD=2x+30, find x.
*Thursday, January 2, 2014 at 6:03pm*

**Geometry**

idk
*Thursday, January 2, 2014 at 2:18pm*

**Geometry**

nigaa
*Tuesday, December 31, 2013 at 1:33pm*

**geometry**

Recall that surface area of cone is SA = pi*r^2 + pi*r*s where r = radius s = slant height Substituting, SA = 3.14 * 15^2 + 3.14 * 15 * 11 SA = 706.5 + 518.1 SA = 1224.6 sq units Hope this helps :3
*Monday, December 30, 2013 at 7:56am*

**geometry**

Find the surface area of a cone with a slant height of 11 and a radius of 15. Use 3.14 for ð. Round your answer to the nearest tenth.
*Sunday, December 29, 2013 at 11:44pm*

**Geometry**

200%
*Friday, December 27, 2013 at 11:50pm*

**geometry**

kkmk
*Friday, December 27, 2013 at 3:23am*

**Geometry**

thanks
*Tuesday, December 24, 2013 at 7:39pm*

**Geometry**

Since DE is parallel to OA, and DB = 1/2 AB and EB = 1/2 OB, DE = 1/2 OA
*Tuesday, December 24, 2013 at 6:50pm*

**Geometry**

a,b,o are points in the triangle Then it wants to know an distance and d Is the midpoint of ab and e is the midpoint of Bo so if who take the two midpoints and connect them it' wants to know the distance of them
*Tuesday, December 24, 2013 at 2:24pm*

**Geometry - eh?**

Hard to say. Are a and b points or distances? Where the heck are d and e? Usually, points are labeled with capital letters, and line segments with small letters, or by the two endpoints, as AB or DE.
*Tuesday, December 24, 2013 at 5:50am*

**Geometry**

If point a is (0,n) and b is (k,0) and point o is (0,0) what is the length of a and b And the length of de Do not solve just wants letters
*Tuesday, December 24, 2013 at 12:23am*

**geometry**

Did you sketch it ? AB= √(0^2 + 8^2) = 8 F is (5/2,8) E is (5/2, 0) FE = √( (5/2-5/2)^2 + (8-0)^2 ) = √(0 + 64) = 8 so AB = EF
*Saturday, December 21, 2013 at 8:37pm*

**geometry**

Given: Rectangle ABCD with coordinates A(0, 0), B(0, 8), C(5, 8), and D(5, 0). E is mdpt. of BC, and F is mdpt. of AD. Prove: EF = AB
*Saturday, December 21, 2013 at 7:24pm*

**Geometry.**

the distance must be less than the longest side The longest side is less than the sum of the other two sides The perimeter is the sum of the longest side and the other two sides. That should help.
*Friday, December 20, 2013 at 12:43am*

**Geometry.**

no; the altitudes always intersect. Nor can you infer anything.
*Friday, December 20, 2013 at 12:39am*

**Geometry.**

can you imply something if you are given that two altitudes intersect in a triangle?
*Thursday, December 19, 2013 at 7:08pm*

**Geometry.**

prove that the distance between any two points inside triangle abc is not greater than half the perimeter of triangle abc my teacher said to try doing this proof by contradiction. can someone please help??
*Thursday, December 19, 2013 at 6:59pm*

**geometry**

I sketched your triangle and concluded that the right angle must be at N It is easy to show that triangle MPN ≈ triangle NPO then MP/PN = PN/PO 15/9 = 9/PO 15PO = 81 PO = 81/15 = 27/5
*Thursday, December 19, 2013 at 11:52am*

**geometry**

M 15 z P 9 x N y O
*Thursday, December 19, 2013 at 9:02am*

**geometry**

In a triangle MNO with altitude NP with MP of 15 and NP of 9 what is the length of PO? MNO creates a right angle!
*Thursday, December 19, 2013 at 8:43am*

**Geometry**

http://www.angelfire.com/electronic2/tyler_geometry/page4.html
*Wednesday, December 18, 2013 at 6:40pm*

**Geometry**

What's the difference between a postulate and a theorem?
*Wednesday, December 18, 2013 at 6:32pm*

**geometry**

tkia
*Wednesday, December 18, 2013 at 6:05pm*

**geometry**

36.5
*Wednesday, December 18, 2013 at 3:07pm*

**geometry**

1+1=
*Tuesday, December 17, 2013 at 7:53pm*

**GEOMETRY TANGENTS**

I assumed AD was a typo, or due to label discrepancy, since the subject is the tangents, which are of length 4,5,11.
*Tuesday, December 17, 2013 at 12:11pm*

**GEOMETRY TANGENTS**

done, look at your previous post of this
*Tuesday, December 17, 2013 at 9:04am*

**GEOMETRY TANGENTS**

Who knows ? You don't mention x in your description. You don't mention where B, D, and F are located, but I must have labeled mine the same way as Steve, since I got his answers as well. What still has to be found is AD In triangle ACE 9^2 = 16^2 + 15^2 - 2(15)(16)cosC...
*Tuesday, December 17, 2013 at 8:58am*

**GEOMETRY TANGENTS **

Triangle ACE is a circumscribed triangle. Points B, D and F are the points of tangency. If AC= 16 and AE is 9 cm and CE = 15 cm . Find AD, FE and CD HELP PLEASE VERY URGENT
*Tuesday, December 17, 2013 at 6:29am*

**GEOMETRY TANGENTS**

but whats x ?
*Tuesday, December 17, 2013 at 6:28am*

**GEOMETRY TANGENTS**

You don't explain the labeling very well, but I get that AC is divided into 5+11 CE is divided into 11+4 AE is divided into 4+5
*Tuesday, December 17, 2013 at 5:27am*

**GEOMETRY TANGENTS **

TRIANGLE ACE IS A CIRCUMSCRIBED TRIANGLE. POINTS B,D AND F ARE POINTS OF TANGENCY . IF AC= 16 AND AE= 9 CM AND CE = 15 CM FIND AD, FE, AND CD. Thank you very mych
*Tuesday, December 17, 2013 at 5:19am*

**MATH - GEOMETRY/TRIGONOMETRY**

since the angle of elevation of the sun is about 30°, all we can say is that it's about 2 hours after sunrise, whenever that was. Or, 2 hours before sunset. Pretty vague question.
*Tuesday, December 17, 2013 at 5:17am*

**College geometry**

x-intercept (15,0)
*Tuesday, December 17, 2013 at 1:27am*

**College geometry**

Find the coordinates of the x-intercept. -x + 5y = -15 (Points : 1)
*Tuesday, December 17, 2013 at 1:13am*

**geometry**

are the angles of x and 5.5y opposite angles or adjacent angles? What do you mean by "the triangle?" Is there a triangle drawn inside the parallelogram?
*Monday, December 16, 2013 at 11:09pm*

**geometry**

The measure of the angles in a parallelogram total 360. Write and solve a system of equations to determine the values of x and y. The parallelogram shows x and 5.5y. The triangle shows x and y. I have the first equation at 180= x + 5.5y. What would be another equation for the ...
*Monday, December 16, 2013 at 10:26pm*

**MATH - GEOMETRY/TRIGONOMETRY**

A 6 foot man stands by a 30 foot radio tower and casts a 10 foot shadow. How long is the shadow cast by the tower and what time is it? I got 50 feet for the length of the shadow. What I cannot find out is, what time is it? I NEED THE TIME OF DAY. **NOT** THE LENGTH OF THE ...
*Monday, December 16, 2013 at 9:06pm*

**Math [Geometry]**

6/30 = 10/x 300 = 6x x = 50 50/30 =x/10
*Monday, December 16, 2013 at 9:01pm*

**Math [Geometry]**

A 6 foot man stands by a 30 foot radio tower and casts a 10 foot shadow. How long is the shadow cast by the tower and what time is it? I got 50 feet for the length of the shadow. What I cannot find out is, what time is it?
*Monday, December 16, 2013 at 8:37pm*

**Geometry**

36
*Monday, December 16, 2013 at 3:45pm*

**geometry**

since ABCD is a rhombus, all the sides are equal. So, we already know that AD=AB=BC=CD Since BD is an angle bisector, <ABD = <DBC So, we have SAS and the triangles are congruent.
*Sunday, December 15, 2013 at 6:09pm*

**geometry**

The two chords intersect to form 2 similar triangles, in this case 2 similar right-angled triangles so we can easily find that ED = 24 by Pythagoras: AC = √20 and BD = 6√20 Does that give you a start ?
*Sunday, December 15, 2013 at 11:13am*

**geometry**

I need to figure out this proof, the figure is two triangles forming a rhombus. Given: segment BD is the angle bisector of triangle ABC and triangle ADC Prove: Triangle ABD is congruent to Triangle CBD So far I have segment BD is the angle bisector of triangle ABC and triangle...
*Sunday, December 15, 2013 at 10:51am*

**geometry **

in circle O, perpendicular chords AB and CD intersect at E so that AE=2, eb=12 and CE=4. find the radius of a circle O and the shortest distance from e to the circle
*Sunday, December 15, 2013 at 8:35am*

**Geometry shapes and designs**

what are the dimensions of all the possible rectangles that can be made with the given number of square tiles
*Saturday, December 14, 2013 at 11:17am*

**Math**

If we lay one side along the x-axis, and one vertex at (0,0), and let the side be of length 1, the triangles sides can be written as y=0 y=√3x y=√3(1-x) Assuming the obvious position where one side of the square lies along one side of the triangle, from (a,0) to (...
*Friday, December 13, 2013 at 9:40pm*

**Geometry**

assuming R,S,T are collinear, we have RS+ST=RT so, 3x+1 + 2x-2=64 5x-1 = 64 5x=65 x=13
*Friday, December 13, 2013 at 3:59pm*

**Geometry**

RS=3X+1,ST=2x-2,And RT=64 Find The Vaule Of X ?
*Friday, December 13, 2013 at 11:43am*

**please help this is a geometry question**

622
*Thursday, December 12, 2013 at 7:00pm*

**Math - Precalculus - Parametric Equations/Geometry**

Well, I would just like to point out to steve that 24/√2 = 12√2, not 6√2. And Samuel? Ahem.
*Thursday, December 12, 2013 at 5:33pm*

**GEOMETRY PLZ HELP!**

See previous post: Tue,12-10-13,8:35 PM.
*Thursday, December 12, 2013 at 4:36pm*

**geometry**

c^2 = a^2 + b^2 c^2 = 6^2 + 6^2 = 72 c = 8.49 Ft. = 101.82 In. 101.82/4 = 25.46 or 25 Wall panels. 0.46 * 4 = 1.8 Inches. .
*Thursday, December 12, 2013 at 4:33pm*

**Math - Precalculus - Parametric Equations/Geometry**

2pi radians per rotation. 3 rotations/sec
*Wednesday, December 11, 2013 at 5:29pm*

**Math - Precalculus - Parametric Equations/Geometry**

Thanks, but I was wondering how you got the horizontal and vertical speeds, sorry I'm not that good at parametric equations yet :( Also how did you get that the angular velocity is 3*2pi / s? Thanks
*Wednesday, December 11, 2013 at 5:24pm*

**Math - Precalculus - Parametric Equations/Geometry**

(a) the horizontal and vertical speeds are 24/√2 = 6√2, so at time t, the center is at (6√2 t, 6√2 t) (b) the puck is a circle of radius 3/2, and angular velocity = 3*2π /s The point (0,3/2) corresponds to θ=0 So, we have, relative to the puck...
*Wednesday, December 11, 2013 at 5:14pm*

**Math - Precalculus - Parametric Equations/Geometry**

A hockey puck of diameter 3 inches is spinning around its center at a speed of 3 counterclockwise rotations per second. The center of the puck is traveling at a speed of 24 inches per second at an angle of 45 degrees to the positive real axis. (a) At time t=0, the center of ...
*Wednesday, December 11, 2013 at 5:07pm*

**chemsitry**

The molecular geometry of POCl3 is tetrahedral. What hybridization does this indicate for the central P atom? Why is the answer spy?
*Wednesday, December 11, 2013 at 3:36pm*

**basic geometry**

AB^2 = (-3-2)^2 + (6-3)^2 = 25+9 = 34 AC^2 = (-4-2)^2 + (-3-3)^2 = 36+36 + 72 BC^2 = (-4+3)^2 + (-3-6)^2 = 1+81 = 82 take square root to get the actual side lengths. If ABC is a right triangle, we'd need 34+72=82, but it does not.
*Wednesday, December 11, 2013 at 12:51pm*

**basic geometry**

as usual, draw a diagram. (a) x^2 + 100^2 = 250^2 (b) 25^2 + x^2 = 250^2 (c) 25^2 + 50^2 = x^2
*Wednesday, December 11, 2013 at 12:33pm*

**basic geometry**

for safety reasons , anyone parasailing behind a boat needs to stay at least 25 feet away from the back of the boat. if you were parasailing using a 250 foot rope , how far from the boat would you be if you were 100 feet high in the air ? round to the nearest tenth. what is ...
*Wednesday, December 11, 2013 at 9:13am*

**basic geometry**

the three verticies of a triangle can be found at the points A(2,3) , B(-3,6), C(-4,-3) how long is each side of the triangle ab,bc,and ac explain how you got the answer. is this triangle a right triangle ? why and explain.
*Wednesday, December 11, 2013 at 9:05am*

**basic geometry**

since it's a right triangle, the 3rd side is just the hypotenuse, so c^2 = 6^2+6^2 only one 4" panel would fit in a 6" wall.
*Tuesday, December 10, 2013 at 11:27pm*

**GEOMETRY PLZ HELP!**

Mrs.sally needed to build a triangular platform in the shape of an isosceles right triangle . The legs of the triangle are 6"long how long was the base side of the triangular platform ? She also made wall panels for the show that were 4" wide . How many panels would ...
*Tuesday, December 10, 2013 at 9:29pm*

**basic geometry**

mrs. sally needed to build a triangular platform in the shape of an isosceles right triangle. the legs of the triangle are 6"long how long was the base side of the triangular platform? she also made well panels for the show that were 4" wide. how many wall panels ...
*Tuesday, December 10, 2013 at 9:20pm*

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