# How many triangles are possible with 10 points on a plane? all points must also be vertices.

Posted by on

• Level 3:

An expert who has achieved level 3 by getting 1000 points

All-Star:

An expert that got 10 achievements.

MVP:

An expert that got 5 achievements.

President:

An expert whose answer got voted for 500 times.

• Master

16 triangles

Posted on May 26, 2010

Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
goodluck!

Posted on Jan 02, 2017

×

my-video-file.mp4

×

## Related Questions:

### Finding the area of a triangle in the coordinate Plane

I find the easiest way to solve these is to sketch them first (I'm a visual learner;) We get a nice right-angled triangle, with the right-angle at B. The formula for the area of a triangle is 1/2 * base* height or (base * height)/2.

We can use BC or AB as the base.

If we use BC as the base, the length is 9-4 or 5. The height is 6-2 or 4.

We can now but the base and the height in the formula to figure out the area.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

Tip

There are two adjustment points (screws). For example left headlight: you will see one point near the end of left wing, at bottom side through a wing hole. This is for vertical plane headlight beam adjustment (up-down). Another one point is against water radiator side. This is for horizontal plane headlight beam adjustment (left-right).

on Dec 18, 2009 | Audi A4 Cars & Trucks

### Finding area of a Triangle in the Coordinate Plane

The area of a triangle is 1/2 times base times height. A sketch of the triangle in the coordinate plane will determine how easy or hard this will be to be. From the sketch, you will see that this is a right-angled triangle with B being the right-angle. This makes it easier because we can easily determine the base and the height to use in the formula.

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

### Right Triangle ABC has vertices A (-4 2)B(-4 6) C (9 6)

The area of a triangle is 1/2 times base times height. A sketch of the triangle in the coordinate plane will determine how easy or hard this will be to be. From the sketch, you will see that this is a right-angled triangle with B being the right-angle. This makes it easier because we can easily determine the base and the height to use in the formula.

We can chose AB or BC to be the base, while the other will be the height. If we choose the base of AB, its length is 4, the 6 - 2. The height is 9-(-4) or 13.

We can now put the length and height into the formula to calculate the area of the triangle.

Good luck.

Paul

Mar 19, 2015 | Office Equipment & Supplies

### Trying to determine if a tangent is present when a triangle overlaps a circle

In which language? This is really more a mathematical problem than a programming/compilation one. Try walking the circumference of the circle, testing for a point along each of the triangle's segments. If only one such point can be found, or a series of consecutive points (since software "circles" are never true circles, they're made up of short straight segments), then you have a tangent.

You can save a lot of time and math by making coarse estimates first - for example, don't bother testing for tangentiality for any circumferential segment that's fully above, to the left or right of, or below, all three points of the triangle - you CANNOT have a tangent in those areas.

An alternate approach is... walk the triangle instead of walking the circle. Test for a single circle crossing point. Again, you can save time by checking each segment first, and walking only segments that actually cross the circle's circumference. If no part of the circle can be found mathematically between any pair of vertices, then discard that segment outright.

Mar 10, 2014 | Computers & Internet

### Draw the developnent of lateral surface of pentagonal pyramid of base side 30mm and axis height 60mm

You need 5 isosceles triangles with a base of 30 mm. To get the height of the triangles (perpendicular to the 30 mm bases) you need to calculate the apothem of the pentagon (assumed to be regular).

If you cut the pyramid by a plane passing through its apex, the center of the base-pentagon,and the midpoint of one side, the plane figure created by the three points (apex, center, and midpoint) is a right triangle. The legs are the apothem, and the altitude of the pyramid. The hypotenuse is the slant height of the pyramid, and is thus the height of the triangles in the development. pyramid (60 mm) form.
Use the Pythagorean Theorem to find that slant height.
s^2=a^2+h^2.

Nov 11, 2013 | Computers & Internet

### I Have an Audi a4 Quatro 2004 model. Adjust Low Beams?

There are two adjustment points (screws). For example left headlight: you will see one point near the end of left wing, at bottom side through a wing hole. This is for vertical plane headlight beam adjustment (up-down). Another one point is against water radiator side. This is for horizontal plane headlight beam adjustment (left-right).

Nov 19, 2009 | 2004 Audi A4

There are two adjustment points (screws). For example left headlight: you will see one point near the end of left wing, at bottom side through a wing hole. This is for vertical plane headlight beam adjustment (up-down). Another one point is against water radiator side. This is for horizontal plane headlight beam adjustment (left-right).

Jul 13, 2009 | 2003 Audi A4 Cabriolet

### Image will not wrap correctly in vanishing point

Hi again,
Ok, I finally got it licked and why you're having the problems and it's fairly simple. It's all based on 2 things but most of all, the order in which you create your planes or grids.

Because the main part of the image is on the front face that is what you want as your Parent face so get that as accurate as you can to start off with. Hold the 'x' key to temp zoom in and get the position as correct as you can.
Once you've created you're primary ctrl-click(cmd-click on mac) the mid top point and pull it out then the same for the mid left point.
Now if it still has the problem when you paste in the image and drag it onto the primary face then undo the paste. We need to adjust the secondary planes. Select the top plane and in the angle parameter (which should show 270°) change it to 269°. Then change the left planes angle to 89°. Try it now and it should work fine. Then put it back to 270° and 90° again.

Let me know how you get on.

All the best
Sean
PiDigital.co.uk

Dec 28, 2008 | Adobe Photoshop CS3 10

### Trig and distance

Debs,

this is solved by starting with the SECOND leg of the flight considered as a triangle.

Find how far DUE east the plane travels (100 km x Cos10 = 83.9 km) using 10 degrees because that is how much is left of the 90 degree quadrent after subtracting the 80 degrres course direction.
Similarly find out how far DUE north the plane travels (100 km x Cos80 = 11 km).

Now get the total flight NORTH = 300 + 11 = 311
- and total EAST = 83.9

Finally, use Pythagoras theorem.to get the total DISTANCE:
square root of ( 311 squared + 83 squared ) = 321 km

Also the DIRECTION from its Tangent of 83 /311 = 14.9 degrees

Dec 19, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

## Open Questions:

#### Related Topics:

22 people viewed this question

Level 3 Expert