Yes. 3+2+6=11. If you divide 11 and 3, your answer is 3r2. You will have two 3 inch sides and one 5 inch side.
You remove right angled triangles whose legs are 3*[2-sqrt(2)] = 1.7574 inches long.
Let's start at the beginning of the taper, where the opening is zero, and go out a foot until it opens to 3/4 inch. We have a long skinny triangle. The line down the middle is 1 ft long, the line that closes off the open end is 1-1/2 inches. The line down the middle divides the long skinny triangle into 2 back-to-back right triangles. Each one has a 12-inch leg and a 3/4-inch leg. The angle back at the pointy starting end of each triangle is half of the angle we're looking for. That [half]-angle is tan-1(.75/12) = tan-1(0.0625) = 3.576 degrees (rounded) The whole angle is 7.153 degrees (rounded)
That is impossible because it can only have 1 perpendicular line so if it has 2 it wouldn't be a triangle.
Then its not a triangle.
Let's assume the triangle has points A, B, and C. Method 1 (3 lines) Draw two lines across the triangle parallel to line segment AB. Now you have two trapezoids and one triangle. Draw another line from C to the any point on the closest of the two lines you just drew, splitting the triangle into two more triangles. Method 2 (2 lines) Draw one line across the triangle parallel to line segment AB. Now you have one trapezoid and one triangle. Draw a second line that passes through C and is perpendicular to AB, splitting the trapezoid into two trapezoids and the triangle into 2 triangles. Method 3 (3 lines) Draw one line from point C to any point on line segment AB. Then draw a line parallel to AC and one parallel to BC, but don't let them cross the line you just drew.
No. A triangle with 2-inch sides is not congruent with a triangle with 3-inch sides.
yes. 8+4>2
if the triangle is a right angle it has 2 perpendicular line segments.
You remove right angled triangles whose legs are 3*[2-sqrt(2)] = 1.7574 inches long.
The formula for area of a triangle is bh/2 b=base h=height /=divide so multiply the base(or the bottom) side by the height of the triangle, and then divide that by two.
Picture a square. now make a line connecting the two diagonal points. If bh is the volume of the rectangle, then we only have half a rectangle for each right triangle giving 1/2 *bh or bh/2. For any other traingle put the base parallel to the ground. Make a perpendicular line to the ground then you have two parts of a triangle divided at a vertex. Each of the these parts has a right angle. Make a second triangle of the same size and if you turn it into the two triangle fromt he first triangle and cut it apart, you can manipulate it to make a rectangle. The two triangles are a rectangle which is bh. divide by two to get one triangle so 1/2*bh or bh/2
No. The three vertices would all lie on one line.
Need the angle between those two sides, or some more info. Imagine taking a 1 inch tooth pick and a 2 inch toothpick. Now line up the tips and you can make tons of angles. If you joined the outer tips of your toothpicks you would have a triangle. Can you see that you can have many different sides lengths depending on the angle between them?
x=y is the diagonal line which runs through 0,0 so all you have to do is reflect the triangle on the diagonal line. hop that helps :)
Let's start at the beginning of the taper, where the opening is zero, and go out a foot until it opens to 3/4 inch. We have a long skinny triangle. The line down the middle is 1 ft long, the line that closes off the open end is 1-1/2 inches. The line down the middle divides the long skinny triangle into 2 back-to-back right triangles. Each one has a 12-inch leg and a 3/4-inch leg. The angle back at the pointy starting end of each triangle is half of the angle we're looking for. That [half]-angle is tan-1(.75/12) = tan-1(0.0625) = 3.576 degrees (rounded) The whole angle is 7.153 degrees (rounded)
the star of david
A triangle with two lines of symmetry does not exist. It can have one line of symmetry (an isosceles triangle) or three (an equilateral triangle), but not two.