No because all right triangles have 2 legs and a hypotenuse. The hypotenuse is always longer than either leg so right triangles can't be equilateral triangles.
isosceles has 2 congruent angles 2 congruent sides right triangles has sides that consist of 2 legs and a hypotenuse. in the Pythagorean therom. a+b are legs...A2 +B2=C2 one right angle(90)
Yes if you mean in degrees
Yes, because a^2 + b^2 = c^2, where a and b are legs and c is the hypotenuse. Therefore, if a and b are equal, so will c.
Only if the two triangles are congruent will they have equal areas. A third fact is required to determine they are congruent (and thus have the same area): 1) The third sides are equal; 2) The angles enclosed between the sides are equal; or 3) The same one of the sides is the hypotenuse of the triangles, which are right angled triangles.
No because all right triangles have 2 legs and a hypotenuse. The hypotenuse is always longer than either leg so right triangles can't be equilateral triangles.
isosceles has 2 congruent angles 2 congruent sides right triangles has sides that consist of 2 legs and a hypotenuse. in the Pythagorean therom. a+b are legs...A2 +B2=C2 one right angle(90)
The interior angles of an octagon are 135 degrees, which means that if you extend one of the sides, the complementary external angle is 45 degrees. So imagine an octagon instead as a square with four cut out right triangles on the corners. Let's start with the length of a side x. Since the right triangles have a hypotenuse x, we know that the legs are length (sqrt(2)/2)x. The area of the triangle is 1/2 * base * height, which means one triangle is x2 / 4. Since there are 4 triangles, the total area of the triangles is x2. Next, we need the area of the square. One side is a leg of the octagon plus two legs of our triangles, which would be x + (sqrt(2)/2)x + (sqrt(2)/2)x = x + sqrt(2)*x. Square this, and you get 3x2 + 2*sqrt(2)*x2. But! Subtract the area of the triangles from this, and you get (after some redistributing): A = 2x2 * (1 + sqrt(2))
Yes if you mean in degrees
Lets work it out:- If the legs are 12 and 16 inches, then a rectangle of that size would be 192 square inches in area. As the the diagonal of the rectangle makes two equal triangles (of legs 12 and 16 inches), the area of one of these is half the area of the triangle - 192/2 = 96 square inches.
1. There are two right triangles. 2. They have congruent hypotenuses. 3. They have one pair of congruent legs.
Yes, because a^2 + b^2 = c^2, where a and b are legs and c is the hypotenuse. Therefore, if a and b are equal, so will c.
Not entirely. You already know that the hypotenuse is only in right triangles, and what it is, right? It's (the two legs)^2=hypotenuse^2. So, for example, if the two legs were to be 3 and 5, it would be 3^2+5^2 which would be 36. But that is the squared answer, so you have to square root it to get 6. Understand!!?
Only if the two triangles are congruent will they have equal areas. A third fact is required to determine they are congruent (and thus have the same area): 1) The third sides are equal; 2) The angles enclosed between the sides are equal; or 3) The same one of the sides is the hypotenuse of the triangles, which are right angled triangles.
It depends on what type of triangle you are dealing with, and also what you are meaning by what are they. If you are just talking in general terms 2 are legs, and one is the hypotenuse. In right triangles the hypotenuse is the side opposite the right angle, and the legs are adjacent to it.
To find the area of a right triangle, you can use the formula A = 0.5 * base * height, where the base and height are the two sides that form the right angle. Alternatively, you can use the Pythagorean theorem to find the lengths of the sides if they are not given, and then use the formula mentioned earlier. Another method is to use trigonometric functions such as sine, cosine, or tangent to find the area based on the given angle and side lengths.
Yes, if two of the angles in a triangle are obtuse or right, theres no logical way for the legs to meet at a point.