Not necessarily.
All one can say about "any rectangle" is that the opposite triangles are of equal areas.... that does not mean that adjacent ones do.
So, in a rectangle ABCD, with diagonals which cross at E
ABE = CDE
and
ADE = BCE
but
ABE may not be equal to ADE
no
We know that diagonals of parallelogram bisect each other. Therefore, O is the mid-point of AC and BD. BO is the median in ΔABC. Therefore, it will divide it into two triangles of equal areas. Area (ΔAOB) = Area (ΔBOC) ... (1) In ΔBCD, CO is the median. Area (ΔBOC) = Area (ΔCOD) ... (2) Similarly, Area (ΔCOD) = Area (ΔAOD) ... (3) From equations (1), (2), and (3), we obtain Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD) Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.
Triangles (four of) and a square.One square as base and four triangles as sides.
Triangular. the base is square, with four other triangular faces.
Since a rectangle is a parallelogram with four right angles and the rhombus is a parallelogram with four equal sides with opposite sides complement to each other, the parallelogram that is both rectange and rhombus is the square.
A=1/2bh The area of a triangle is 1/2bh. If the base of it is a triangle and all 4 of the triangles aren't the same, then you have to find the area of the base triangle and then the three other triangles (which should all have the same area). If all four of the triangles have the same area, then just find the area of one of the triangles and multiply that by four. A triangular pyramid that has four equal triangles is also called a tetrahedron.
One square and four triangles.
Yes, they do.
In a rectangle, the diagonals divide it into four triangles. Each diagonal connects two opposite corners, creating two triangles for each diagonal. Therefore, by drawing both diagonals, you can form a total of four distinct triangles within the rectangle.
To create five triangles from the four triangles formed by nine matchsticks, you can move two matchsticks from one of the triangles to create a new triangle. Specifically, remove one matchstick from the base of one triangle and one from the apex of an adjacent triangle, and reposition them to form an additional triangle that shares sides with the existing triangles. This reconfiguration will yield a total of five triangles.
When you multiple the area of the small triangle by four it equals the area of the large triangle.
when you multiply the area of the small triangle by four it equals the area of the large triangle.
Easy way to work it out: Break the square down into four equilateral triangles, formed by the border of the square and two diagonal lines intersecting it. Each of those triangles will be right isosceles triangles, with their sides adjacent to the right angle each having a length that is equal to the radius of the circle. Their area then would be the square of that radius, divided by two (a = wh/2). Each of those four triangles then, has an area of 242 / 2. There are four of those triangles forming the square, so it's area would be 2 * (24cm)2, or 1152 cm2.
1. you misspelled parallelogram( no E) 2.Either a square or rectange.
A shape formed by four triangles would have to be a tetrahedron. But I believe that a tetrahedron can have at most three right angled triangles. One with four of them is, I think, impossible.
Yes. 4 Triangles Can Turn Into 8 Triangles.
Four triangles can be arranged in a square. Area of square built upon hypotenuse of right angle is equal to the sum of the area of the squares upon the remaining sides.