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The length of the shorter side at the right angle measures 2 units in the original shape but only 1.8 units in the rectangle. This means, the real triangles of the original shape overlap in the rectangle. The overlapping area is a parallelogram, the diagonals and sides of which can be computed via the Pythagorean theorem.
A parallelogram with base b and height h can be divided into a trapezoid and a right triangle, and rearranged into a rectangle, as shown in the figure to the left. This means that the area of a parallelogram is the same as that of a rectangle with the same base and height: =.
Firstly it works for arbitrary triangles rather than only for right angled ones and secondly it uses parallelograms rather than squares. For squares on two sides of an arbitrary triangle it yields a parallelogram of equal area over the third side and if the two sides are the legs of a right angle the parallelogram over the third side will be ...
From the first proof, one can see that the sum of the diagonals is equal to the perimeter of the parallelogram formed. Also, we can use vectors 1/2 the length of each side to first determine the area of the quadrilateral, and then to find areas of the four triangles divided by each side of the inner parallelogram.
Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids. The centroid of the shape must lie on this line . Divide the shape into two other rectangles, as shown in fig 3. Find the centroids of these two rectangles by drawing the diagonals. Draw a line joining the centroids.
The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
By this usage, the area of a parallelogram or the volume of a prism or cylinder can be calculated by multiplying its "base" by its height; likewise, the areas of triangles and the volumes of cones and pyramids are fractions of the products of their bases and heights. Some figures have two parallel bases (such as trapezoids and frustums), both ...
The perimeter of the medial triangle equals the semiperimeter of the original triangle, and the area is one quarter of the area of the original triangle. This can be proven by the midpoint theorem of triangles and Heron's formula. The orthocenter of the medial triangle coincides with the circumcenter of the original triangle.