Q: Do the diagonals of a rhombus divide it into four triangles of equal areas?

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If you multiply the lengths of the two diagonals, and divide by 2, you get the area of a rhombus. How does this work: Call the diagonals A & B for clarity. Diagonal A will split the rhombus into 2 congruent triangles. Looking at one of these triangles, its base is the diagonal A, and its height is 1/2 of diagonal B. So the area of one of the triangles is (1/2)*base*height = (1/2)*A*(B/2) = A*B/4. The other triangle has the same area, so the two areas together make up the whole rhombus = 2*(A*B/4) = A*B/2.

Because in both cases their diagonals cross at right angles So their areas are: 0.5*product of diagonals

The diagonals (drawn from a point) help in dividing the regular polygon into smaller triangles. The sum of the areas of these smaller triangles help in determining the total area of the polygon.

Select any one of the vertices and draw all the diagonals from that vertex. This will divide the polygon (with n sides) into n-2 triangles. Use the coordinates of the vertices of each triangle to calculate its area, and then add the areas of these triangles together.

Divide the polygon into triangles. Calculate the areas of the triangles and then sum these.

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If you multiply the lengths of the two diagonals, and divide by 2, you get the area of a rhombus. How does this work: Call the diagonals A & B for clarity. Diagonal A will split the rhombus into 2 congruent triangles. Looking at one of these triangles, its base is the diagonal A, and its height is 1/2 of diagonal B. So the area of one of the triangles is (1/2)*base*height = (1/2)*A*(B/2) = A*B/4. The other triangle has the same area, so the two areas together make up the whole rhombus = 2*(A*B/4) = A*B/2.

Because in both cases their diagonals cross at right angles So their areas are: 0.5*product of diagonals

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.

The diagonals (drawn from a point) help in dividing the regular polygon into smaller triangles. The sum of the areas of these smaller triangles help in determining the total area of the polygon.

Select any one of the vertices and draw all the diagonals from that vertex. This will divide the polygon (with n sides) into n-2 triangles. Use the coordinates of the vertices of each triangle to calculate its area, and then add the areas of these triangles together.

In general, you divide up the polygon into triangles, calculate the areas of the triangles and then sum these.

Divide the polygon into triangles. Calculate the areas of the triangles and then sum these.

The only general way is to divide the pentagon into three triangles, calculate the areas of the triangles and add them together.

You would need to divide it into triangles, find the area of each triangle and sum these areas together.

I'll be happy to help you, but in order for me to compare the areas of those triangles, you have to tell me the areas of those triangles.

There is no simple formula. You need to divide it into triangles, calculate the areas of each one and sum the results.

One method is to divide it into regular shapes - rectangles, triangles, etc. - and measure the areas of those shapes.