Find the area of rhombus whose perimeter is 80m and one of whose diagonal is 24m

The rhombus has a perimeter of 80m.

A rhombus has 4 equal sides, each would be 20m.

One of the diagonals is 24m ... this is bisected by the other diagonal into 12m segs.

The diagonals of a rhombus intersect at right angles. Use the Pythagorean Theorem to find the length of the other diagonal ...

12² + b² = 20²

144 + b² = 400

b² = 256

b = 16

This is of course the length of just one segment, the diagonal is 32m.

Now we can use our area formula with the diagonals.

A = ½d¹d²

A = ½ (24) (32)

A = 384 m²

384 square meters

There is a formula for finding area of a rhombus given diagonal (d) and perimeter (P) : 1/4d * √ (P² - 4d²), which would be 1/4 * 24 * sqrt (80² - 4*24²), which gives 384

You can break the rhombus up into triangles. In a rhombus the diagonals are perpendicular bisectors of each other, so you have 4 right triangles, and each of them have a side length 12 for one side (since its half of the diagonal). In a rhombus, all sides are equal, so one side is a quarter of the perimeter. This gives a hypotenuse of 20. Using Pythagorean theorem (c² = a² + b²) we get 20² = 12² + x². We get from simplifying that that x² = 256, and x = 16. Since we now have the base and height we can solve for the triangle (bh/2), and we get 12*16/2 = 96. Since that triangle was only a quarter of the rhombus, we multiply by 4 to get the answer of 384