Question

In the rhombus below, AC = 14 cm and BC = 25 cm. Find the area

Answer 1

The area of rhombus ABCD, with diagonals AC and BC measuring
cm and
cm, is
`175` square centimeters.

In a rhombus, the diagonals bisect each other at right angles, forming four congruent right-angled triangles. The area (A) of a rhombus can be found using the formula:

`\[ A = (1)/(2) * d_1 * d_2 \]`

where
`\(d_1\)` and
`\(d_2\)` are the lengths of the diagonals.

Given
`\(AC = 14 \, \text{cm}\)` and
`\(BC = 25 \, \text{cm}\)`, these represent the diagonals
`\(d_1\)` and
`\(d_2\)` respectively.

`\[ A = (1)/(2) * 14 * 25 \]\[ A = (1)/(2) * 350 \]\[ A = 175 \, \text{cm}^2 \]`

Therefore, the area of the rhombus ABCD is
`\(175 \, \text{cm}^2\)`.

Answer 2

In order to calculate the area of the rhombus, first let's draw the diagonal BD.

This diagonal and the diagonal AC will intersect at point M, which is the midpoint of each diagonal.

Since M is midpoint of AC, we have AM = MC = 7 cm.

Now, let's calculate the length of BM, using the Pythagorean theorem in the right triangle BMC:

`\begin{gathered} BC^2=BM^2+MC^2\\ \\ 25^2=BM^2+7^2\\ \\ BM^2=625-49\\ \\ BM^2=576\\ \\ BM=24 \end{gathered}`

Calculating the area of this triangle, we have:

The triangles BMC, BMA, DMC and DMA are all congruent, so the area of the rhombus is: