In the rhombus ABCD, the side AB measures 5 cm, and the diagonal AC is 8 cm. Find the area of the rhombus.

Question

Accepted Answer

The diagonals of a rhombus bisect each other at right angles. Given that diagonal AC = 8 cm, half of this diagonal is:

$$
OC = \frac{AC}{2} = \frac{8}{2} = 4 \, 	ext{cm}.
$$

Since the diagonals are perpendicular, applying the Pythagoras theorem in triangle ODC:

$$
OD^2 + OC^2 = CD^2.
$$

Substituting the values:

$$
OD^2 + 4^2 = 5^2,
$$

$$
OD^2 + 16 = 25,
$$

$$
OD^2 = 9 \implies OD = 3 \, 	ext{cm}.
$$

Thus, the full length of diagonal BD is:

$$
BD = 2 	imes OD = 2 	imes 3 = 6 \, 	ext{cm}.
$$

The area of the rhombus is given by:

$$
	ext{Area} = \frac{1}{2} 	imes AC 	imes BD,
$$

$$
	ext{Area} = \frac{1}{2} 	imes 8 	imes 6 = 24 \, 	ext{cm}^2.
$$

Hence, the area of the rhombus is **24 cm²**.

Answer

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