Question

1. what is the area of rhombus abcd? round your answer to the nearest hundredth.

97.62 cm²
24.57 cm²
78.00 cm²

Explanation:

Step1: Recall the formula for the area of a rhombus

The area formula for a rhombus is \(A = ab\sin\theta\), where \(a\) and \(b\) are the lengths of two adjacent sides and \(\theta\) is the included angle. In a rhombus, the diagonals bisect the angles. Here, if we consider two - adjacent sides (we can use the side \(AD = 10\ cm\) and the length from a vertex to the intersection of diagonals. But another formula for the area of a rhombus is \(A = 2\times\frac{1}{2}ab\sin\theta\) (sum of the areas of two congruent triangles formed by a diagonal). The formula for the area of a triangle is \(A_{\triangle}=\frac{1}{2}xy\sin\alpha\). If we consider the two triangles formed by diagonal \(BD\), and use the formula for the area of a rhombus \(A = d_1\times d_2\times\frac{1}{2}\) (where \(d_1\) and \(d_2\) are the lengths of the diagonals) is not directly applicable here. Another approach:
We know that the area of a parallelogram (a rhombus is a special parallelogram) is \(A = base\times height\). But we can also use the formula \(A = 2\times\frac{1}{2}\times AC\times AD\times\sin40^{\circ}\) (since the diagonal \(AC = 7.8\ cm\), \(AD = 10\ cm\) and the angle between them is \(40^{\circ}\) in one of the triangles formed by the diagonal \(AC\)).
The formula for the area of a triangle with two sides \(x\) and \(y\) and included angle \(\alpha\) is \(A=\frac{1}{2}xy\sin\alpha\). The area of the rhombus \(ABCD\) (composed of two congruent triangles \(\triangle ABC\) and \(\triangle ADC\)) is \(A = 2\times\frac{1}{2}\times AC\times AD\times\sin40^{\circ}\).

Step2: Calculate the area

Substitute \(AC = 7.8\ cm\), \(AD = 10\ cm\) and \(\sin40^{\circ}\approx0.6428\) into the formula \(A=AC\times AD\times\sin40^{\circ}\).

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Answer:

\(97.62\ cm^{2}\)