Question

1. what is the area of $delta aob$?

in the quadrilateral figure:
$ad = 10$ cm, $oc = 7.8$ cm, $angle ocd = 40^circ$
angles marked: $a_1^circ$, $a_2^circ$, $b_1^circ$, $b_2^circ$, $c_1^circ$, $d_1^circ$, $d_2^circ$

Explanation:

Step1: Identify congruent triangles

In parallelogram \(ABCD\), \(\triangle AOB \cong \triangle COD\), so \(AO = 7.8\ \text{cm}\), \(AB = CD = 10\ \text{cm}\), and \(\angle OAB = c_1 = 40^\circ\).

Step2: Calculate height of \(\triangle AOB\)

The height \(h\) from \(B\) to \(AO\) is \(AB \cdot \sin(40^\circ)\).
\(h = 10 \cdot \sin(40^\circ) \approx 10 \cdot 0.6428 = 6.428\ \text{cm}\)

Step3: Compute area of \(\triangle AOB\)

Area formula: \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times 7.8 \times 6.428\)

Answer:

\(\approx 25.07\ \text{square centimeters}\)