What is the area of a quadrilateral in which angle 109 degrees is between sides 0.38cm and 0.69cm when adjacent angle 123 degrees is between sides 0.38cm and 0.42cm?

Answer 1

Here's how I solve it using the Sine and Cosine rules, and area of a triangle based on sine of angle between two given lengths:

Let the quadrilateral be ABCD with ADC = 109°, DCB = 128°, AD = 0.69cm, DC = 0.38cm, CB = 0.42cm

Draw in diagonal AC. Area quadrilateral = area ACD + area ABC.

Length AC can be found from the cosine rule on triangle ADC:

AC = √(0.69² + 0.38² - 2 × 0.69 × 0.38 × cos 109°) cm ≈ 0.89 cm

Angle ACD can be found using the sine rule:

ACD = arc sin(0.69/0.89 × sin109°) ≈ 47.18°
→ BCD = 128 - ACD ≈ 80.82°

→ area quadrilateral ≈ ½ × 0.69 cm × 0.38 cm × sin 109° + ½ × 0.42 cm × 0.89 cm × sin 80.82°
≈ 0.308 cm²

Another Answer: Using triangulation and trigonometry the area of the 4 sided quadrilateral works out as 0.305 square cm rounded to three decimal places.