In \( \Delta \mathrm{FGH}, g=920 \mathrm{~cm}, h=950 \mathrm{~cm} \) and \( \angle \mathrm{F}=110^{\circ} \). Find the area of \( \Delta \mathrm{FGH} \), to the nearest square centimeter.

To find the area of \( \Delta \mathrm{FGH} \) with sides \( g = 920 \) cm, \( h = 950 \) cm, and angle \( \angle \mathrm{F} = 110^\circ \), we can use the formula for the area of a triangle when two sides and the included angle are known: \[ \text{Area} = \frac{1}{2} \times g \times h \times \sin(\angle \mathrm{F}) \] Plugging in the given values: \[ \text{Area} = \frac{1}{2} \times 920 \, \text{cm} \times 950 \, \text{cm} \times \sin(110^\circ) \] First, calculate the sine of \( 110^\circ \): \[ \sin(110^\circ) \approx 0.9396926 \] Now, substitute this value back into the area formula: \[ \text{Area} = \frac{1}{2} \times 920 \times 950 \times 0.9396926 \] \[ \text{Area} = 437000 \times 0.9396926 \approx 410646 \, \text{cm}^2 \] **Therefore, the area of \( \Delta \mathrm{FGH} \) is approximately \( 410,646 \) square centimeters.** Answer: The area of ΔFGH is approximately 410 646 cm².