- A triangular building is bounded by three streets. The building measures approximately 93 feet on the first street, 195 feet on the second street, and 180 feet on the third street. Approximate the ground area K covered by the building. \( \mathrm{K} \approx \square \) square feet (Round to two decimal places as needed.)

To approximate the area \( K \) of a triangular building with sides measuring 93 feet, 195 feet, and 180 feet, we can use Heron's formula. Heron's formula states that the area of a triangle with sides \( a \), \( b \), and \( c \) is given by: \[ K = \sqrt{s(s-a)(s-b)(s-c)} \] where \( s \) is the semi-perimeter of the triangle, calculated as: \[ s = \frac{a + b + c}{2} \] In this case, let: - \( a = 93 \) feet - \( b = 195 \) feet - \( c = 180 \) feet First, we calculate the semi-perimeter \( s \): \[ s = \frac{93 + 195 + 180}{2} = \frac{468}{2} = 234 \text{ feet} \] Next, we can calculate the area \( K \) using Heron's formula: \[ K = \sqrt{s(s-a)(s-b)(s-c)} \] Calculating each term: \[ s - a = 234 - 93 = 141 \] \[ s - b = 234 - 195 = 39 \] \[ s - c = 234 - 180 = 54 \] Now substituting these values into the formula: \[ K = \sqrt{234 \times 141 \times 39 \times 54} \] Calculating the product inside the square root: \[ 234 \times 141 = 33054 \] \[ 39 \times 54 = 2106 \] \[ 33054 \times 2106 = 69604964 \] Now, taking the square root: \[ K = \sqrt{69604964} \approx 8356.56 \] Thus, the approximate ground area \( K \) covered by the building is: \[ \mathrm{K} \approx 8356.56 \text{ square feet} \] So, rounding to two decimal places, we have: \[ \mathrm{K} \approx 8356.56 \text{ square feet} \]