I1. Solve the following problems. Without solurion is considered wrong. 2 points each. 11. The area of a triangular garden is \( 1530 \mathrm{dm}^{2} \) and the base is 45 dm . Find the measure of its altitude. 12. The \( \triangle A B C \) is an isosceles triangle, \( \overline{A B} \cong \overline{B C} \) and point \( D \) is the median point of \( \overline{A C} \). What is the measure of \( \overline{A D} \) if the perimeter of \( \triangle A B C \) is 330 m and the half measure of \( \overline{B C} \) is 61.5 m ? what is the \( m \angle G \) ? 14. What is the perimeter of an equilateral triangle whose the measure of a side is 4.7 decameters? Express the perimeter in millimeters. 15. Find the area of a triangle whose the base and height are 2.3 dm and 45 cm . Express the area in meters.

Let's solve each problem step by step. ### Problem 11 **Given:** - Area of the triangle \( A = 1530 \, \text{dm}^2 \) - Base \( b = 45 \, \text{dm} \) **Formula for the area of a triangle:** \[ A = \frac{1}{2} \times b \times h \] where \( h \) is the height (altitude). **Substituting the known values:** \[ 1530 = \frac{1}{2} \times 45 \times h \] **Solving for \( h \):** \[ 1530 = 22.5h \] \[ h = \frac{1530}{22.5} = 68 \, \text{dm} \] **Answer:** The measure of its altitude is \( 68 \, \text{dm} \). --- ### Problem 12 **Given:** - Perimeter of triangle \( ABC = 330 \, \text{m} \) - Half of \( BC = 61.5 \, \text{m} \) (thus \( BC = 123 \, \text{m} \)) **Let \( AB = AC = x \). Then:** \[ x + x + 123 = 330 \] \[ 2x + 123 = 330 \] \[ 2x = 330 - 123 = 207 \] \[ x = \frac{207}{2} = 103.5 \, \text{m} \] **Now, to find \( AD \):** Since \( D \) is the median, it divides \( AC \) into two equal parts. The length of \( AD \) can be found using the median formula: \[ AD = \sqrt{\frac{2AB^2 + 2AC^2 - BC^2}{4}} \] Substituting the values: \[ AD = \sqrt{\frac{2(103.5^2) + 2(103.5^2) - 123^2}{4}} \] Calculating: \[ AD = \sqrt{\frac{2(10713.25) + 2(10713.25) - 15129}{4}} = \sqrt{\frac{42853 - 15129}{4}} = \sqrt{\frac{27724}{4}} = \sqrt{6931} \approx 83.3 \, \text{m} \] **Answer:** The measure of \( \overline{AD} \) is approximately \( 83.3 \, \text{m} \). --- ### Problem 14 **Given:** - Side of the equilateral triangle \( s = 4.7 \, \text{decameters} \) **Perimeter of an equilateral triangle:** \[ P = 3s \] Calculating: \[ P = 3 \times 4.7 = 14.1 \, \text{decameters} \] **Convert to millimeters:** \[ 1 \, \text{decameter} = 1000 \, \text{millimeters} \] \[ P = 14.1 \times 1000 = 14100 \, \text{mm} \] **Answer:** The perimeter is \( 14100 \, \text{mm} \). --- ### Problem 15 **Given:** - Base \( b = 2.3 \, \text{dm} \) - Height \( h = 45 \, \text{cm} = 4.5 \, \text{dm} \) **Area of the triangle:** \[ A = \frac{1}{2} \times b \times h \] Calculating: \[ A = \frac{1}{2} \times 2.3 \times 4.5 = \frac{1}{2} \times 10.35 = 5.175 \, \text{dm}^2 \] **Convert to square meters:** \[ 1 \, \text{dm}^2 = 0.01 \, \text{m}^2 \] \[ A = 5.175 \times 0.01 = 0.05175 \, \text{m}^2 \] **Answer:** The area of the triangle is \( 0.05175 \, \text{m}^2 \).