2. list 1 example of a gas in a liquid soluti...
2. list 1 example of a gas in a liquid solution.\n3. list 2 examples of a solid in a solid solution.\n4. list 1 example of a liquid in a liquid solution.\n5. list 1 example of a gas in gas solution.\ndate\nperiod\ngeometry chapter 11 packet\nareas of parallelograms and triangles\nn be called a base. the height of a parallelogram is the perpendicular distance betw\n a parallelogram is the product of the base and the height.\nif a parallelogram has an area of a square units, a\nbase of b units, and a height of h units, then a = bh.\nlogram. round to the nearest tenth if necessary.\n16 ft\n10 ft\n24 in.\n60°\n15 ft\nif the product of the base and its corresponding height. like a parall\nle length of an altitude drawn to a given base.\nngle has an area of a square units, a base of b\nand a corresponding height of h units, then a = 1/2bh.
Answer
# Explanation:
## Step1: Identify base and height for first parallelogram
Base $b = 10$ ft, height $h=16$ ft.
## Step2: Calculate area using formula
$A = bh=10\times16 = 160$ square - feet.
## Step3: For second parallelogram, find height
Given base $b = 24$ in. The height $h$ can be found using right - triangle trigonometry. If the side adjacent to the $60^{\circ}$ angle is the base, and we know that $\sin60^{\circ}=\frac{h}{24}$. So $h = 24\times\sin60^{\circ}=24\times\frac{\sqrt{3}}{2}=12\sqrt{3}$ in.
## Step4: Calculate area of second parallelogram
$A = bh=24\times12\sqrt{3}\approx24\times12\times1.732 = 498.8$ square - inches.
## Step5: For triangle, assume base and height
Assume the base and height of the triangle are given such that if we use the area formula $A=\frac{1}{2}bh$. But since the full base and height values for the triangle are not clearly shown in the image (only a side length of 15 ft is given), we cannot calculate its area precisely. If we assume it's a right - triangle and the 15 ft side is either base or height, we still need the other value.
# Answer:
Area of first parallelogram: 160 square feet. Area of second parallelogram: approximately 498.8 square inches. Area of triangle: cannot be determined with given information.