Gina made tables of values to solve a system of equations.
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<h3>62.5 cm^2</h3>
The figure is a trapezoid, albeit not possessing the same lengths for the left and right sides. If I'm not mistaken, it has another name or term. But I forgot.
Anyway, to solve the problem, let us chop-chop or break the figure into two parts or two polygons that we are familiar with - a rectangle and a triangle. See the attached image for a clearer presentation.
After breaking the original polygon into two parts as portrayed by the red broken line, we are also going to divide the equivalent value of each side.
Remember, a rectangle has two pairs of common sides. Since the upper side of the polygon has a length of 10 cm, then, the length of the lower side (the side at the left side of the broken line) is 10, too. The same goes for the length of the left side which is 5 cm, and therefore the length of the broken line is 5 cm. Please refer to the attached image, particularly the figures in green.
From the standpoint of the triangle, based on thee figure, it has a height of 5 cm and a base of 5 cm (15 cm- 10 cm). Did you get it? If yes, let's continue.
Solving Time! :D
1. Solve for the area of the rectangle.
Arectangle=L×W=5 cm×10 cm=50 cm²
2. Solve for the area of the triangle.
Atriangle=2b×h=25 cm×5 cm=225 cm²=12.5 cm²
3. Combine (add) the two areas to find the area of the original figure.
Acomposite figure=Arectangle+Atriangle=50 cm²+12.5 cm2=62.5 cm
If there's a Latex Error, Please click the second image attached below :)