Question

applying it

• use what you learned to solve these problems.

1. what is the area of the polygon in square units? show your work.
2. to find the area of a trapezoid, badru decomposes it into two triangles as shown. he says the area of the trapezoid is two - times the area of the top triangle. is badru correct? why or why not?
3. amber meets with mr. aba for speech therapy. the polygon represents the top of mr. abas desk. what is the area of the top of the desk? show your work.

Explanation:

Step 1: Analyze the first polygon problem

Since the image of the first polygon is not clear - enough to determine its dimensions precisely from the provided picture, we will focus on the trapezoid problem (second problem) for a general - case example of polygon area calculation. The formula for the area of a trapezoid is $A=\frac{(b_1 + b_2)h}{2}$, where $b_1$ and $b_2$ are the lengths of the parallel sides and $h$ is the height.

Step 2: Analyze Badru's claim

For the trapezoid with parallel sides $b_1 = 12$ in and $b_2=14$ in and height $h = 6$ in. If we decompose the trapezoid into two non - congruent triangles. Let's assume the top triangle has base $b_1 = 12$ in and height $h = 6$ in. The area of the top triangle $A_1=\frac{1}{2}\times b_1\times h=\frac{1}{2}\times12\times6 = 36$ square inches. The area of the trapezoid $A=\frac{(b_1 + b_2)h}{2}=\frac{(12 + 14)\times6}{2}=\frac{26\times6}{2}=78$ square inches.

Step 3: Evaluate Badru's claim

Badru says the area of the trapezoid is two times the area of the top triangle. Since $2\times A_1=2\times36 = 72
eq78$, Badru is not correct. The two triangles formed by decomposing the trapezoid are not equal in area, and the trapezoid's area formula is based on the sum of the parallel sides, not just the top - side length.

For the third problem (the desk - shaped polygon), we can decompose the polygon into two rectangles. One rectangle has dimensions $50$ cm by $120$ cm and the other rectangle has dimensions (assuming the overall shape and dimensions can be decomposed this way) such that we can find its area and sum the areas of the two rectangles. Let's assume the second rectangle also has a relevant side length of $50$ cm. The area of the first rectangle $A_1=l_1\times w_1=120\times50 = 6000$ square cm. If the second rectangle has appropriate dimensions to complete the polygon (for example, if its length is also $50$ cm and width is some value that can be determined from the figure), we sum the areas of the two rectangles.

Answer:

1. For the trapezoid problem: Badru is not correct. The area of the trapezoid with $b_1 = 12$ in, $b_2 = 14$ in and $h = 6$ in is $A=\frac{(12 + 14)\times6}{2}=78$ square inches, while two times the area of the top - triangle (with base $b_1 = 12$ in and height $h = 6$ in) is $2\times\frac{1}{2}\times12\times6=72$ square inches.
2. For the desk - shaped polygon: Decompose it into two rectangles. If one rectangle has dimensions $120$ cm by $50$ cm with area $A_1=120\times50 = 6000$ square cm. Find the area of the second rectangle (depending on its dimensions from the figure) and sum the two areas. Without clear dimensions from the figure, we can't give a numerical answer, but the method is to decompose into rectangles and sum their areas.