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area of compound figures page 2 keep going! find the area of each figur…

Question

area of compound figures
page 2
keep going! find the area of each figure.


25 m
28 yd.
9 yd.
17 yd.
9 yd.
8 m
21 yd.
11 m
11 m
11 m
17 yd.
6 m
11 m


6 cm
6 ft.
15 cm
18 ft.
7 ft.
12 cm
5 ft.
12 ft.
12 cm
5 ft.
7 ft.
think about it! go back to problem 5 at the top of this page. try enclosing the compound figure in a larger rectangle and subtracting the areas of the missing pieces. do you get the same answer as before? explain why you think that happens.

Explanation:

Problem 5

Step1: Find base of outer trapezoid

Base = $6 + 8 + 11 = 25$ m

Step2: Calculate area of outer trapezoid

Area = $\frac{1}{2} \times (11 + 25) \times 25 = 450$ m²

Step3: Calculate area of inner rectangle

Area = $8 \times 11 = 88$ m²

Step4: Subtract inner from outer area

Area = $450 - 88 = 362$ m²

Problem 6

Step1: Split into rectangle + trapezoid

Rectangle: $17 \times 21$, Trapezoid bases: $17+9=26$, $28$

Step2: Calculate rectangle area

Area = $17 \times 21 = 357$ yd²

Step3: Calculate trapezoid area

Area = $\frac{1}{2} \times (26 + 28) \times 9 = 243$ yd²

Step4: Sum the two areas

Area = $357 + 243 = 600$ yd²

Problem 7

Step1: Split into rectangle + trapezoid

Rectangle: $12 \times 12$, Trapezoid bases: $6$, $12$, height: $15$

Step2: Calculate rectangle area

Area = $12 \times 12 = 144$ cm²

Step3: Calculate trapezoid area

Area = $\frac{1}{2} \times (6 + 12) \times 15 = 135$ cm²

Step4: Sum the two areas

Area = $144 + 135 = 279$ cm²

Problem 8

Step1: Split into trapezoid + 2 rectangles

Trapezoid bases: $12-5=7$, $18$, height: $5$; Rect1: $7 \times 7$, Rect2: $5 \times 7$

Step2: Calculate trapezoid area

Area = $\frac{1}{2} \times (7 + 18) \times 5 = 62.5$ ft²

Step3: Calculate rectangle areas

Area1 = $7 \times 7 = 49$ ft², Area2 = $5 \times 7 = 35$ ft²

Step4: Sum all areas

Area = $62.5 + 49 + 35 = 146.5$ ft²

Think about It (Problem 5)

Step1: Enclose in large rectangle

Dimensions: $25 \times 25 = 625$ m²

Step2: Calculate missing areas

2 triangles + inner rectangle:
Tri1: $\frac{1}{2} \times 14 \times 25 = 175$ m², Tri2: $\frac{1}{2} \times 14 \times 25 = 175$ m², Inner: $88$ m²
Total missing: $175+175+88=438$ m²

Step3: Subtract missing from large rectangle

Area = $625 - 438 = 362$ m²

Step4: Explain equivalence

Both methods calculate the same enclosed space, just using addition of parts vs subtraction of missing parts from a whole.

Answer:

  1. Problem 5: $\boldsymbol{362}$ square meters
  2. Problem 6: $\boldsymbol{600}$ square yards
  3. Problem 7: $\boldsymbol{279}$ square centimeters
  4. Problem 8: $\boldsymbol{146.5}$ square feet
  5. Think about It: Yes, the answer is the same (362 m²). Both methods measure the exact same shaded region—one adds the areas of the two outer sections, the other subtracts the unshaded regions from a larger enclosing shape, so they yield identical results.