9514 1404 393
Answer:
- area: 51.59 in²; perimeter: 30 in
- area: 54 cm²; perimeter: 32.49 cm
- area: 30 ft²; perimeter: 28 ft
Explanation:
1. The figure can be decomposed into a square and an equilateral triangle. Both have the same side length: 6 in. For an equilateral triangle of side length s, the area is ...
A = (√3)/4·s²
For a square with side length s, the area is ...
A = s²
The total of the two areas is ...
A = ((√3)/4 + 1)s² ≈ 1.433013×(6 in)²
A ≈ 51.59 in²
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The perimeter is the sum of the 5 equal side lengths:
P = 5(6 in)
P = 30 in
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2. The hypotenuse of the isosceles right triangle at the top of the figure tells you the side length of the triangle is (6√2)/√2 = 6 cm. The area of the triangle is half that of the square below it, so the total area is ...
A = (1 +1/2)(6 cm)²
A = 54 cm²
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The perimeter is the equivalent of 4 sides of 6 cm and one of 6√2 cm, so is ...
P = (4 +√2)(6 cm)
P ≈ 32.49 cm
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3. The triangle at the left end of the figure has a hypotenuse of 5 and a height of 3. It is recognizable as a 3-4-5 right triangle, so the bottom leg of it is 4 ft long. That makes the bottom side of the trapezoid have a length of 8+4 = 12 ft.
The area of the trapezoid is ...
A = 1/2(b1 +b2)h
A = (1/2)(12 ft +8 ft)(3 ft)
A = 30 ft²
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The perimeter is the sum of the side lengths. Starting at left and working clockwise, that sum is ...
P = (5 + 8 + 3 + 12) ft
P = 28 ft