Compute perimeter and area of common plane figures - TACHS Math

1. Refer to the figure. What is the area of the rectangle?
   Answer: $40\text{ cm}^2$ Area of a rectangle is found with $A = \text{length}\times\text{width} = 8\text{ cm}\times5\text{ cm}=40\text{ cm}^2$. B: 26 is the sum of the two sides, not the product. C: 16 comes from subtracting, not multiplying, the sides. D: 13 is half of the perimeter of the shorter sides, not the area.
2. Use the diagram. A regular hexagon is inscribed in a circle of radius $6$ cm. What is the perimeter of the hexagon?
   Answer: $36$ cm In a regular hexagon inscribed in a circle, each side equals the radius: side $= 6$ cm. Perimeter $= 6 \times 6 = 36$ cm. (A) uses 3 sides. (B) uses side 4. (D) uses side 12 (diameter).
3. Refer to the diagram. An equilateral triangle has a perimeter of $30$ cm. What is its area, to the nearest tenth? (Use $\sqrt{3} \approx 1.732$)
   Answer: $43.3$ sq cm Side $= 30/3 = 10$. Height $= (10\sqrt{3})/2 = 5\sqrt{3} \approx 8.66$. Area $= \tfrac{1}{2}(10)(8.66) = 43.3$ sq cm. Or use $A=\tfrac{\sqrt{3}}{4}s^2 = \tfrac{1.732}{4}(100)=43.3$. (A) uses base × height/4. (C) uses half side squared. (D) omits the $\tfrac{1}{2}$.
4. Use the figure. An L-shaped figure is formed by removing a $3$ ft by $4$ ft rectangle from the corner of a $9$ ft by $7$ ft rectangle. What is the perimeter of the L-shape?
   Answer: $32$ ft The perimeter of an L-shape formed by removing a corner rectangle equals the perimeter of the original rectangle: $2(9+7)=32$ ft. The two cut edges replace equal-length portions of the original sides. (A) subtracts the cut. (C) adds cut lengths. (D) adds perimeter of both rectangles.
5. Based on the figure, a kite has diagonals that meet at right angles. One diagonal is divided into segments of $3$ cm and $9$ cm, and the other diagonal has total length $8$ cm. What is the area of the kite?
   Answer: $48$ sq cm Total length of first diagonal $= 3+9=12$. Kite area $= \tfrac{1}{2}d_1 d_2 = \tfrac{1}{2}(12)(8)=48$ sq cm. (A) uses only the $3$-cm segment. (B) uses the $9$-cm segment. (D) omits the $\tfrac{1}{2}$.
6. Refer to the figure. A parallelogram has a base of $14$ in and one slanted side of $10$ in that makes the slant so the height (perpendicular distance between bases) is $6$ in. What is the perimeter and area of the parallelogram?
   Answer: Perimeter $48$ in; Area $84$ sq in Perimeter $= 2(14+10)=48$ in. Area $=$ base $\times$ height $= 14 \times 6 = 84$ sq in. (B) uses slant as height. (C) miscomputes perimeter. (D) uses both wrong.
7. Use the figure. A running track consists of a rectangle with a semicircle on each short end. The rectangle is $100$ m long and $60$ m wide (the short ends, which are the diameters of the semicircles). What is the total distance around the track (perimeter)? (Use $\pi \approx 3.14$)
   Answer: $388.4$ m Perimeter $= 2(\text{length}) + \pi(\text{diameter}) = 2(100) + 3.14(60) = 200 + 188.4 = 388.4$ m. (A) is only the circular part. (B) uses one length plus circle. (C) doubles everything wrongly.
8. Refer to the diagram. A square has side length $10$ cm. Inside the square, four identical quarter-circles are drawn, each centered at a corner of the square with radius $5$ cm. What is the area of the region INSIDE the square but OUTSIDE all four quarter-circles?
   Answer: $21.5$ sq cm Four quarter-circles of radius $5$ combine to one full circle: area $= \pi(5^2) = 78.5$ sq cm. Square area $= 100$. Remaining area $= 100 - 78.5 = 21.5$ sq cm. (B) is $100/2$. (C) is the circle area alone. (D) is the square's full area.
9. Refer to the circle shown. A circle has a circumference of $31.4$ cm. What is its area?
   Answer: $78.5$ sq cm Circumference $= 2\pi r$, so $r = 31.4/(2 \cdot 3.14) = 5$ cm. Area $= \pi r^2 = 3.14(25) = 78.5$ sq cm. (A) uses $r=5$ but multiplies by $\pi/2$. (B) confuses circumference with area. (D) uses $r=10$ (diameter).
10. Refer to the figure. A rectangle measuring $12$ cm by $8$ cm has a semicircle removed from one of its longer sides (the semicircle's diameter lies along the $12$-cm side and equals $8$ cm). What is the area of the remaining shaded region?
    Answer: $70.88$ sq cm Rectangle area: $12 \times 8 = 96$. Semicircle radius: $8/2 = 4$, area: $\tfrac{1}{2}\pi r^2 = \tfrac{1}{2}(3.14)(16) = 25.12$. Shaded area: $96 - 25.12 = 70.88$ sq cm. (B) subtracts a full circle instead of a semicircle. (C) uses radius $8$ incorrectly. (D) adds instead of subtracts a quarter circle.
11. Use the figure. Two circles share the same center. The outer circle has radius $10$ cm and the inner circle has radius $6$ cm. What is the area of the ring (annulus) between them?
    Answer: $200.96$ sq cm Area $= \pi(R^2-r^2)=3.14(100-36)=3.14(64)=200.96$ sq cm. (A) uses $(R-r)^2=16$ with wrong factor. (B) uses $2\pi(R-r)$. (C) uses $(R-r)^2 \pi$.

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