Solve measurement problems involving length, area, and perimeter - TACHS Math

1. Use the figure shown. A trapezoid has parallel sides of $14$ inches and $22$ inches, and a height of $9$ inches. A rectangle has the same area as this trapezoid and a width of $6$ inches. What is the length of the rectangle?
   Answer: $27$ inches Trapezoid area $=\tfrac12(14+22)(9)=\tfrac12(36)(9)=162$. Rectangle length $=162/6=27$. (A) uses only $14$ as average. (C) uses the average $18\times... /6$ incorrectly. (D) forgets the $\tfrac12$: $36\times9/6...$.
2. Refer to the figure. A rectangular living-room floor measures $15$ feet by $12$ feet. Carpet is sold in square yards. How many square yards of carpet are needed to exactly cover the floor?
   Answer: $20$ square yards Area $=15\times12=180$ sq ft. $1$ sq yd $=9$ sq ft, so $180/9=20$ sq yd. (A) divides perimeter by something. (C) divides by $3$ not $9$. (D) leaves in square feet.
3. Refer to the figure. The hexagonal figure is made by attaching a square to an equilateral triangle along one side. If each side of the square is $8$ cm, what is the perimeter of the combined figure?
   Answer: $40$ cm The shared side is internal. Perimeter uses 3 sides of square $+$ 2 sides of triangle $=3(8)+2(8)=40$ cm. (A) only the square. (C) doesn't remove the shared sides. (D) adds both shapes' full perimeters minus one side.
4. Use the figure to answer the question. A parallelogram has a base of $16$ cm and a corresponding height of $7$ cm. A triangle has the same area as this parallelogram and a base of $14$ cm. What is the height of the triangle?
   Answer: $16$ cm Parallelogram area $=16\times7=112$. Triangle: $112=\tfrac12(14)h\Rightarrow h=16$. (A) forgets factor of $2$ in reverse. (B) equals the base. (D) doubles unnecessarily.
5. Refer to the figure. Rectangle $ABCD$ has a perimeter of $34\text{ m}$. Two opposite sides are each $11\text{ m}$ long, and the other two opposite sides are each $x\text{ m}$ long. What is the value of $x$?
   Answer: $6\text{ m}$ The perimeter of a rectangle is $2(\text{length}+\text{width})$. Here, $2(11+x)=34$, so $11+x=17$ and $x=6\text{ m}$. Distractors: B. $5\text{ m}$ comes from subtracting incorrectly: $34-11=23\rightarrow x=5\text{ m}$. C. $12\text{ m}$ mistakes perimeter for the missing side by doubling the given length. D. $13\text{ m}$ results from dividing the perimeter by 2 and then adding the length instead of subtracting.
6. Refer to the figure. A rectangular poster is $2$ feet $6$ inches wide and $3$ feet $4$ inches tall. A decorative ribbon will be placed along the entire border of the poster. How many inches of ribbon are needed?
   Answer: $140$ inches Width $=30$ in, height $=40$ in. Perimeter $=2(30+40)=140$ in. (A) adds $30+40$ only. (B) mixes units wrong. (D) computes $2(30)+2(50)$.
7. Refer to the figure. A right triangle has legs of $9$ cm and $12$ cm. A square is drawn with one side equal to the hypotenuse of this triangle. What is the perimeter of the square?
   Answer: $60$ cm Hypotenuse $=\sqrt{81+144}=\sqrt{225}=15$. Perimeter of square $=4\times15=60$. (A) uses $9+12+21$. (C) uses $4\times21$ (sum of legs). (D) is $15^2$, the area not perimeter.
8. Refer to the figure. A flower bed is composed of a rectangle and a semicircle. The rectangle measures $12\text{ m}$ by $8\text{ m}$, and the semicircle is attached along the $12\text{ m}$ side, using it as the diameter. What is the total area of the flower bed to the nearest whole square meter?
   Answer: $153\text{ m}^2$ Rectangle area $=12\times8=96\text{ m}^2$. Semicircle radius $r=6\text{ m}$, so area $=\tfrac12\pi r^2=\tfrac12\pi(36)=18\pi\approx56.5\text{ m}^2$. Total $\approx96+56.5=152.5\text{ m}^2\approx153\text{ m}^2$. Distractors: A. Rounds the semicircle area too low. B. Calculates using $\pi\approx3$ instead of $3.14$. C. Uses $\pi=3.14$ but mistakenly uses radius 5.
9. The L-shaped figure shown is made up of two rectangles. Use the figure to find the perimeter of the entire L-shape.
   Answer: $52$ cm The L-shape has outer dimensions $16\times10$ with a $6\times4$ rectangle removed from the top-right corner. Sides: $16+10+10+4+6+6=52$ cm. (A) forgets one segment. (C) adds the removed rectangle's perimeter twice. (D) uses $2(16+10)+2(6+... )$ over-counting.
10. Refer to the figure. A square lawn $ABCD$ has side length $18\text{ m}$. A uniform walkway surrounds the lawn on all four sides, forming a larger square $EFGH$ whose side length is $22\text{ m}$. What is the area of just the walkway?
    Answer: $160\text{ m}^2$ Area walkway $=(22)^2-(18)^2=484-324=160\text{ m}^2$. Distractors: B. Uses outer area minus one incorrect inner dimension 17 m (484-289=195) then rounds. C. Adds the two areas instead of subtracting. D. Squares an average side length (20 m) then subtracts inner area incorrectly.
11. Use the figure to solve. A rectangular swimming pool is $25$ meters long and $10$ meters wide. It is surrounded by a tiled deck of uniform width. The total area of the pool and deck together is $594$ square meters. What is the width of the deck?
    Answer: $4$ meters Let $w$ = deck width. $(25+2w)(10+2w)=594$. Expand: $250+70w+4w^2=594$, so $4w^2+70w-344=0$, $2w^2+35w-172=0$. Using quadratic: $w=\frac{-35+\sqrt{1225+1376}}{4}=\frac{-35+\sqrt{2601}}{4}=\frac{-35+51}{4}=4$. (A), (B), (D) don't satisfy equation.
12. Refer to the figure. Rhombus $ABCD$ has diagonals $\overline{AC}=12\text{ cm}$ and $\overline{BD}=10\text{ cm}$. What is the area of the rhombus?
    Answer: $60\text{ cm}^2$ Area of a rhombus $=\tfrac12(d_1\times d_2)=\tfrac12(12\times10)=60\text{ cm}^2$. Distractors: B. Multiplies the diagonals without halving. C. Halves only one diagonal before multiplying ($\tfrac12\times12\times8$). D. Halves both diagonals separately then multiplies.
13. Refer to the figure. Each square in the grid has side length $1\text{ m}$. What is the area of the shaded garden?
    Answer: $22\text{ m}^2$ Counting the shaded unit squares gives 22, so the area is $22\text{ m}^2$. Distractors: A. Counts one fewer column (20 squares). C. Adds an extra 2 squares that are not shaded (24 squares). D. Miscounts by adding 4 nonexistent squares (26 squares).
14. Refer to the figure. A square and an equilateral triangle have the same perimeter. The side of the square is $9$ inches. How long is each side of the triangle?
    Answer: $12$ inches Square perimeter $=36$ in. Triangle side $=36/3=12$ in. (A) equates sides not perimeters. (C) divides by $2.67$ or confuses with half. (D) divides by $2$.
15. Refer to the figure. A rectangular garden measures $24$ feet by $18$ feet. A uniform path of width $3$ feet is built around the outside of the garden. What is the area of the path alone, in square feet?
    Answer: $288$ square feet Outer rectangle: $(24+6)(18+6)=30\times24=720$. Inner garden: $24\times18=432$. Path area $=720-432=288$. (A) results from forgetting to double the width on each dimension ($27\times26-432=270$... close miscount). (C) comes from using width $3$ only once on each side incorrectly. (D) is the perimeter times width ($84\times3/...$) style error without corners.
16. Refer to the figure. Right triangle $XYZ$ has legs $XY=9\text{ cm}$ and $YZ=12\text{ cm}$. What is the perimeter of the triangle?
    Answer: $36\text{ cm}$ Hypotenuse $XZ=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15\text{ cm}$. Perimeter $=9+12+15=36\text{ cm}$. Distractors: A. Adds only the legs and one miscalculated hypotenuse (9+12+9=30). B. Uses $\sqrt{225}\approx15$ correctly but then rounds down one leg. D. Adds an over-rounded hypotenuse of 17 cm.

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