Perimeter of Polygons - SHSAT Math

1. Refer to the coordinate plane shown. Quadrilateral $PQRS$ has vertices $P(-3,-2)$, $Q(5,-2)$, $R(8,2)$, $S(0,2)$. What is the perimeter of $PQRS$?
   Answer: $26$ $PQ=|5-(-3)|=8$; $QR=\sqrt{(8-5)^2+(2-(-2))^2}=\sqrt{9+16}=5$; $RS=|8-0|=8$; $SP=\sqrt{(0-(-3))^2+(2-(-2))^2}=\sqrt{9+16}=5$. Perimeter $=8+5+8+5=26$. Distractor A miscounts horizontal. C forgets Pythagoras (uses horizontal diff only). D doubles diagonals.
2. Refer to the figure. Trapezoid $ABCD$ has $AB\parallel CD$, with $AB=14$, $CD=30$, and legs $AD=BC$. The height of the trapezoid is 6. What is the perimeter of trapezoid $ABCD$?
   Answer: $64$ Since the trapezoid is isosceles, each leg's horizontal projection is $(30-14)/2=8$. Each leg $=\sqrt{8^2+6^2}=\sqrt{100}=10$. Perimeter $=14+30+10+10=64$. Distractor A uses leg 5; C uses leg 12; D uses leg 15.
3. Refer to the figure. A rectangular garden measuring 24 ft by 16 ft is surrounded on all four sides by a uniform walkway 3 ft wide. What is the perimeter of the outer boundary of the walkway?
   Answer: $104$ ft The outer rectangle has dimensions $(24+6)\times(16+6)=30\times 22$. Perimeter $=2(30+22)=104$ ft. Distractor A is the garden's perimeter. B adds only 3 ft once per dimension. D adds 6 ft twice.
4. A trapezoid has parallel sides of length $14$ cm and $22$ cm. The non-parallel sides each make a $60°$ angle with the longer base. If the height of the trapezoid is $4\sqrt{3}$ cm, what is the perimeter?
   Answer: $52$ cm The height is $4\sqrt{3}$ cm and each non-parallel side makes a $60°$ angle with the base. Using trigonometry: $\sin(60°) = \frac{\text{height}}{\text{side length}} = \frac{4\sqrt{3}}{\text{side length}}$. Since $\sin(60°) = \frac{\sqrt{3}}{2}$, we have $\frac{\sqrt{3}}{2} = \frac{4\sqrt{3}}{\text{side length}}$, so side length = $\frac{4\sqrt{3} \times 2}{\sqrt{3}} = 8$ cm. Perimeter = $14 + 22 + 8 + 8 = 52$ cm. Choice A (44) omits one side. Choice B (48) uses incorrect side calculation. Choice D (56) adds an extra 4 cm error.
5. In the figure shown, all angles that appear to be right angles are right angles. What is the perimeter, in inches, of the polygon?
   Answer: $58$ For a rectilinear polygon, the sum of all horizontal segments on the top equals the sum on the bottom; similarly for vertical segments. The bounding box is 18 in wide and 11 in tall, so the perimeter equals $2(18+11)=58$ in. Distractor A forgets one jog. C adds an interior edge twice. D double-counts a segment.
6. A regular polygon has an exterior angle of $24°$. If each side length is $6$ cm, what is the perimeter of this polygon?
   Answer: $90$ cm The sum of exterior angles of any polygon is $360°$. If each exterior angle is $24°$, then the number of sides = $\frac{360°}{24°} = 15$. With each side length = 6 cm, perimeter = $15 \times 6 = 90$ cm. Choice A (72) uses 12 sides instead of 15. Choice B (84) uses 14 sides. Choice D (96) uses 16 sides.
7. Refer to the figure. A track consists of a rectangle with two semicircular ends. The straight portions are each 80 m long, and the width of the rectangle (equal to the diameter of each semicircle) is 50 m. What is the perimeter (total length around) of the track, to the nearest meter? (Use $\pi\approx 3.14159$.)
   Answer: $317$ m The two semicircles combine to form a full circle of diameter 50, circumference $50\pi\approx 157.08$ m. Add the two straight sides: $2(80)=160$. Total $\approx 157.08+160=317.08\approx 317$ m. Distractor A forgets one straight; B uses radius 50 in circumference formula incorrectly; D includes the diameter segments (which aren't part of the track).
8. A rhombus has diagonals of length $16$ cm and $12$ cm. What is the perimeter of the rhombus?
   Answer: $40$ cm In a rhombus, the diagonals bisect each other at right angles. Half-diagonals have lengths 8 cm and 6 cm. Using the Pythagorean theorem, each side length = $\sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ cm. Perimeter = $4 \times 10 = 40$ cm. Choice A (28) would result from adding the half-diagonals incorrectly. Choice C (56) might come from $2(16 + 12)$. Choice D (64) might come from $4 \times 16$.
9. In the figure shown, a right triangle with legs 9 and 12 shares its hypotenuse with one side of a rectangle. The opposite side of the rectangle has length equal to the hypotenuse, and the rectangle's other two sides have length 8. What is the perimeter of the combined figure (triangle plus rectangle)?
   Answer: $52$ The hypotenuse is $\sqrt{9^2 + 12^2} = \sqrt{225} = 15$. The hypotenuse is shared by the triangle and the rectangle, so it lies inside the combined figure and is not part of the perimeter. Walk around the outside: the two legs of the triangle, $9$ and $12$, then the three remaining sides of the rectangle, $8$, $15$ and $8$. The perimeter is $9 + 12 + 8 + 15 + 8 = 52$. The value $46$ is the perimeter of the rectangle alone, $2(15 + 8)$, which leaves out the triangle's legs. Adding the shared hypotenuse as well would give $67$; any total above $52$ counts some interior segment or repeats a side.
10. A regular hexagon is inscribed in a circle of radius 10 cm, and an equilateral triangle is inscribed in the same circle. Using the figure shown, what is the positive difference between the perimeter of the hexagon and the perimeter of the triangle?
    Answer: $60-30\sqrt{3}$ cm A regular hexagon inscribed in a circle of radius $r$ has side length $r$, so perimeter $=6(10)=60$ cm. An equilateral triangle inscribed in a circle of radius $r$ has side length $r\sqrt{3}$, so perimeter $=3(10\sqrt{3})=30\sqrt{3}$ cm. Since $30\sqrt{3}\approx 51.96 < 60$, the hexagon has the larger perimeter. The positive difference is $60-30\sqrt{3}$ cm. Distractor B reverses the order. C uses incorrect side length formula. D uses diameter instead of proper calculation.
11. A regular octagon is inscribed in a circle of radius $10$ cm. What is the perimeter of the octagon, rounded to the nearest whole number?
    Answer: $61$ cm For a regular octagon inscribed in a circle of radius $r$, each side length equals $2r\sin(\frac{\pi}{8})$. The central angle for each side is $\frac{360°}{8} = 45°$, so each side subtends $22.5°$ at the center. Using $\sin(22.5°) = \frac{\sqrt{2-\sqrt{2}}}{2} \approx 0.3827$, each side length $\approx 2(10)(0.3827) = 7.654$ cm. Perimeter $= 8 \times 7.654 \approx 61.2$ cm, which rounds to $61$ cm.

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