SHSAT Math Quiz: Perimeter Of Polygons

11 questions · exam conditions

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Perimeter Of PolygonsQuestion 1 of 11

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 $PQRS$ ?

$Question graphic$

21

26

28

32

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SHSAT Math Quiz

SHSAT Math Quiz: Perimeter Of Polygons

Practice Perimeter Of Polygons in SHSAT Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Perimeter Of Polygons, giving you a quick way to practice the rules, question types, and explanations that matter most for SHSAT Math.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 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 $PQRS$ ?

$21$
$26$ (correct answer)
$28$
$32$

Explanation:

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.

Question 2

In the figure shown, all angles that appear to be right angles are right angles. What is the perimeter, in inches, of the polygon?

$54$
$58$ (correct answer)
$62$
$66$

Explanation: 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.

Question 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?

$80$ ft
$92$ ft
$104$ ft (correct answer)
$116$ ft

Explanation: 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.

Question 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?

$44$ cm
$48$ cm
$52$ cm (correct answer)
$56$ cm

Explanation: 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.

Question 5

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?

$59$ cm
$61$ cm (correct answer)
$63$ cm
$65$ cm

Explanation: 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.

Question 6

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?

$60-30\sqrt{3}$ cm (correct answer)
$30\sqrt{3}-60$ cm
$60\sqrt{3}-60$ cm
$30$ cm

Explanation: 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.

Question 7

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 $ABCD$ ?

$54$
$64$ (correct answer)
$68$
$74$

Explanation: 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.

Question 8

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)?

$46$ (correct answer)
$52$
$56$
$62$

Explanation: Hypotenuse

=\sqrt{9^2+12^2}=15

. The shared edge (hypotenuse = 15) is interior and not counted. Perimeter = two legs (9 + 12) + three sides of rectangle excluding shared side (15 + 8 + 8) =

9+12+15+8+8=52

. Wait recount: triangle contributes its two legs = 9+12=21 (hypotenuse shared). Rectangle contributes 3 sides: the far 15-side and two 8-sides = 15+8+8=31. Total = 21+31=52. Answer B. Distractor A forgets one short side; C adds hypotenuse; D adds hypotenuse twice.

Question 9

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 $\pi\approx 3.14159$ .)

$257$ m
$307$ m
$317$ m (correct answer)
$417$ m

Explanation: 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).

Question 10

A rhombus has diagonals of length $16$ cm and $12$ cm. What is the perimeter of the rhombus?

$28$ cm
$40$ cm (correct answer)
$56$ cm
$64$ cm

Explanation: 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

Question 11

A regular polygon has an exterior angle of $24°$ . If each side length is $6$ cm, what is the perimeter of this polygon?

$72$ cm
$84$ cm
$90$ cm (correct answer)
$96$ cm

Explanation: 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.