Solved Examples: Polygons and Curves



Samuel Dominic Chukwuemeka (SamDom For Peace) Formulas:
(1.) Geometry Formulas
(2.) Mensuration Formulas

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
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(1.) A student made several mathematical statements/sentences or asked questions regarding polygons and curves.
As a teacher, how would you respond?

(a.) All squares are rectangles but not all rectangles are squares.

(b.) All figures with four congruent sides are squares.

(c.) A regular polygon is a polygon with congruent sides or a polygon with congruent angles.

(d.) Number 1d

If ABCD is a valid name for the accompanying​ rectangle, then is ACBD also​ valid.

(e.) A square is just an equiangular rhombus.

(f.) A square is just an equilateral rectangle.

(g.) A rhombus is regular because all of its sides are congruent.

(h.) Can a triangle can have two right angles?

(i.) I want to make tiles in the shape of a convex polygon with all interior angles acute.
I am wondering if there are any such polygons besides triangles.

(j.) Is there a convex decagon with exactly four right angles?

(k.) The symbol: "SOS" is a good choice for the international distress​ symbol.
What geometric property makes it so?


(a.) The statement is correct.
A rectangle is a parallelogram with four right angles.
A square has four right angles and four equal​ sides, so a square is a rectangle.
A rectangle does not have to have four equal sides.

(b.) The statement is incorrect.
A figure must contain four congruent sides and four right angles to be a square.

(c.) A regular polygon is a polygon in which all the angles are congruent and all the sides are congruent.
Both conditions are necessary to describe a regular polygon.
In general neither condition implies the​ other, and hence neither is sufficient to describe a regular polygon.

(d.) When naming​ polygons, the order of the letters represent consecutive​ vertices, either clockwise or counterclockwise.
It is agreed that ABCD is a valid​ name, but ACBD is not valid as A and C are not consecutive vertices.

(e.) The statement is correct.
A rhombus with four congruent angles is a square.

(f.) The statement is correct.
A rectangle with sides the same length has four right​ angles. This satisfies the definition of a square.

(g.) To be​ regular, all sides must be​ congruent and all angles must be congruent.
All angles are not congruent unless the rhombus is a square.

(h.) A triangle cannot have two right angles because the sum of the interior angles of the triangle would be greater than 180°.

(i.) The only possible convex polygons with all acute interior angles are triangles.

(j.) The four right angles would have exterior angles that add up to 360°.
If such a decagon​ existed, the sum of the measures of its exterior angles would have to be greater than 360°​, so such a decagon does not exist.

(k.) "SOS" has point​ symmetry, so it looks the same when rotated by 180°.
(2.) For each of the following questions, name the required properties.
If this is not​ possible, state the reason(s).

(a.) Two properties that hold true for all parallelograms but not for all squares.

(b.) Two properties that hold true for all rectangles but not for all rhombuses.

(c.) Two properties that hold true for all squares but not for all isosceles trapezoids.


(a.) This is not possible.
All squares are parallelograms.

(b.) Two properties that hold true for all rectangles but not for all rhombuses:
(i.) All angles must be right angles.
(ii.) All diagonals are the same length.

(c.) Two properties that hold true for all squares but not for all isosceles trapezoids are:
(i.) All sides are the same length.
(ii.) All angles must be right angles.
(3.)

(4.) The relationship between the sum of the measures of the interior angles of a polygon, S, and the number of sides of the polygon, n, is given by the equation S = 180(n − 2).
The sum of the interior angle measures of a certain polygon is 1,440°.
How many sides does this polygon have?

$ A.\;\; 6 \\[3ex] B.\;\; 8 \\[3ex] C.\;\; 10 \\[3ex] D.\;\; 12 \\[3ex] E.\;\; 16 \\[3ex] $

$ S = 180(n - 2) \\[3ex] 1440 = 180(n - 2) \\[3ex] 180(n - 2) = 1440 \\[3ex] n - 2 = \dfrac{1440}{180} \\[5ex] n - 2 = 8 \\[3ex] n = 8 + 2 \\[3ex] n = 10 \\[3ex] $ The polygon has 10 sides.
(5.) Given these quadrilaterals, answer the following questions.

Number 5

Which figure(s) have:
(a.) No parallel sides
(b.) No congruent sides
(c.) No right angles


(a.) Figure E: Kite

(b.) Figure F: Trapezoid

(c.) Figures A, C, E, F
Rhombus, Parallelogram, Kite, Trapezoid
(6.) ACT How long, in centimeters, is 1 side of a square whose perimeter is equal to the circumference of a circle with a radius of 2 centimeters?

$ F.\;\; \pi \\[3ex] G.\;\; \dfrac{\pi}{2} \\[5ex] H.\;\; 4\pi \\[3ex] J.\;\; 16\pi \\[3ex] K.\;\; 4 \\[3ex] $

$ \underline{Square} \\[3ex] Let\;\;side = x \\[3ex] Perimeter = 4 * side \\[3ex] P = 4x \\[3ex] \underline{Circle} \\[3ex] radius = 2\;cm \\[3ex] Circumference = 2 * \pi * radius \\[3ex] C = 2 * \pi * 2 \\[3ex] C = 4\pi \\[3ex] \underline{Question} \\[3ex] P = C \\[3ex] 4x = 4\pi \\[3ex] x = \dfrac{4\pi}{4} \\[5ex] x = \pi \\[3ex] $ The length of a side of the square is π centimeters
(7.) CSEC The diagram below, not drawn to scale, shows the plan of a swimming pool in the shape of a rectangle and two semicircles.
The rectangle has dimensions 8 metres by 3.5 metres.
$ \left[Use\:\: \pi = \dfrac{22}{7}\right] \\[3ex] $ Number 7

(i) State the length of the diameter of the semicircle, $AFE$
(ii) Calculate the perimeter of the swimming pool.


The diameter of the semicircle is the width of the rectangle

$ (i) \\[3ex] diameter\:\:of\:\:semicircle = 3.5\:m \\[3ex] $ The perimeter of the swimming pool is the sum of:
Circumference of the semicircle $AFE$ and
Length of the rectangle $ED$ and
Circumference of the semicircle $DCB$ and
Length of the rectangle $BA$

$ (ii) \\[3ex] Circumference\:\:of\:\:semicircle\:\:AFE = \dfrac{\pi d}{2} \\[5ex] Circumference\:\:of\:\:semicircle\:\:DCB = \dfrac{\pi d}{2} \\[5ex] Length\:\:of\:\:rectangle\:\:ED = 8\:m \\[3ex] Length\:\:of\:\:rectangle\:\:BA = 8\:m \\[3ex] Perimeter\:\:of\:\:the\:\:swimming\:\:pool \\[3ex] = \dfrac{\pi d}{2} + \dfrac{\pi d}{2} + 8 + 8 \\[5ex] = \pi d + 16 \\[3ex] = \dfrac{22}{7} * 3.5 + 16 \\[5ex] = \dfrac{22}{7} * \dfrac{7}{2} + 16 \\[5ex] = 11 + 16 \\[3ex] = 27\:m $
(8.)


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(15.) Given these quadrilaterals, answer the following questions.

Number 15

Which figure(s) have:
(a.) No parallel sides
(b.) No congruent sides
(c.) No right angles


(a.) Figure C: Kite

(b.) Figure E: Trapezoid

(c.) Figures B, C, D, E, F
Rhombus, Kite, Parallelogram, Trapezoid, Isosceles Trapezoid
(16.)

(17.)

(18.)


(19.)


(20.) Explain whether the following patterns have rotational symmetry (turn​ symmetry)
If it has turn symmetry, identify the turn center.

(a.) Number 20a

(b.) Number 20b

(c.) Number 20c

(d.) Number 20d

(e.) Number 20e

(f.) Number 20f

(g.) Number 20g

(h.) Number 20h


(a.) The figure does not have rotational symmetry at any point.

(b.) The figure does not have rotational symmetry at any point.

(c.) The figure has rotational symmetries of 90°, 180°, and 270° about the center of the large square.

(d.) The figure does not have rotational symmetry at any point.

(e.) The figure has rotational symmetry of 180° about the center of the large square.

(f.) The figure does not have rotational symmetry at any point.

(g.) The figure has rotational symmetries of 90°, 180°, and 270° about the center of the large square.

(h.) The figure has rotational symmetry of 180° about the center of the large square.





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(21.)

(22.) For the following figures: identify and sketch the lines of symmetry if any exist.

(a.) Number 22a

(b.) Number 22b

(c.) Number 22c

(d.) Number 22d

(e.) Number 22e

(f.) Number 22f

(g.) Number 22g

(h.) Number 22h

(i.) Number 22i

(j.) Number 22j

(k.) Number 22k

(l.) Number 22l


(a.) Number 22a
There​ is 0 ​line of symmetry.

(b.) Number 22b
There​ are 2 lines of symmetry.

(c.) Number 22c
There​ are 2 lines of symmetry.

(d.) Number 22d
There​ is 0 ​line of symmetry.

(e.) Number 22e
There​ is 0 ​line of symmetry.

(f.) Number 22f
There​ is 1 ​line of symmetry.

(g.) Number 22g
There​ are 3 lines of symmetry (because it is an equilateral triangle).

(h.) Number 22h
There​ is 0 ​line of symmetry.

(i.) Number 22i
There​ are 2 lines of symmetry.

(j.) Number 22j
There​ is 1 ​line of symmetry.

(k.) Number 22k
There​ are 2 ​lines of symmetry.

(l.) Number 22l
There​ are 6 ​lines of symmetry.
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(39.) In the figure​ shown, triangle ABC is a right triangle with right angle C and semicircles are drawn with centers at​ E, O, and P.

Number 39

Show that the area of the shaded portion is equal to the area of the right triangle by doing the following tasks:
(a.) Determine the area of triangle ABC
(b.) In​ general, how can the area of the shaded regions be​ found?
(c.) Determine the area of semicircle AB
(d.) Determine the area of the semicircle about E outside the triangle
(e.) Determine the area of the semicircle about O
(f.) Determine the area of the semicircle about P
(g.) Determine the area of the shaded region using the previous results from (a.) thorugh (f.)
(h.) AB needs to be rewritten in terms of AC and BC to allow for the previous expression to be simplified.
How can this be​ done?
(i.) Substitute the result from the previous step for AB.
How can it be shown that the area of the shaded region is equal to the area of the​ triangle?


Triangle ABC is a right triangle where:
$\overline{BC}$ is the base and
$\overline{AC}$ is the perpendicular height
A = area
d = diameter
r = radius

$ (a.) \\[3ex] Area\;\;of\;\;\triangle ABC = \dfrac{1}{2} \cdot base \cdot \perp height \\[3ex] = \dfrac{1}{2} \cdot \overline{BC} \cdot \overline{AC} \\[5ex] = \dfrac{\overline{BC} \cdot \overline{AC}}{2} \\[5ex] $ (b.) In​ general, how can the area of the shaded regions be​ found?
Determine the area of the semicircles centered about O and P and subtract the area of the semicircle about E that is outside the triangle.

(c.) Area of semicircle AB

$ d = \overline{AB} \\[3ex] A = \dfrac{\pi d^2}{8} \\[5ex] A = \dfrac{\pi \overline{AB}^2}{8} \\[5ex] $ (d.) Area of the semicircle about E outside the triangle =
Area of semicircle AB
−
Area of triangle ABC

$ A = \dfrac{\pi \overline{AB}^2}{8} - \dfrac{1}{2} \cdot \overline{BC} \cdot \overline{AC} \\[5ex] A = \dfrac{\pi \overline{AB}^2}{8} - \dfrac{\overline{BC} \cdot \overline{AC}}{2} \\[5ex] A = \dfrac{\pi \overline{AB}^2 - 4 \cdot \overline{BC} \cdot \overline{AC}}{8} \\[5ex] $ (e.) Determine the area of the semicircle about O

$ d = \overline{AC} \\[3ex] A = \dfrac{\pi \overline{AC}^2}{8} \\[5ex] $ (f.) Determine the area of the semicircle about P

$ d = \overline{BC} \\[3ex] A = \dfrac{\pi \overline{BC}^2}{8} \\[5ex] $ (g.) The area of the shaded region using the previous results from (a.) thorugh (f.) =
Area of the semicircle about O
+
Area of the semicircle about P
−
Area of the semicircle about E outside the triangle

$ A = \dfrac{\pi \overline{AC}^2}{8} + \dfrac{\pi \overline{BC}^2}{8} - \dfrac{\pi \overline{AB}^2 - 4 \cdot \overline{BC} \cdot \overline{AC}}{8} \\[5ex] A = \dfrac{\pi \overline{AC}^2 + \pi \overline{BC}^2 - (\pi \overline{AB}^2 - 4 \cdot \overline{BC} \cdot \overline{AC})}{8} \\[5ex] A = \dfrac{\pi \overline{AC}^2 + \pi \overline{BC}^2 - \pi \overline{AB}^2 + 4 \cdot \overline{BC} \cdot \overline{AC}}{8} \\[5ex] $ (h.) AB needs to be rewritten in terms of AC and BC to allow for the previous expression to be simplified. How can this be​ done?
Use the Pythagorean theorem to establish the relationship:

$ \underline{\triangle ABC} \\[3ex] hyp^2 = leg^2 + leg^2 ...Pythagorean\;\;Theorem \\[3ex] \overline{AB}^2 = \overline{AC}^2 + \overline{BC}^2 \\[3ex] $ (i.) Substitute the result from the previous step for AB.
How can it be shown that the area of the shaded region is equal to the area of the​ triangle?

$ A = \dfrac{\pi \overline{AC}^2 + \pi \overline{BC}^2 - \pi \overline{AB}^2 + 4 \cdot \overline{BC} \cdot \overline{AC}}{8} \\[5ex] Substitute\;\;for\;\; \overline{AB}^2 \\[3ex] A = \dfrac{\pi \overline{AC}^2 + \pi \overline{BC}^2 - \pi \overline{AB}^2 + 4 \cdot \overline{BC} \cdot \overline{AC}}{8} \\[5ex] A = \dfrac{\pi \overline{AC}^2 + \pi \overline{BC}^2 - \pi (\overline{AC}^2 + \overline{BC}^2) + 4 \cdot \overline{BC} \cdot \overline{AC}}{8} \\[5ex] A = \dfrac{\pi \overline{AC}^2 + \pi \overline{BC}^2 - \pi \overline{AC}^2 - \pi \overline{BC}^2 + 4 \cdot \overline{BC} \cdot \overline{AC}}{8} \\[5ex] A = \dfrac{4 \cdot \overline{BC} \cdot \overline{AC}}{8} \\[5ex] A = \dfrac{\overline{BC} \cdot \overline{AC}}{2} \\[5ex] = Area\;\;of\;\;\triangle ABC $
(40.)






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