geometry of 2d shapes chapter 12, page 140
TRANSCRIPT
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Geometry of 2D shapesChapter 12, Page 140
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TrianglesRules:1) The sum of the interior angles of a triangle = 180°
(<‘s of a ∆ = 180°)2) The exterior angle of a triangle is equal to the sum of the interior opposite angles
(ext < of a ∆)3) In an isos. triangle, the ANGLES opposite equal SIDES are equal to each other.
In an isos. triangle, the SIDES opposite equal ANGLES are equal to each other.
(isos. ∆)4) In an equilateral triangle, all sides are equal in length.
(equil. ∆)
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Examples:
1)
+ 65° + 45°= 180° (<‘s of a ∆ = 180°) = 180° – 65° – 45° = 70°
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2)
= 73° + 32° (ext < of a ∆) = 105°
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3)
= 58° (isos. ∆) = 7cm (isos. ∆)
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4)
= 5cm (equil. ∆) = 5cm (equil. ∆), and = 60°(equil. ∆)
DO: Exercise 12.2 Page 145 ALLLLL!!!!!
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Exercise 12 .2: Answers + 65°+ 58° = 180° (<‘s of a ∆ = 180°)
= 180° - 65°- 58° = 57°
+ 36° + 73° = 180° (<‘s of a ∆ = 180°)
= 180° – 36° - 73° = 71°
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Exercise 12 .2: Answers + 25°+ 68° = 180° (<‘s of a ∆ = 180°)
= 180° - 25°- 68° = 87°
= 47° + 63° (ext < of a ∆) = 110°
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Exercise 12 .2: Answers + 42°= 125° (ext < of a ∆)
= 125° – 42° = 83°
+ 90° + 35° = 180° (<‘s of a ∆ = 180°)
= 180° - 90° - 35° =55°
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Exercise 12 .2: Answers + 21°= 58° (ext < of a ∆)
= 58° – 21° = 37°
`````````````` + 37° + 21°= 180° (<‘s of a ∆ = 180°) = 180° – 37° – 21° = 122°
+ 2 + 132° = 180° (<‘s of a ∆ = 180°)3 = 180° - 132° = 48° = 16 °
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Exercise 12 .2: Answers + 90° + 71°= 180° (<‘s of a ∆ = 180°)
= 180° – 90° – 71° = 19° + 71° + 19° + 24° = 180° (<‘s of a ∆ = 180°) = 180° - 71° - 19° - 24° = 66°
+ 2 + 2 = 180° (<‘s of a ∆ = 180°)5 = 180° = 36°
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Exercise 12 .2: Answers 2 – 10° + 52° + 36°= 180° (<‘s of a ∆ = 180°)
2 = 180° + 10° – 52° - 36°2 = 102° = 51°
2 + 10° + 68° + 56° = 180° (<‘s of a ∆ = 180°)
2 = 180° - 10° – 68° - 56°2 = 46°= 23°
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Theorem of PythagorasIn a RIGHT angled ∆, the side opposite the 90° angle is called the HYPOTENUSE. This is the longest side of the ∆.
Hypotenuse
Hypotenuse
Hypotenuse
DO: Exercise 12.3 Page 146 #1
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Exercise 12.3: Answers
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Pythagorean Theorem: The square of the hypotenuse is equal to the sum of the squares of the other two sides.h2 = s2 + s2
Example 1:
h2 = s2 + s2 (pythag)h2 = (3)2 + (6)2
h2 = 9 + 36h2 = 45h = h = 6,71cm
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Example 2:h2 = s2 + s2 (pythag)h2 = (3)2 + (4)2
h2 = 9 + 16h2 = 25h = h = 5cm
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Example 3:h2 = s2 + s2 (pythag)(9)2 = (4)2 + s2
81 = 16 + s2
81 - 16 = s2
65 = s2
= 8,06cm
Formula to calculate a side:s2 = h2 - s2
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Example 4:s2 = h2 - s2 (pythag)s2 = (5)2 – (3)2
s2 = 25 - 9s2 = 16s = s = 4cm
Do: Exercise 12.4 Page 149ALL
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Exercise 12.4: Answers1a) h2 = s2 + s2
(pythag)= (4)2 + (3)2
= 16 + 9= 25 h = 5cm
b) h2 = s2 + s2
(pythag)= (7)2 + (24)2
= 49 + 576= 625h = 25cm
c) h2 = s2 + s2
(pythag)= (6)2 + (8)2
= 36 + 64 = 100 h = 10cm
d) h2 = s2 + s2
(pythag)= (40)2 + (9)2
= 1600 + 81= 1681h = 41cm
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Exercise 12.4: Answerse) h2 = s2 + s2
(pythag)= (3)2 + (7)2
= 9 + 49= 58 h = 7,6cm
f) h2 = s2 + s2
(pythag)= (1,4)2 + (2)2
= 1,96 + 4= 5,96h = 2,4cm
2a) s2 = h2 - s2 (pythag)s2 = (50)2 – (40)2
s2 = 2500 - 1600s2 = 900s = 30mm
b) s2 = h2 - s2 (pythag)s2 = (13)2 – (12)2
s2 = 169 - 144s2 = 25s = 5cm
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Exercise 12.4: Answersc) s2 = h2 - s2
(pythag)s2 = (14)2 – (8)2
s2 = 196 - 64s2 = 132s = 11,5cm
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Pythagoras worksheet: Answers 1) 2 = s2 + s2 (pythag) = (4)2 + (3)2
= 16 + 9= 25
= 52) 2 = s2 + s2 (pythag) = (5)2 + (12)2
= 25 + 144= 169
= 13
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Pythagoras worksheet: Answers 3) 2 = s2 + s2 (pythag) = (15)2 + (8)2
= 225 + 64= 289 = 17
4) 2 = h2 - s2 (pythag)= (25)2 – (24)2
= 625 - 576= 49= 7
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Pythagoras worksheet: Answers 5) 2 = h2 - s2 (pythag)
= (17)2 – (8)2
= 289 - 64= 225
= 15
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Pythagoras worksheet: Answers 6) 2 = s2 + s2 (pythag) = (12)2 + (9)2
= 225 = 15
2 = h2 - s2 (pythag)= (25)2 – (15)2
= 400 = 20
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Pythagoras worksheet: Answers 7) AC2 = h2 - s2 (pythag)
= (17)2 – (8)2
= 225 AC = 15
2 = s2 + s2 (pythag) = (15)2 + (20)2
= 625 = 25
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QuadrilateralsPage 149
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DEF: A quadrilateral is any closed shape with four straight sides.
The angles of a quadrilateral add up to 360
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Square
Both pairs of opposite sides parallel All 4 sides = in length All 4 angles = 90 Diagonals bisect each other at 90 Diagonals bisect the angles
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Rhombus
Both pairs of opposite sides parallel All 4 sides = in length Opposite angles = Diagonals bisect each other at 90 Diagonals bisect the angles Diagonal bisect each other
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Rectangle
Both pairs of opposite sides parallel Both pairs of opposite sides equal All 4 angles = 90 Diagonals are equal in length Diagonals bisect each other
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Parallelogram
Both pairs of opposite sides parallel Both pairs of opposite sides equal Both pairs of opposite angles equal Diagonals bisect each other
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Trapezium / Trapezoid
Only one pair of opposite sides parallel
Trapezium Isosceles Trapezium
Only one pair of opposite sides parallel
One pair of opposite sides equal
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Kite
2 pairs of adjacent sides = Diagonals bisect at 90 The longer diagonal bisects the angles The longer diagonal bisects the shorter
diagonal
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Definitions:Quadrilateral: Any closed shape with 4 straight sides
Trapezium: Quadrilateral with one pair of opposite sides equal
Kite: Quadrilateral with two pairs of adjacent sides equal in length
Parallelogram: Quadrilateral with both pairs of opposite sides parallel
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Rectangle: A parallelogram with one interior angle equal to 90°
Rhombus: A parallelogram with one pair of adjacent sides equal in length (therefore all 4
sides are equal in length)
Square: A rectangle with a pair of adjacent sides equal in length
orA rhombus with one angle equal to 90°
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Example 1:
What kind of quadrilateral is this?Parallelogram
Calculate and = 135 (opp. angles of a parm =) = 8cm (opp. sides of a parm =)
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Example 2:
Calculate and QPS = 68° (opp. angles of a rhombus =) = 34 (diagonal bisects angles) = 9cm (sides of rhombus =) = 90 (diag. of rhombus bisect at 90)
What kind of quadrilateral is this?Rhombus
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Example 3:
a) = 90(diag. of rhombus bisect at 90)b) AC = 6cm (diag. of rhombus bisect each other)c) AB2 = 32 + 42 (pythag)
AB2 = 25AB = 5cm
d) 5 + 5 + 5 + 5 = 20cm DO: Ex 12.5 Page 153ALL
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Exercise 12.5: Answers1a) Parallelogram
= 50°(opp <‘s of parm =) = 9cm (opp sides parm =)
b) Square<ASQ = 90° (<‘s of square = 90°) = 45° (diags of a square bisect the <‘s) = 90° (diags of a square bisect at 90°) = 5cm (all sides of square =)
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c) Rhombus = 98°(opp <‘s of a rhombus =) = 6cm (all sides of rhombus =) = 82° (opp <‘s of a rhombus =)
d) Quadrilateral = 360° – 85° – 47° – 90° (<‘s of a quad = 360°) = 138°
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e) Kite = 102° (opp <‘s kite =) = 4cm (adj sides kite =) = 7cm (adj sides kite =)
f) Rectangle2 = 102 + 62 (pythag)
= 136 = = 11.66cm = 10cm (opp sides rectangle =) = 6cm (opp sides rectangle =)
y
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2a) <S = 90°(<‘s rectangle = 90°)SR = 5cm (opp sides rectangle =) PR2= 122 + 52 (pythag) = 169 PR= PR = 13cm
b) PM = 6,5cm (diagonals of rectangle bisect each other)c) <MSR = 90°- 24° (<‘s rectangle = 90°)
= 66°d) SM = MR (diag. of rectangle bisect eachother)
∆ SMR is isos. (2 sides =) <MRS = 66° (isos ∆)<SMR = 180° - 66° - 66° (<‘s of ∆ = 180°)= 48°
5cm
12cm 24°66°
66°5cm
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3a) DG, FG and ED = 8cm (all sides rhombus =) b) i) <EHF = 90°(diags of rhombus bisect at 90°)
ii) <EDH = 32° (diags of rhombus bisect the <‘s)iii) <EFG = 64°(opp <‘s rhombus =)iv) <F1 = 32° (diags of rhombus bisect the <‘s)v) In ∆ EDF:
<E = 180° - 32° - 32° (<‘s of a ∆ = 180°)= 116°
vi) <G = 116 ° (opp angles of rhombus =) <G1 = 58° (diags of rhombus bisect angles)
8cm
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4a) <M1 = 90° (diags of kite bisect at 90°)b) PQ2= 32 + 42 (pythag)
= 25 PQ= PQ= 5cm
c) PS = 5cm (adj. sides kite =)d) MR2 = 92 - 42 (pythag)
= 65 PQ= PR = 8,1cm
e) <Q1 = 180°- 90° - 26° (<‘s of a ∆)= 64°
f) < PSR = < PQR = 101° (opp <‘s kite =)
g) <P1 = 180° - 90°- 37° (<‘s of a ∆) = 53°
<P2 = 53° (long diagonal bisects the <‘s) <QPS= 53° + 53°
= 106°
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SimilarityDefinition:1) All pairs off corresponding
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Exercise 12.12: Answers1a) Parallelogramb) Equal and parallelc) Equal d) Bisect
2. Congruent shapes have identical sides and angles. Similar shapes have the same angles, but the sides are in proportion
3a) Sometimes trueb) Always truec) Always trued) Sometimes true
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4a) = 180° – 52° – 77° (<‘s of a ∆)= 51°
b) = 48° + 56° (ext < of a ∆) = 104°
c) + 33° = 84° (ext < of a ∆) = 84° - 33° = 51°
d) = 180° - 72° - 72° (<‘s of a ∆) = 36°
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e) ++ 72° = 180° (isos ∆ ; <‘s of a ∆)2 = 180° - 72°2 = 108° = 54°
f) = 8 cm (sides of a square =)
g) = 68° (opp <‘s of a parm =) h) + 90° + 90° + 75° = 360° (<‘s of a quad)
+ 225° = 360° = 360° – 225° = 105°
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i) ² + 24² = 25² (pythag)² + 576 = 625² = 625 – 576² = 49 = 7cm
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5) = 180° – 125° - 36°(<‘s of a ∆) = 19° = 180° - 125° - 19° (<‘s of a ∆)y = 36° 1) <A = <P (proven above)2) <B = <Q (given)3) <C = <R (proven above)∆ABC lll ∆PQR (AAA) = =
7 = 455 = 6535 = 630 = 18
90
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6) 15² = 12² + 9²225 = 144 + 81225 = 225 ABD is a right <‘d ∆ (converse of Pythagoras)
7a) Not congruent OR similar
b) ∆ABD CDB (RHS) or (SSS)
78°
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c) <B
= 70°(<‘s of a ∆)∆
ABC lll ∆ XYZ
d) ∆ABD ∆ CDB (RHS)
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Extend your skills: Ex. 12.13 Pg. 1671a)
AC² = 3² + 4² (pythag)
= 9 + 16
= 25 AC=
= 5 ² = 13² - 5² (pythag)
= 169 - 25= 144= = 12
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b) <ABC = 180° - 120° (str line)= 60°
3 = – 10° + 60° (ext. <‘s of a ∆)3 –= 50°2 = 50° = 25°
A
B C60°
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c)
ABCD is a Rhombus (adj. sides =) <DEC = 90° (diag. bisect at 90°) <ECD = 180° - 90° - 28° (<‘s of a ∆)
= 62° = 62° (diag. bisect the <‘s)
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Class test 1Column A Column B1. The sum of the interior angles of a quadrilateral
A. Angle, Angle, Angle
2. Any closed shape with 4 straight sides B. 180°3. A case for congruency C. Isosceles4. A type of triangle with no equal sides or angles
D. Quadrilateral
5. Opposite angles are equal E. Kite6. Diagonals bisect each other F. Side, Side, Side7. The sum of the interior angles of a triangle G. Rhombus8. Two pairs of adjacent sides equal H. Scalene9. A case for similarity I. 360°10. A type of triangle with two angles equal. J. Rectangle
A. Angle, Angle, Angle
B. 180°
C. Isosceles
D. Quadrilateral
E. Kite
F. Side, Side, Side
G. RhombusH. Scalene
I. 360°
J. Rectangle
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Class test 2Find the missing lengths:1.
2. Prove that the ∆’s below are congruent.
65° 60°
65°
60°
A
B CD
E
F
2cm
4cm3cm𝑥
𝑦
7cm
1. <B = <E (given)2. <C = <D (given)3. <A = <F = 55°(<‘s of a
∆) ∆ ABC lll ∆ FED (AAA)
= =
= =
2 = 12cm = 6cm
4 = 14cm = 3,5cm
A
B
C
D
1 2