Geometry of 2D shapesChapter 12, Page 140

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. ∆)

Examples:

1)

+ 65° + 45°= 180° (<‘s of a ∆ = 180°) = 180° – 65° – 45° = 70°

2)

= 73° + 32° (ext < of a ∆) = 105°

3)

= 58° (isos. ∆) = 7cm (isos. ∆)

4)

= 5cm (equil. ∆) = 5cm (equil. ∆), and = 60°(equil. ∆)

DO: Exercise 12.2 Page 145 ALLLLL!!!!!

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°

Exercise 12 .2: Answers + 25°+ 68° = 180° (<‘s of a ∆ = 180°)

= 180° - 25°- 68° = 87°

= 47° + 63° (ext < of a ∆) = 110°

Exercise 12 .2: Answers + 42°= 125° (ext < of a ∆)

= 125° – 42° = 83°

+ 90° + 35° = 180° (<‘s of a ∆ = 180°)

= 180° - 90° - 35° =55°

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 °

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°

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°

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

Exercise 12.3: Answers

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

Example 2:h2 = s2 + s2 (pythag)h2 = (3)2 + (4)2

h2 = 9 + 16h2 = 25h = h = 5cm

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

Example 4:s2 = h2 - s2 (pythag)s2 = (5)2 – (3)2

s2 = 25 - 9s2 = 16s = s = 4cm

Do: Exercise 12.4 Page 149ALL

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

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

Exercise 12.4: Answersc) s2 = h2 - s2

(pythag)s2 = (14)2 – (8)2

s2 = 196 - 64s2 = 132s = 11,5cm

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

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

Pythagoras worksheet: Answers 5) 2 = h2 - s2 (pythag)

= (17)2 – (8)2

= 289 - 64= 225

= 15

Pythagoras worksheet: Answers 6) 2 = s2 + s2 (pythag) = (12)2 + (9)2

= 225 = 15

2 = h2 - s2 (pythag)= (25)2 – (15)2

= 400 = 20

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

QuadrilateralsPage 149

DEF: A quadrilateral is any closed shape with four straight sides.

The angles of a quadrilateral add up to 360

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

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

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

Parallelogram

Both pairs of opposite sides parallel Both pairs of opposite sides equal Both pairs of opposite angles equal Diagonals bisect each other

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

Kite

2 pairs of adjacent sides = Diagonals bisect at 90 The longer diagonal bisects the angles The longer diagonal bisects the shorter

diagonal

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

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°

Example 1:

What kind of quadrilateral is this?Parallelogram

Calculate and = 135 (opp. angles of a parm =) = 8cm (opp. sides of a parm =)

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

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

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 =)

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°

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

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

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

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°

SimilarityDefinition:1) All pairs off corresponding

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

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°

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°

i) ² + 24² = 25² (pythag)² + 576 = 625² = 625 – 576² = 49 = 7cm

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

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°

c) <B

= 70°(<‘s of a ∆)∆

ABC lll ∆ XYZ

d) ∆ABD ∆ CDB (RHS)

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

b) <ABC = 180° - 120° (str line)= 60°

3 = – 10° + 60° (ext. <‘s of a ∆)3 –= 50°2 = 50° = 25°

A

B C60°

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)

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

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