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Special Properties, SMB-031

In the diagram, `ABCDE` is a regular pentagon. The diagonals `AC` and `BD` intersect at `F`.

  1. Show that the size of `/_ABC` is 108°.  (1 mark)

  2. Find the size of `/_BAC`. Give reasons for your answer.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `36°`
Show Worked Solution
i.    

`text(Sum of all internal angles)`

`= (n-2) xx 180°`

`= (5-2) xx 180°`

`= 540°`
 

`:. /_ABC= 540/5= 108°`
 

ii.  `BA = BC\ \ text{(equal sides of a regular pentagon)}`

`:. Delta BAC\ text(is isosceles)`

`/_BAC= 1/2 (180-108)=36^{\circ} \ \ \ text{(base angle of}\ Delta BAC text{)}`

Special Properties, SMB-030

A regular decagon is pictured below.
 

  1. What is the value of `x`?   (2 marks)

  2. What is the size of an internal angle of a decagon?   (2 marks)

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i.    `36^@`

ii.   `144^@`

Show Worked Solution

i.    `text{Sum of exterior angles = 360°}`

`text{Since the decagon is regular, all external angles are equal.}`

`:.x^{\circ}= 360/10 = 36^{\circ}`

  
ii.    `text{Method 1: Using exterior angle}`

`text{Internal angle}` `=180-\text{exterior angle}`
  `=180-36`
  `=144^{\circ}`

  
`text{Method 2: Using Internal angle sum formula}`

`text{Sum of internal angles}` `=(n-2) xx 180`
  `=(10-2) xx 180`
  `=1440^{\circ}`

  
`:.\ text{Internal angle}\ = 1440/10 = 144^{\circ}`

Special Properties, SMB-018

A six sided figure is drawn below.
  

What is the sum of the six interior angles?   (2 marks)

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`720^@`

Show Worked Solution

`\text{Method 1}`

`text(Reflex angle) = 360-90 = 270^@`

`:.\ text(Sum of interior angles)`

`= (270 xx 2) + (30 xx 2) + (60 xx 2)`

`= 720^@`
 

`\text{Method 2}`

`text{Sum of interior angles (formula)}`

`= (n-2) xx 180`

`=4 xx 180`

`= 720^@`

Special Properties, SMB-016

Two identical quadrilaterals fit together to make this regular pentagon.
 

What is the value of `x`?   (2 marks)

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`108^@`

Show Worked Solution

`text(Consider regular pentagon:)`

`text{Sum of internal angles (formula)}`

`= (n-2) xx 180`

`= 3 xx 180`

`= 540^@`
 

`:. x` `= 540/5`
  `= 108^@`
TIP: Two quadrilaterals joining does not make the internal angle sum = 2 × 360°!

Special Properties, SMB-015

A star is drawn on the inside of a regular pentagon, as shown below.
 

What is the size of the angle marked `x`?   (3 marks)

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`36^@`

Show Worked Solution

`text(Consider the triangle)\ ABC\ \ text(in the pentagon:)`

STRATEGY: The internal angle sum is  3 × 180 = 540 (since 3 triangles can be drawn internally from one point).

`text(Total degrees in a pentagon)`

`= 3 xx 180`

`= 540^@`

 
`text{Internal angle}\ =540/5 = 108^@\ \ \text{(regular pentagon)}`
  
`DeltaABC\ text(is isosceles)`

`:. x + x + 108` `= 180`
`2x` `= 72`
`x` `= 36^@`

Special Properties, SMB-014

The sum of the interior angles of a 6 sided polygon can be found by first dividing it into triangles from one vertex.
 

 

What is the sum of the interior angles of this polygon?   (2 marks)

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`720\ text(degrees)`

Show Worked Solution

`text{Since the polygon can be divided into 4 separate triangles:}`

`text(Sum of interior angles)`

`= 4 xx 180`

`= 720\ text(degrees)`

Special Properties, SMB-008

The sum of the internal angles of a polygon can be calculated by drawing triangles from any given vertex as shown below. 
 

What is the size of the angle marked `x` in the diagram below?   (2 marks)

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`107°`

Show Worked Solution

`text{Since the quadrilateral was divided into two triangles}`

`=>\ \text{Sum of internal angles}\ = 2 xx 180 = 360^{\circ}`

`:. x` `= 360-(103 + 88 + 62)`
  `= 360-253`
  `= 107°`

Plane Geometry, 2UA 2011 HSC 6a

The diagram shows a regular pentagon `ABCDE`. Sides `ED` and `BC` are produced to meet at `P`.
  

  1. Find the size of `/_CDE`.    (1 mark)

  2. Hence, show that `Delta EPC` is isosceles.    (2 marks)

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  1. `108°`
  2. `text(Proof)\ \ text{(see Worked Solutions)}`
Show Worked Solution
i.  

`text(Angle sum of pentagon)=(5-2) xx 180°=540°`

`:.\ /_CDE` `= 540/5\ \ \ text{(regular pentagon has equal angles)}`
  `= 108°`
MARKER’S COMMENT: Very few students solved part (i) efficiently. Remember the general formula for the sum of internal angles equals (# sides – 2) x 90°.

 
ii.  `text(Show)\ Delta EPC\ text(is isosceles)`

`text(S)text(ince)\ ED=CD\ \ text{(sides of a regular pentagon)}`

`Delta ECD\ text(is isosceles)`

`/_DEC=1/2 xx (180-108)= 36^{\circ}\ \ \ text{(Angle sum of}\ Delta DEC text{)}`

`/_CDP=72^@\ \ \ (\angle PDE\ \text{is a straight angle})`

`/_DCP=72^@\ \ \ (\angle PCB\ \text{is a straight angle})`

`=> /_CPD= 180-(72 + 72)=36^{\circ}\ \ \ text{(angle sum of}\ Delta CPD text{)}`

`:.\ Delta EPC\ \text(is isosceles)\ \ \ text{(2 equal angles)}`