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Right-angled Triangles, SM-Bank 010 MC

Which of the following is true of the hypotenuse in a right-angled triangle?

  1. The hypotenuse is the side opposite to the right angle.
  2. The hypotenuse is the side adjacent to the right angle.
  3. The hypotenuse is one of the shorter sides in a right angled triangle.
  4. The hypotenuse is equal to the sum of the square of the other two sides.
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\(A\)

Show Worked Solution

\(\text{The hypotenuse is the side opposite to the right angle.}\)

  \(\Rightarrow A\)

Right-angled Triangles, SM-Bank 001 MC

Henry flies a kite attached to a long string, as shown in the diagram below.
 

The horizontal distance of the kite to Henry’s hand is 8 m.

The vertical distance of the kite above Henry’s hand is 15 m.

The length of the string, in metres, is

  1.  13
  2.  17
  3.  23
  4.  289
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\(B\)

Show Worked Solution

\(\text{Using Pythagoras}\)

\(s^2\) \(= 8^2 + 15^2\)
  \(= 289\)
\(\therefore\ s\) \(=\sqrt{289}\)
  \(= 17\ \text{metres}\)

  \(\Rightarrow B\)

Composite Figures, SM-Bank 033

The design below is made up of one sector with an angle of \(\theta^\circ\) and one equilateral triangle.

Calculate the the value of \(\large \theta\) if the perimeter of the shape in terms of \(\large \pi\) is \((27+3\pi)\) metres.  (3 marks)

NOTE: \(\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r\)

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\(60^\circ\)

Show Worked Solution

\(\text{Radius of sector}\ r=9\ \text{m, Sector angle}=\theta^\circ\)

\(\text{Perimeter}=27+3\pi\)

\(\text{Perimeter}\) \(=3\ \text{straight edges}+1\ \text{arc length}\)
\(27+3\pi\) \(=3\times 9+\Bigg(\dfrac{\theta}{360}\times 2\pi \times 9\Bigg)\)
\(27+3\pi\) \(=27+\Bigg(\dfrac{\theta \times 18\pi}{360}\Bigg)\)
\(\dfrac{\theta\times \pi}{20}\) \(=3\pi\)
\(\theta\) \(=3\times 20=60^\circ\)

 
\(\therefore\ \theta=60^\circ\)

Composite Figures, SM-Bank 032

The design below is made up of one \(106^\circ\) sector arc and two right angled triangles.

Calculate the perimeter of the design, correct to two decimal place.  (2 marks)

NOTE: \(\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r\)

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\(23.25\ \text{mm  (2 d.p.)}\)

Show Worked Solution

\(\text{Radius of sector}\ r=5\ \text{m, Sector angle }\theta=106^\circ\)

\(\text{Perimeter}\) \(=3\ \text{straight edges}+1\ \text{arc length}\)
  \(=2\times 3+8+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)\)
  \(=14+\Bigg(\dfrac{106}{360}\times 2\pi\times 5\Bigg)\)
  \(=14+\dfrac{53}{18}\times \pi\)
  \(=23.2502\dots\)
  \(\approx 23.25\ \text{mm  (2 d.p.)}\)

Composite Figures, SM-Bank 031

The design below is made up of one \(40^\circ\) sector arc and a right angled triangle.

Calculate the perimeter of the design, correct to one decimal place.  (2 marks)

NOTE: \(\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r\)

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\(29.1\ \text{cm  (1 d.p.)}\)

Show Worked Solution

\(\text{Radius of sector}\ r=13\ \text{m, Sector angle }\theta=40^\circ\)

\(\text{Perimeter}\) \(=3\ \text{straight edges}+1\ \text{arc length}\)
  \(=12+5+13+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)\)
  \(=30+\Bigg(\dfrac{40}{360}\times 2\pi\times 13\Bigg)\)
  \(=30+\dfrac{26}{9}\times \pi\)
  \(=29.075\dots\)
  \(\approx 29.1\ \text{cm  (1 d.p.)}\)

Composite Figures, SM-Bank 030

The design below is made up of one \(120^\circ\) sector arc and three identical rhombuses.

Calculate the perimeter of the design, correct to one decimal place.  (2 marks)

NOTE: \(\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r\)

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\(34.0\ \text{cm  (1 d.p.)}\)

Show Worked Solution

\(\text{Radius of sector}\ r=4.2\ \text{m, Sector angle }\theta=120^\circ\)

\(\text{Perimeter}\) \(=6\ \text{straight edges}+1\ \text{arc length}\)
  \(=6\times 4.2+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)\)
  \(=25.2+\Bigg(\dfrac{120}{360}\times 2\pi\times 4.2\Bigg)\)
  \(=25.2+\dfrac{1}{3}\times 8.4\pi\)
  \(=33.996\dots\)
  \(\approx 34.0\ \text{cm  (1 d.p.)}\)

Composite Figures, SM-Bank 029

Jonti is constructing the stage for the local music festival. The design is made up of one \(60^\circ\) sector arc and two equilateral triangles.

Calculate the perimeter of Jonti's stage. Give your answer correct to one decimal place.  (2 marks)

NOTE: \(\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r\)

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\(25.2\ \text{m  (1 d.p.)}\)

Show Worked Solution

\(\text{Radius of sector}\ r=5\ \text{m, Sector angle }\theta=60^\circ\)

\(\text{Perimeter}\) \(=4\ \text{straight edges}+1\ \text{arc length}\)
  \(=4\times 5+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)\)
  \(=20+\Bigg(\dfrac{60}{360}\times 2\pi\times 5\Bigg)\)
  \(=20+\dfrac{5}{3}\pi\)
  \(=25.235\dots\)
  \(\approx 25.2\ \text{m  (1 d.p.)}\)

Composite Figures, SM-Bank 028

Pixie is designing a new company logo. The design is made up of three identical sectors with the radius of each sector being 18 millimitres.

Calculate the perimeter of Pixie's. Give your answer correct to the nearest millimetre.  (3 marks)

NOTE: \(\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r\)

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\(127\ \text{cm  (nearest mm)}\)

Show Worked Solution

\(\text{Radius of sector}\ r=18\ \text{mm, Sector angle }\theta=60^\circ\)

\(\text{Perimeter}\) \(=6\ \text{straight edges}+3\times \text{arc length}\)
  \(=6r+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)\)
  \(=6\times 18+\Bigg(\dfrac{60}{360}\times 2\pi\times 18\Bigg)\)
  \(=108 +\Bigg(\dfrac{60\times 36\pi}{360}\Bigg)\)
  \(=108+6\pi\)
  \(=126.849\dots\)
  \(\approx 127\ \text{cm  (nearest mm)}\)

Composite Figures, SM-Bank 027

Pedro had one piece of pizza left to eat. The piece of pizza was one-eighth of the whole pizza and had a radius of 12 centimetres.

Calculate the perimeter of Pedro's piece of pizza. Give your answer correct to the nearest centimetre.  (2 marks)

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\(89\ \text{cm  (nearest cm)}\)

Show Worked Solution

\(\text{Radius of sector}=12\ \text{cm}\)

\(\text{Perimeter}\) \(=2\ \text{straight edges}+\dfrac{1}{8}\times \text{circumference} \)
  \(=2r+\Bigg(\dfrac{1}{8}\times 2\pi r\Bigg)\)
  \(=2\times 12 +\Bigg(\dfrac{1}{8}\times 2\pi \times 12\Bigg)\)
  \(=24+\dfrac{24\pi}{8}\)
  \(=80+3\pi\)
  \(=89.424\dots\)
  \(\approx 89\ \text{cm  (nearest cm)}\)

Composite Figures, SM-Bank 026

Find the exact perimeter of the composite shape below.  (2 marks)

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\(80+10\pi)\ \text{cm}\)

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\(\text{Radius of quadrant}=29\ \text{mm}\)

\(\text{Perimeter}\) \(=4\ \text{straight edges}+\dfrac{1}{4}\times \text{circumference} \)
  \(=4\times 20+\Bigg(\dfrac{1}{4}\times 2\pi r\Bigg)\)
  \(=80 +\Bigg(\dfrac{1}{4}\times 2\pi \times 20\Bigg)\)
  \(=80+\dfrac{40\pi}{4}\)
  \(=(80+10\pi)\ \text{cm}\)

Composite Figures, SM-Bank 025

A guitar plectrum was made using a semicircle and a quadrant as shown in the diagram below.

Calculate the perimeter of the plectrum, correct to the one decimal place.  (2 marks)

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\(8.3\ \text{cm  (1 d.p.)}\)

Show Worked Solution

\(\text{Radius of quadrant}=2\ \text{cm,  Diameter of semicircle}=2\ \text{cm}\)

\(\text{Perimeter}\) \(=1\ \text{straight edge}+\dfrac{1}{4}\times \text{circumference(1)} +\dfrac{1}{2}\times \text{circumference(2)}\)
  \(=2+\Bigg(\dfrac{1}{4}\times 2\pi r\Bigg)+\Bigg(\dfrac{1}{2}\times \pi d\Bigg)\)
  \(=2 +\Bigg(\dfrac{1}{4}\times 2\pi \times 2\Bigg)+\Bigg(\dfrac{1}{2}\times \pi\times 2\Bigg)\)
  \(=2+\dfrac{4\pi}{4}+\dfrac{2\pi}{2}\)
  \(=2+2\pi\)
  \(=8.283\dots\)
  \(=8.3\ \text{cm  (1 d.p.)}\)

Composite Figures, SM-Bank 024

James ate one-quarter of his extra large pizza. Calculate the perimeter of the remaining pizza, correct to the nearest centimetre.  (2 marks)

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\(107\ \text{cm  (nearest cm)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=2\ \text{straight sides} +\dfrac{3}{4}\times \text{circumference}\)
  \(=2\times 16 +\Bigg(\dfrac{3}{4}\times 2\pi r\Bigg)\)
  \(=32 +\Bigg(\dfrac{3}{4}\times 2\pi\times 16\Bigg)\)
  \(=32+24\pi\)
  \(=107.398\dots\)
  \(=107\ \text{cm  (nearest cm)}\)

Composite Figures, SM-Bank 023

Calculate the perimeter of the following shape correct to 1 decimal place.  (2 marks)

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\(72.8\ \text{cm  (1 d.p.)}\)

Show Worked Solution

\(\text{Perimeter}\) \(=4\ \text{straight sides} +\dfrac{1}{4}\times \text{circumference}\)
  \(=2\times 15+12+7.5+4.5 +\Bigg(\dfrac{1}{4}\times 2\pi\times 12\Bigg)\)
  \(=54+6\pi\)
  \(=72.849\dots\)
  \(=72.8\ \text{cm  (1 d.p.)}\)

Composite Figures, SM-Bank 022

The diagram below shows the design for a chicken pen at Gayle's farm.

  1. Gayle wishes to construct a fence around the pen. Calculate the amount of fencing required, correct to the next metre.  (2 marks)

  2. Gayle has chosen fencing that costs $22 per metre, with an additional cost of $240 for the installation of a gate. Calculate the total cost of fencing the pen.  (2 marks)

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a.    \(17\ \text{m  (next metre)}\)

b.    \($614\)

Show Worked Solution
a.    \(\text{Perimeter}\) \(=2\ \text{radii}+\dfrac{1}{4}\ \text{circumference}\)
    \(=2\times 4.5+\Bigg(\dfrac{1}{4}\times 2\pi r\Bigg)\)
    \(=9+\Bigg(\dfrac{1}{4}\times 2\pi \times 4.5\Bigg)\)
    \(=9 +\dfrac{1}{2}\times 4.5\pi\)
    \(=16.068\dots\)
    \(\approx 17\ \text{m  (next metre)}\)

 

b.    \(\text{Total cost}\) \(=$22\times 17 +$240\)
    \(=$614\)

Composite Figures, SM-Bank 021

Write an expression in terms of \(\large \pi\) for the perimeter of the shape below. Give your answer in simplest form.  (2 marks)

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\(2r +\dfrac{3\pi r}{2}\)

Show Worked Solution

\(\text{Radius}=r\)

\(\text{Perimeter}\) \(=2\ \text{straight sides}+\dfrac{3}{4}\ \text{circumference}\)
  \(=2\times r+\Bigg(\dfrac{3}{4}\times 2\pi r\Bigg)\)
  \(=2r+\Bigg(\dfrac{6}{4}\times \pi \times r\Bigg)\)
  \(=2r +\dfrac{3\pi r}{2}\)

Composite Figures, SM-Bank 020

Calculate the perimeter of the composite shape below correct to the nearest centimetre.  (2 marks)

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\(39\ \text{cm  (nearest centimetre)}\)

Show Worked Solution

\(\text{Radius}=8\ \text{cm}\)

\(\text{Perimeter}\) \(=4\ \text{straight sides}+1\ \text{quadrant}\)
  \(=2\times 5+10+ 6+\Bigg(\dfrac{1}{4}\times 2\pi r\Bigg)\)
  \(=26+\Bigg(\dfrac{1}{4}\times 2\pi \times 8\Bigg)\)
  \(=26+4\pi\)
  \(=38.566\dots\)
  \(\approx 39\ \text{cm  (nearest centimetre)}\)

Composite Figures, SM-Bank 019 MC

The courtyard shown below incorporates quadrants and rectangles in its design.

The landscaping company needs to calculate the perimeter of the courtyard so they can provide estimates for fencing.

Which answer represents the approximate length of fencing required, to the nearest metre?

  1. \(23\ \text{m}\)
  2. \(33\ \text{m}\)
  3. \(37\ \text{m}\)
  4. \(47\ \text{m}\)
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\(D\)

Show Worked Solution

\(\text{Radius}=4+2=6\ \text{m}\)

\(\text{Perimeter}\) \(=6\ \text{straight sides}+2\ \text{quadrants}\)
  \(=2\times 4+2\times 8+ 2\times 2+\Bigg(2\times \dfrac{1}{4}\times 2\pi r\Bigg)\)
  \(=28+\pi\times 6\)
  \(=46.849\dots\)
  \(\approx 47\ \text{m  (nearest metre)}\)

\(\Rightarrow D\)

Composite Figures, SM-Bank 018 MC

A pool area is in the shape of a rectangle and a semicircle, as shown below.

A fence is to be constructed around the perimeter of the pool area.

Which expression represents the length of fencing required, in terms of \(\large \pi\)?

  1. \(13+3\pi\)
  2. \(13+6\pi\)
  3. \(20+3\pi\)
  4. \(20+6\pi\)
Show Answers Only

\(C\)

Show Worked Solution
\(\text{Perimeter}\) \(=3\ \text{straight sides}+1\ \text{semicircle}\)
  \(=2\times 7+6+\Bigg(\dfrac{1}{2}\times \pi d\Bigg)\)
  \(=20+\Bigg(\dfrac{1}{2}\times \pi\times 6\Bigg)\)
  \(=20 +3\pi \)

\(\Rightarrow C\)

Composite Figures, SM-Bank 017

Jules cuts away one-half of the circle shown below.

  1. Draw a sketch of Jules' new shape. Include the length of the diameter in your sketch.  (1 mark)

  2. What is the perimeter of the new shape, to the nearest metre?  (2 marks)

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a. 

b.    \(26\ \text{m  (nearest m)}\)

Show Worked Solution

a.

b.    \(\text{Perimeter}\) \(=\text{diameter}+\dfrac{1}{2}\times \text{circumference}\)
    \(=10 +\Bigg(\dfrac{1}{2}\times \pi d\Bigg)\)
    \(=10 +\Bigg(\dfrac{1}{2}\times \pi\times 10\Bigg)\)
    \(=10+5\pi\)
    \(=25.707\dots\)
    \(=26\ \text{m  (nearest m)}\)

Composite Figures, SM-Bank 015

Vinnie cuts away two-thirds of a circle, leaving the shape shown below.   

If the circles' radius is 10 cm, what is the perimeter of the remaining shape, to the nearest centimetre?   (2 marks)

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\( 41\ \text{cm  (nearest cm)}\)

Show Worked Solution

\(\text{Fraction of circle}=\dfrac{360-240}{360}=\dfrac{120}{360}=\dfrac{1}{3}\)

\(\text{Perimeter}\) \(=2\ \text{radii}+\dfrac{1}{3}\times \text{circumference}\)
  \(=2\times 10 +\Bigg(\dfrac{1}{3}\times 2\pi\times r\Bigg)\)
  \(=20 +\Bigg(\dfrac{1}{3}\times 2\pi\times 10\Bigg)\)
  \(=20+\dfrac{20\pi}{3}\)
  \(= 40.943\dots\)
  \(= 41\ \text{cm  (nearest cm)}\)

Composite Figures, SM-Bank 016 MC

A one-on-one basketball court is a composite shape made up of a rectangle and a semicircle, as shown below.

A boundary line is painted around the perimeter of the shape.

The total length of the boundary line, in metres, is closest to

  1. \(38.8\)
  2. \(57.7\)
  3. \(66.8\)
  4. \(76.5\)
Show Answers Only

\(A\)

Show Worked Solution
\(\text{Perimeter}\) \(=\Bigg(\dfrac{1}{2}\times \pi\times 12\Bigg) + 4 + 12 + 4\)
  \(=6\pi + 20\)
  \(= 38.849\dots\ \text{m}^2\)

\(\Rightarrow A\)

Composite Figures, SM-Bank 014

Pedro cuts a sector from a circle so that  \(\dfrac{3}{8}\)  of the area of the circle remains.
 
 
 If the circles' radius is 5 cm, what is the perimeter of the shape, to the nearest centimetre?  (2 marks)

Show Answers Only

\(22\ \text{cm}\)

Show Worked Solution
\(\text{Perimeter}\) \(=\dfrac{3}{8}\times 2 \pi r + 2 r\)
  \(=\dfrac{3}{8}\times 2 \pi \times 5 + 10\)
  \(= 21.78\dots\)
  \(= 22\ \text{cm  (nearest cm)}\)

Composite Figures, SM-Bank 013 MC

Grant cut two semicircles from a rectangle to create Shape 1.

He then joins the semicircles to each end of an identical rectangle to create Shape 2.

Which of the following statements is true about Shape 1 and Shape 2?

  1. They have the same area and the same perimeter.
  2. They have different areas and the same perimeter.
  3. They have the same area and different perimeters.
  4. They have different areas and different perimeters.
Show Answers Only

\(B\)

Show Worked Solution

\(\text{The semi-circles in both shapes have the same diameters}\)

\(\text{but have been turned inward in Shape 2.}\)

\(\therefore\ \text{They have different areas and the same perimeter.}\)

\(\Rightarrow B\)

Composite Figures, SM-Bank 012

Find the perimeter of the figure below giving your answer as an exact value in terms of \(\large \pi\).  (2 marks)

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\((84+70\pi)\ \text{cm}\)

Show Worked Solution

\(\text{Fraction of circle}=\dfrac{360-60}{360}=\dfrac{300}{360}=\dfrac{5}{6}\)

\(\text{Perimeter}\) \(=2\ \text{radii}+\dfrac{5}{6}\times\text{circumference}\) 
  \(=2\times 42+\dfrac{5}{6}\times 2\pi r\)
  \(=84+\dfrac{5}{6}\times 2\pi \times 42\)
  \(=(84+70\pi)\ \text{cm}\)

 

Composite Figures, SM-Bank 011

Explain using mathematical reasoning why the rule for finding the perimeter of the figure below is \(p=4\pi r\).  (2 marks)

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\(\text{See worked solution}\)

Show Worked Solution

\(\text{Diameter}=2r\ \longrightarrow\ \text{Radius}=r\)

\(\text{Perimeter}\) \(=4\ \text{semi-circles}=2\ \text{circles}\) 
  \(=2\times 2\pi r\)
  \(=4\pi r\ \text{units}\)

 

Composite Figures, SM-Bank 010

Find the perimeter of the figure below giving your answer as an exact value in terms of \(\large \pi\).  (2 marks)

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\((64+8\pi)\ \text{m}\)

Show Worked Solution
\(\text{Perimeter}\) \(=4\ \text{straight sides}+2\ \text{quarter-circles (ie one semi-circle)}\) 
  \(=4\times 16+\dfrac{\pi d}{2}\)
  \(=64+\dfrac{\pi\times 16}{2}\)
  \(=(64+8\pi)\ \text{m}\)

 

Composite Figures, SM-Bank 009

Find the perimeter of the figure below correct to one decimal place.  (2 marks)

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\(36.6\ \text{m  (1 d.p.)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=2\ \text{straight sides}+2\ \text{semi-circles (ie one circle)}\) 
  \(=2\times 12+\pi d\)
  \(=24+\pi \times 4\)
  \(=36.5663\dots\)
  \(=36.6\ \text{m  (1 d.p.)}\)

 

Composite Figures, SM-Bank 008

Find the perimeter of the figure below correct to one decimal place.  (2 marks)

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\(114.2\ \text{m  (1 d.p.)}\)

Show Worked Solution

\(\text{Large semi-circle: }\ d_{1}=40,\ \text{Small semi-circle: }\ d_{2}=20\)

\(\text{Perimeter}\) \(=2\ \text{straight sides}+2\ \text{semi-circles}\)
  \(=2\times 10+\dfrac{\pi d_{1}}{2}+\dfrac{\pi d_{2}}{2}\)
  \(=20+\dfrac{\pi \times 40}{2}+\dfrac{\pi \times 20}{2}\)
  \(=114.2477\dots\)
  \(=114.2\ \text{m  (1 d.p.)}\)

 

Composite Figures, SM-Bank 007

Find the perimeter of the figure below correct to one decimal place.  (2 marks)

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\(21.9\ \text{cm  (1 d.p.)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=2\ \text{straight sides}+1\ \text{semi-circle}\)
  \(=6+8+\dfrac{\pi d}{2}\)
  \(=14+\dfrac{\pi \times 5}{2}\)
  \(=21.8539\dots\)
  \(=21.9\ \text{cm  (1 d.p.)}\)

 

Composite Figures, SM-Bank 006

Find the perimeter of the figure below correct to two decimal places.  (2 marks)

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\(164.25\ \text{cm  (2 d.p.)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=2\ \text{straight sides}+2\ \text{semi-circles (ie 1 circle)}\)
  \(=2\times 35+\pi d\)
  \(=70+\pi \times 30\)
  \(=164.2477\dots\)
  \(=164.25\ \text{cm  (2 d.p.)}\)

 

Composite Figures, SM-Bank 005

Find the perimeter of the figure below correct to two decimal places.  (2 marks)

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\(585.39\ \text{mm  (2 d.p.)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=2\ \text{straight sides}+\dfrac{280}{360}\times \ \text{circumference}\)
  \(=2\times 85+\Bigg(\dfrac{7}{9}\times 2\pi r\Bigg)\)
  \(=170+\Bigg(\dfrac{7}{9}\times 2\pi \times 9.8\Bigg)\)
  \(=585.3883\dots\)
  \(=585.39\ \text{mm  (2 d.p.)}\)

 

Composite Figures, SM-Bank 004

Find the perimeter of the figure below correct to one decimal place.  (2 marks)

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\(905.2\ \text{cm  (1 d.p.)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=2\ \text{straight sides}+\dfrac{3}{4}\ \text{circle}\)
  \(=2\times 9.8+\Bigg(\dfrac{3}{4}\times 2\pi r\Bigg)\)
  \(=19.6+\Bigg(\dfrac{3}{4}\times 2\pi \times 9.8\Bigg)\)
  \(=905.1556\dots\)
  \(=905.2\ \text{cm  (1 d.p.)}\)

 

Composite Figures, SM-Bank 003

Find the perimeter of the figure below correct to one decimal place.  (2 marks)

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\(67.7\ \text{cm  (1 d.p.)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=5\ \text{straight sides}+1\ \text{semi-circle}\)
  \(=2\times 12+20+2\times 4+\Bigg(\dfrac{\pi \times d}{2}\Bigg)\)
  \(=52+\Bigg(\dfrac{\pi \times 10}{2}\Bigg)\)
  \(=67.7079\dots\)
  \(=67.7\ \text{cm  (1 d.p.)}\)

 

Composite Figures, SM-Bank 002

Find the perimeter of the figure below correct to one decimal place.  (2 marks)

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\(51.4\ \text{cm  (1 d.p.)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=2\ \text{straight sides}+2\ \text{semi-circles}\)
  \(=2\times 10+\Bigg(2\times\dfrac{\pi \times d}{2}\Bigg)\)
  \(=20+\pi \times 10\)
  \(=51.4159\dots\)
  \(=51.4\ \text{cm  (1 d.p.)}\)

 

Composite Figures, SM-Bank 001

Find the perimeter of the figure below correct to one decimal place.  (2 marks)

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\(88.3\ \text{cm  (1 d.p.)}\)

Show Worked Solution
\(\text{Perimeter}\) \(=3\ \text{straight sides}+\text{semi-circle}\)
  \(=2\times 21 +18+\Bigg(\dfrac{\pi \times d}{2}\Bigg)\)
  \(=60+\Bigg(\dfrac{\pi \times 18}{2}\Bigg)\)
  \(=88.2743\dots\)
  \(=88.3\ \text{cm  (1 d.p.)}\)

 

Circles, SM-Bank 055

A plane is circumnavigating the earth at the equator. It is cruising at a constant altitude of 10 kilometres above the earths' surface.

Given the earths' radius is approximately 6400 kilometres at the equator, calculate the distance, \(d\), that the plane has travelled after completing one lap of the earth.

Give your answer correct to the nearest whole kilometre.  (3 marks)

Show Answers Only

\(40\ 275\ \text{km  (nearest whole kilometre)}\)

Show Worked Solution

\(\text{Radius of planes’ path}=6400+10=6410\ \text{km}\)

\(d\) \(=2\pi r\)
  \(=2\pi \times 6410\)
  \(=40\ 275.2178\dots\)
  \(=40\ 275\ \text{km  (nearest whole kilometre)}\)

 
\(\therefore\ \text{The distance the plane travels around the earth}\)

\(\text{at the equator is approximately }40\ 275\ \text{kilometres.}\)

Circles, SM-Bank 054

Given the earths' radius is approximately 6400 kilometres, calculate the distance, \(d\), around the earth at the equator. Give your answer correct to the nearest whole kilometre.  (2 marks)

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\(40\ 212\ \text{km  (nearest whole kilometre)}\)

Show Worked Solution
\(d\) \(=2\pi r\)
  \(=2\pi \times 6400\)
  \(=40\ 212.3859\dots\)
  \(=40\ 212\ \text{km  (nearest whole kilometre)}\)

 
\(\therefore\ \text{The distance around the earth at the equator is approximately }40\ 212\ \text{kilometres.}\)

Circles, SM-Bank 053

Town A is 135\(^\circ\) east of Town B along the equator, as shown on the diagram below.

Given the earths' radius is approximately 6400 kilometres, calculate the distance \(d\), between the two towns. Give your answer correct to the nearest whole kilometre.  (2 marks)

NOTE:  \(\text{Arc length}=\dfrac{\theta}{360}\times 2\pi r\)

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\(15\ 080\ \text{km  (nearest whole kilometre)}\)

Show Worked Solution
\(d\) \(=\dfrac{\theta}{360}\times 2\pi r\)
  \(=\dfrac{135}{360}\times 2\pi \times 6400\)
  \(=15\ 079.6447\dots\)
  \(=15\ 080\ \text{km  (nearest whole kilometre)}\)

 
\(\therefore\ \text{Towns A and B are }15\ 080\ \text{kilometres apart.}\)

Circles, SM-Bank 052

Calculate the total perimeter of the sector below, giving your answer correct to the nearest whole number.  (2 marks)

NOTE:  \(\text{Perimeter}=2\times \text{radius}+\text{arc length}\)

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\(1128\ \text{km  (nearest whole number)}\)

Show Worked Solution

\(\text{Perimeter}=2\times \text{radius}+\text{arc length}\)

\(\text{Perimeter}\) \(=2r+\dfrac{\theta}{360}\times 2\pi r\)
  \(=2\times 300+\Bigg(\dfrac{120}{360}\times 2\pi \times 300\Bigg)\)
  \(=1128.3185\dots\)
  \(=1128\ \text{km  (nearest whole number)}\)

Circles, SM-Bank 051

Calculate the total perimeter of the sector below, giving your answer correct to one decimal place.  (2 marks)

NOTE:  \(\text{Perimeter}=2\times \text{radius}+\text{arc length}\)

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\(39.0\ \text{cm  (1 d.p.)}\)

Show Worked Solution

\(\text{Perimeter}=2\times \text{radius}+\text{arc length}\)

\(\text{Perimeter}\) \(=2r+\dfrac{\theta}{360}\times 2\pi r\)
  \(=2\times 14+\Bigg(\dfrac{45}{360}\times 2\pi \times 14\Bigg)\)
  \(=38.9955\dots\)
  \(=39.0\ \text{cm  (1 d.p.)}\)

Circles, SM-Bank 050

Calculate the total perimeter of the sector below, giving your answer correct to one decimal place.  (2 marks)

NOTE:  \(\text{Perimeter}=2\times \text{radius}+\text{arc length}\)

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\(1523.6\ \text{mm  (1 d.p.)}\)

Show Worked Solution

\(\text{Perimeter}=2\times \text{radius}+\text{arc length}\)

\(\text{Perimeter}\) \(=2r+\dfrac{\theta}{360}\times 2\pi r\)
  \(=2\times 500+\Bigg(\dfrac{60}{360}\times 2\pi \times 500\Bigg)\)
  \(=1523.5987\dots\)
  \(=1523.6\ \text{mm  (1 d.p.)}\)

Circles, SM-Bank 049

Calculate the total perimeter of the quadrant below, giving your answer correct to one decimal place.  (2 marks)

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\(105.3\ \text{km  (1 d.p.)}\)

Show Worked Solution

\(\text{Perimeter}=2\times \text{radius}+\dfrac{\text{circumference}}{4}\)

\(\text{Perimeter}\) \(=2r+\dfrac{2\pi r}{4}\)
  \(=2\times 29.5+\Bigg(\dfrac{2\pi \times 29.5}{4}\Bigg)\)
  \(=105.3384 \dots\)
  \(=105.3\ \text{km  (1 d.p.)}\)

Circles, SM-Bank 048

Calculate the total perimeter of the quandrant below, giving your answer as an exact value in terms of \(\large \pi\).  (2 marks)

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\(14+\dfrac{7\pi}{2}\ \text{m}\)

Show Worked Solution

\(\text{Perimeter}=2\times \text{radius}+\dfrac{\text{circumference}}{4}\)

\(\text{Perimeter}\) \(=2r+\dfrac{2\pi r}{4}\)
  \(=2\times 7+\Bigg(\dfrac{2\pi \times 7}{4}\Bigg)\)
  \(=14+\dfrac{7\pi}{2}\ \text{m}\)

Circles, SM-Bank 046

Calculate the total perimeter of the sector below, giving your answer as an exact value in terms of \(\large \pi\).  (2 marks)

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\(100+50\pi\ \text{cm}\)

Show Worked Solution

\(\text{Perimeter}=\text{diameter}+\dfrac{\text{circumference}}{2}\)

\(\text{Perimeter}\) \(=d+\dfrac{\pi d}{2}\)
  \(=100+\Bigg(\dfrac{\pi \times 100}{2}\Bigg)\)
  \(=100+50\pi\ \text{cm}\)

Circles, SM-Bank 047

Calculate the total perimeter of the sector below, correct to one decimal place.  (2 marks)

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\(17.2\ \text{cm  (1 d.p.)}\)

Show Worked Solution

\(\text{Perimeter}=\text{diameter}+\dfrac{\text{circumference}}{2}\)

\(\text{Perimeter}\) \(=d+\dfrac{\pi d}{2}\)
  \(=6.7+\Bigg(\dfrac{\pi \times 6.7}{2}\Bigg)\)
  \(=17.2243\dots\)
  \(=17.2\ \text{cm  (1 d.p.)}\)

Circles, SM-Bank 045

Calculate the total perimeter of the sector below, correct to one decimal place.  (2 marks)

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\(442.2\ \text{cm  (1 d.p.)}\)

Show Worked Solution

\(\text{Perimeter}=\text{diameter}+\dfrac{\text{circumference}}{2}\)

\(\text{Perimeter}\) \(=d+\dfrac{\pi d}{2}\)
  \(=172+\Bigg(\dfrac{\pi \times 172}{2}\Bigg)\)
  \(=442.1769\dots\)
  \(=442.2\ \text{cm  (1 d.p.)}\)

Circles, SM-Bank 044

Calculate the arc length of the sector below, correct to two decimal places.  (2 marks)

\(l=\dfrac{\theta}{360}\times 2\pi r\)

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\(270.18\ \text{m  (2 d.p.)}\)

Show Worked Solution

\(\text{Diameter}=172 \text{cm}\)

\(\therefore\ \text{Radius}=\dfrac{172}{2}=86 \text{cm}\)

\(\text{Method 1}\)

\(l\) \(=\dfrac{\theta}{360}\times 2\pi r\)
  \(=\dfrac{180}{360}\times 2\pi \times 86\)
  \(=270.1769\dots\)
  \(=270.18\ \text{m  (2 d.p.)}\)

 

\(\text{Method 2}\)

\(l\) \(=\dfrac{1}{2}\times \text{circumference}\)
  \(=\dfrac{1}{2}\times \pi d\)
  \(=\dfrac{1}{2}\times \pi \times 172\)
  \(=270.1769\dots\)
  \(=270.18\ \text{m  (2 d.p.)}\)

Circles, SM-Bank 043

Use the arc length formula below to calculate the arc length of the sector, correct to two decimal places.  (2 marks)

\(l=\dfrac{\theta}{360}\times 2\pi r\)

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\(7.54\ \text{m  (2 d.p.)}\)

Show Worked Solution
\(l\) \(=\dfrac{\theta}{360}\times 2\pi r\)
  \(=\dfrac{90}{360}\times 2\pi \times 4.8\)
  \(=7.5398\dots\)
  \(=7.54\ \text{m  (2 d.p.)}\)

Circles, SM-Bank 042

Use the arc length formula below to calculate the arc length of the sector, correct to one decimal place.  (2 marks)

\(l=\dfrac{\theta}{360}\times 2\pi r\)

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\(12.4\ \text{mm  (1 d.p.)}\)

Show Worked Solution
\(l\) \(=\dfrac{\theta}{360}\times 2\pi r\)
  \(=\dfrac{45}{360}\times 2\pi \times 15.8\)
  \(=12.4092\dots\)
  \(=12.4\ \text{mm  (1 d.p.)}\)

Circles, SM-Bank 041

Use the arc length formula below to calculate the arc length of the sector, correct to the nearest whole number.  (2 marks)

\(l=\dfrac{\theta}{360}\times 2\pi r\)

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\(628\ \text{km  (nearest whole number)}\)

Show Worked Solution
\(l\) \(=\dfrac{\theta}{360}\times 2\pi r\)
  \(=\dfrac{120}{360}\times 2\pi \times 300\)
  \(=628.3185\dots\)
  \(=628\ \text{km  (nearest whole number)}\)