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Right-angled Triangles, SM-Bank 053

  1. Use Pythagoras' Theorem to calculate the length of the hypotenuse in the isosceles triangle below. Give your answer correct in exact surd form.  (2 marks)
     
           

  2. Using your result from part (a) above, calculate the perimeter of the shape below, correct to 1 decimal place.  (2 marks)
     
         

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a.    \(\sqrt{50}\ \text{cm (exact surd form)}\)

b.    \(30.6\ \text{cm (1 d.p.)}\)

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a.    \(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=5\ \text{and }b=5\)

\(\text{Then}\ \ c^2\) \(=5^2+5^2\)
\(c^2\) \(=50\)
\(c\) \(=\sqrt{50}\ \text{cm (exact surd form)}\)

 

b.    \(\text{Perimeter}\) \(=\text{chord (a)}\ +\dfrac{3}{4}\times\text{circumference}\)
    \(=\sqrt{50}+\dfrac{3}{4}\times 2\pi r\)
    \(=\sqrt{50}+\dfrac{3}{4}\times 2\pi\times 5\)
    \(=30.633\dots\)
    \(\approx 30.6\ \text{cm (1 d.p.)}\)

Right-angled Triangles, SM-Bank 052

Use Pythagoras' Theorem to calculate the perimeter of the isosceles triangle below, correct to the nearest centimetre.  (3 marks)

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\(29\ \text{cm}\ (\text{nearest cm})\)

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\(\text{Find length of the equal sides of the triangle.}\)

\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=5\ \text{and }b=8\)

\(\text{Then}\ \ c^2\) \(=5^2+8^2\)
\(c^2\) \(=89\)
\(c\) \(=\sqrt{89}\)
\(c\) \(=9.433\dots\)

 
\(\text{Perimeter}\)

\(=2\times 9.433\dots+10\)

\(=28.867\dots\approx 29\ \text{cm}\ (\text{nearest cm})\)

Right-angled Triangles, SM-Bank 051

Use Pythagoras' Theorem to calculate the perimeter of the trapezium below, correct to 1 decimal place.  (3 marks)

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\(31.1\ \text{mm}\ (1\ \text{d.p.})\)

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\(\text{Find length of sloped side of trapezium.}\)

\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=4\ \text{and }b=7)

\(\text{Then}\ \ c^2\) \(=4^2+7^2\)
\(c^2\) \(=65\)
\(c\) \(=\sqrt{65}\)
\(c\) \(=8.062\dots\)

 
\(\text{Perimeter}\)

\(=6+7+10+8.062\dots\)

\(=31.062\dots\approx 31.1\ \text{mm}\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 049

  1. Calculate the length of the hypotenuse in the following right-angled triangle. Give your answer correct to the nearest metre.   (2 marks)
     

  2. Find the perimeter of the triangle.  (1 mark)
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a.    \(7\ \text{m}\)

b.    \(18\ \text{m}\)

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a.    \(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=5\ \text{and }b=6\)

\(\text{Then}\ \ c^2\) \(=5^2+6^2\)
\(c^2\) \(=61\)
\(c\) \(=\sqrt{61}\)
\(c\) \(=7.141\dots\approx 7\)

 
\(\text{The hypotenuse is }7\ \text{metres (nearest metre})\)

b.    \(\text{Perimeter}\)

\(=7+5+6\)

\(=18\ \text{m}\)

Right-angled Triangles, SM-Bank 048

The size of a computer monitor is determined by the length of the diagonal of the screen. Use Pythagoras' Theorem to calculate the size of the monitor in the diagram below.  (2 marks)

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\(51\ \text{centimetres}\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=24\ \text{and }b=45\)

\(\text{Then}\ \ c^2\) \(=24^2+45^2\)
\(c^2\) \(=2601\)
\(c\) \(=\sqrt{2601}\)
\(c\) \(=51\)

 
\(\text{The size of the monitor is }51\ \text{centimetres}\)

Right-angled Triangles, SM-Bank 047

Use Pythagoras' Theorem to calculate the length of the diagonal in the rectangle below, correct to 1 decimal place.  (2 marks)

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\(8.4\ \text{centimetres}\ (1\ \text{d.p.})\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=4.2\ \text{and }b=7.3\)

\(\text{Then}\ \ c^2\) \(=4.2^2+7.3^2\)
\(c^2\) \(=70.98\)
\(c\) \(=\sqrt{70.98}\)
\(c\) \(=8.4219\dots\approx 8.4\ (1\ \text{d.p.})\)

 
\(\text{The length of the string is }8.4\ \text{centimetres}\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 046

Lewis is holding a string attached to a kite. Lewis is 8 metres from the kite and the kite is flying 4 metres above his hand. Use Pythagoras' Theorem to calculate the length of the string attached to the kite, correct to 2 decimal places.  (2 marks)

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\(8.94\ \text{metres}\ (2\ \text{d.p.})\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=4\ \text{and }b=8\)

\(\text{Then}\ \ c^2\) \(=4^2+8^2\)
\(c^2\) \(=80\)
\(c\) \(=\sqrt{80}\)
\(c\) \(=8.9442\dots\approx 8.94\ (2\ \text{d.p.})\)

 
\(\text{The length of the string is }8.94\ \text{metres}\ (2\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 042

Find the value of the pronumeral in the right-angled triangle below, giving your answer in exact surd form.  (2 marks)

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\(c=\sqrt{277}\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=9,\ \text{and }b=14\)

\(\text{Then}\ \ c^2\) \(=9^2+14^2\)
\(c^2\) \(=277\)
\(c\) \(=\sqrt{277}\)

Right-angled Triangles, SM-Bank 039

Find the value of the pronumeral in the right-angled triangle below, giving your answer correct to 1 decimal place.  (2 marks)

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\(c\approx 6.5\ (1\ \text{d.p.})\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=4.8,\ \text{and }b=4.4\)

\(\text{Then}\ \ c^2\) \(=4.8^2+4.4^2\)
\(c^2\) \(=42.4\)
\(c\) \(=\sqrt{42.4}\)
\(c\) \(=6.511\dots\)
\(c\) \(\approx 6.5\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 038

Find the value of the pronumeral in the right-angled triangle below, giving your answer correct to 1 decimal place.  (2 marks)

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\(c\approx 16.2\ (1\ \text{d.p.})\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=15.4,\ \text{and }b=5.1\)

\(\text{Then}\ \ c^2\) \(=15.4^2+5.1^2\)
\(c^2\) \(=263.17\)
\(c\) \(=\sqrt{263.17}\)
\(c\) \(=16.222\dots\)
\(c\) \(\approx 16.2\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 037

Find the value of the pronumeral in the right-angled triangle below, giving your answer correct to 1 decimal place.  (2 marks)

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\(c\approx 12.5\ (1\ \text{d.p.})\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=10.4,\ \text{and }b=6.9\)

\(\text{Then}\ \ c^2\) \(=10.4^2+6.9^2\)
\(c^2\) \(=155.77\)
\(c\) \(=\sqrt{155.77}\)
\(c\) \(=12.480\dots\)
\(c\) \(\approx 12.5\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 036

Find the value of the pronumeral in the right-angled triangle below.  (2 marks)

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\(c=70\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=42,\ \text{and }b=56\)

\(\text{Then}\ \ c^2\) \(=42^2+56^2\)
\(c^2\) \(=4900\)
\(c\) \(=\sqrt{4900}\)
\(c\) \(=70\)

Right-angled Triangles, SM-Bank 034

Find the value of the pronumeral in the right-angled triangle below.  (2 marks)

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\(x=25\)

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\(\text{Pythagoras’ Theorem states:  }c^2=a^2+b^2\)

\(\text{Let }a=24,\ b=7\ \text{and }c=x\)

\(\text{Then}\ \ x^2\) \(=24^2+7^2\)
\(x^2\) \(=625\)
\(x\) \(=\sqrt{625}\)
\(x\) \(=25\)

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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\(B\)

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\(\text{Using Pythagoras}\)

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

  \(\Rightarrow B\)

Measurement, STD2 M6 2011 HSC 9 MC

Two trees on level ground, 12 metres apart, are joined by a cable. It is attached 2 metres above the ground to one tree and 11 metres above the ground to the other.

What is the length of the cable between the two trees, correct to the nearest metre? 

  1.  `9\ text(m)`
  2. `12\ text(m)`
  3. `15\ text(m)`
  4. `16\ text(m)`
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`C`

Show Worked Solution

`text(Using Pythagoras)`

`c^2` `=12^2+9^2`
  `=144+81`
  `=225`
`:.c` `=15,\ \ c>0`

 
`=>C`