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Volume, SM-Bank 066

Gavin is going camping in the summer holidays and purchased the two-person tent shown below.

  1. Given the triangular face of the tent is isosceles, use Pythagoras' Theorem to calculate the perpendicular height of the tent.  (2 marks)
  2. Using your answer from (a), calculate the volume of the tent in cubic metres.  (2 marks)
  3. What is the capacity of the tent in litres?  (1 mark)
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a.    \(2\ \text{m}\)

b.    \(12 \text{m}^3\)

c.    \(12\ 000\ \text{L}\)

Show Worked Solution
a.    \(a^2+b^2\) \(=c^2\)
  \(x^2+1.5^2\) \(=2.5^2\)
  \(x^2\) \(=2.5^2-1.5^2\)
  \(x^2\) \(=4\)
  \(x\) \(=\sqrt{4}=2\)

 
\(\therefore\ \text{The perpendicular height of the tent is }2\ \text{metres.}\)
 

b.    \(V\) \(=Ah\)
    \(=\Big(\dfrac{1}{2}\times 3\times 2\Big)\times 4\)
    \(=3\times 4\)
    \(=12\ \text{m}^3\)

 

c.    \(1\ \text{m}^3\) \(=1000\ \text{L}\)
  \(\therefore\ 12\ \text{m}^3\) \(=12\ 000\ \text{L}\)

  
\(\therefore\ \text{The capacity of the tent is }12\ 000\ \text{litres.}\)

Volume, SM-Bank 065

  1. Given the triangular face of the prism above is isosceles, use Pythagoras' Theorem to calculate its perpendicular height. Give your answer correct to the nearest whole centimetre.  (2 marks)
  2. Using your answer from (a), calculate the volume of the prism in cubic centimetres.  (2 marks)
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a.    \(7\ \text{cm}\)

b.    \(168\ \text{cm}^3\)

Show Worked Solution
a.    \(a^2+b^2\) \(=c^2\)
  \(a^2+3^2\) \(=7.6^2\)
  \(a^2\) \(=7.6^2-3^2\)
  \(a^2\) \(=48.76\)
  \(a\) \(=\sqrt{48.76}=6.982\dots\)
  \(a\) \(\approx 7\)

 
\(\therefore\ \text{The perpendicular height of the triangle is }7\ \text{cm, (nearest whole centimetre).}\)
 

b.    \(V\) \(=Ah\)
    \(=\Big(\dfrac{1}{2}\times 6\times 7\Big)\times 8\)
    \(=21\times 8\)
    \(=168\ \text{cm}^3\)

Volume, SM-Bank 064

  1. For the triangular prism above, use Pythagoras' Theorem to calculate the perpendicular height, \(x\), of the triangular face.  (2 marks)
  2. Using your answer from (a), calculate the volume of the prism in cubic millimetres.  (2 marks)
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a.    \(5\ \text{mm}\)

b.    \(480\ \text{mm}^3\)

Show Worked Solution
a.    \(a^2+b^2\) \(=c^2\)
  \(x^2+12^2\) \(=13^2\)
  \(x^2\) \(=13^2-12^2\)
  \(x^2\) \(=25\)
  \(x\) \(=\sqrt{25}=5\)

 
\(\therefore\ \text{The perpendicular height of the triangle is }5\ \text{mm}\)
 

b.    \(V\) \(=Ah\)
    \(=\Big(\dfrac{1}{2}\times 12\times 5\Big)\times 16\)
    \(=30\times 16\)
    \(=480\ \text{mm}^3\)

Volume, SM-Bank 063

Calculate the volume of the triangular prism below in cubic metres.  (2 marks)

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\(648\ \text{m}^3\)

Show Worked Solution
\(V\) \(=Ah\)
  \(=\Big(\dfrac{1}{2}\times 8\times 9\Big)\times 18\)
  \(=36\times 18\)
  \(=648\ \text{m}^3\)

Volume, SM-Bank 062

Calculate the volume of the triangular prism below in cubic centimetres.   (2 marks) 
 

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\(102.362\ \text{cm}^3\)

Show Worked Solution
\(V\) \(=Ah\)
  \(=\Big(\dfrac{1}{2}\times 3.1\times 5.2\Big)\times 12.7\)
  \(=8.06\times 12.7\)
  \(=102.362\ \text{cm}^3\)

Volume, SM-Bank 061

Calculate the volume of the triangular prism below in cubic metres.  (2 marks)

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\(105\ \text{m}^3\)

Show Worked Solution
\(V\) \(=Ah\)
  \(=\Big(\dfrac{1}{2}\times 6\times 5\Big)\times 7\)
  \(=15\times 7\)
  \(=105\ \text{m}^3\)

Volume, SM-Bank 031 MC

A timber door wedge is pictured below.
 

The wedge is in the shape of a triangular prism.

What is the volume of the wedge in cubic centimetres?

  1. \(7.5\ \text{cm}^3\)
  2. \(37.5\ \text{cm}^3\)
  3. \(75\ \text{cm}^3\)
  4. \(375\ \text{cm}^3\)
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\(B\)

Show Worked Solution

\(\text{10 mm =1 cm}\)

\(\text{Volume}\) \(=Ah\)
  \(=\bigg(\dfrac{1}{2}\times 10\times 3\bigg)\times 2.5\)
  \(=15\times 2.5\)
  \(=37.5\ \text{cm}^3\)

 
\(\Rightarrow\ B\)

Volume, SM-Bank 030 MC

A wheelchair ramp is pictured below.
  

 
The ramp is in the shape of a triangular prism.

What is the volume of the ramp?

  1. \(2.8\ \text{m}^3\)
  2. \(5.6\ \text{m}^3\)
  3. \(0.28\ \text{m}^3\)
  4. \(0.56\ \text{m}^3\)
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\(A\)

Show Worked Solution
\(\text{Volume}\) \(=Ah\)
  \(=(\dfrac{1}{2}\times 7\times 0.4)\times 2\)
  \(=1.4\times 2\)
  \(=2.8\ \text{m}^3\)

 
\(\Rightarrow\ A\)

Volume, SM-Bank 010

A rectangular block of land has width 50 metres and length 85 metres.

In order to build a house, the builders dig a hole in the block of land.

The hole has the shape of a right-triangular prism,  \(ABCDEF\).

The width \(AD\) = 20 m, length \(DC\) = 25 m and height \(EC\) = 4 m are shown in the diagram below.
 

Calculate the volume of the right-triangular prism, \(ABCDEF\), giving your answer in cubic metres.  (2 marks)

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\(1000\ \text{m}^3\)

Show Worked Solution

\(\text{Cross-section is a triangle}\)

\(A\) \(=\dfrac{1}{2}\times bh\)
  \(=\dfrac{1}{2}\times 25\times 4\)
  \(= 50\ \text{m}^2\)

 

\(\therefore\ V\) \(=50\times 20\)
  \(=1000\ \text{m}^3\)

Volume, SM-Bank 002 MC

A right triangular prism has a volume of 320 cm3.

A second right triangular prism is made with the same width, twice the height and three times the length of the prism shown.

The volume of the second prism (in cm3) is

  1. \(640\)
  2. \(960\)
  3. \(1280\)
  4. \(1920\)
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\(D\)

Show Worked Solution

\(\text{Volume of existing prism}\ (V)\)

\(=A\times l\)
\(=\Bigg(\dfrac{1}{2}\times b\times h\Bigg)\times l\)
\(=320\ \text{cm}^3\)

 

\(\text{Volume of new prism}\ (V_1)\)

\(=\Bigg(\dfrac{1}{2}\times b\times 2h\Bigg)\times 3l\)
\(=6\times\dfrac{1}{2}\times b\times h\times l\)
\(=6\times V\)
\(=6\times 320\)
\(=1920 \ \text{cm}^3\)

 
\(\Rightarrow D\)