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Area, SM-Bank 039

The game of squash is played indoors on a court with a front wall, a back wall and two side walls, as shown in the image below.
 

 
Each side wall has the following dimensions.
 

The shaded region in the diagram above is considered part of the playing area.

Calculate the area, in square metres, of the shaded region in the diagram above. Round your answer to two decimal places.  (2 marks)

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\(32.66 \ \text{m}^2\)

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\(\text{Shaded Area (trapezium)}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{9.75}{2}\times (4.57 + 2.13)\)
  \(=32.6625\)
  \(= 32.66\ \text{m}^2 \ (2\ \text{d.p.)}\)

Area, SM-Bank 035

\(PQRS\) is a square of side length 4 m as shown in the diagram below.

The distance \(ST\) is 1 m.

Calculate the shaded area \(PQTS\) in square metres.  (2 marks)

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\(10\ \text{m}^2\)

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\(\text{Method 1:}\)

\(\text{Area of}\ \Delta QRT\) \(=\dfrac{1}{2}\times RT\times QR\)
  \(=\dfrac{1}{2}\times 3\times 4\)
  \(=6\ \text{m}^2\)

 
\(\therefore\ \text{Shaded Area}\ =4\times 4-6 =10\ \text{m}^2\)
 

\(\text{Method 2:}\)

\(\text{Area of Trapezium }PSQT\) \(=\dfrac{PS}{2}(ST+PQ)\)
  \(=\dfrac{4}{2}(1+4)\)
  \(=10\ \text{m}^2\)

Area, SM-Bank 017

Sequoia owns a farm with a rectangular paddock.

She increases the area of the paddock by adding land that changes it into the shape of a trapezium.
 

 What is the area of Sequoia's new paddock, in square metres?  (2 marks)

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\(2599\ \text{m}^2\)

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\(\text{Area trapezium}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{46}{2}(71+42)\)
  \(=2599\ \text{m}^2\)

Area, SM-Bank 016 MC

A trapezium is constructed on a grid of 10 rectangles.

Each rectangle measures  3 cm × 7 cm.
 


   

What is the area of the trapezium?

  1. \(150\ \text{cm}^2\)
  2. \(168\ \text{cm}^2\)
  3. \(189\ \text{cm}^2\)
  4. \(210\ \text{cm}^2\)
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\(B\)

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\(\text{Method 1: Composite}\)

\(\therefore\ \text{Total Area}\) \(=\text{Area 1 rectangle}+2\times\ \text{Area of triangle}\)
  \(=6\times 21+2\times\Bigg(\dfrac{1}{2}\times 3\times 14\Bigg)\)
  \(=126+42\)
  \(=168\ \text{cm}^2\)

  
\(\text{Method 2: Trapezium}\)

\(\text{Area}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{14}{2}(15+9)\)
  \(=168\ \text{cm}^2\)

 
\(\Rightarrow B\)

Area, SM-Bank 015

The floor of a chicken coop is in the shape of a trapezium.

The floor, \(ABCD\), and the chicken coop are shown below.

 

\(AB = 3\ \text{m}, BC = 2\ \text{m and}\ \ CD = 5\ \text{m.}\)
 

  1. What is the area of the floor of the chicken coop?

     

    Write your answer in square metres.  (2 marks)

  2. What is the perimeter of the floor of the chicken coop?

     

    Write your answer in metres, correct to one decimal place.  (2 marks)

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a.    \(8\ \text{m}^2\)

b.    \(12.8\ \text{m}\)

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a.    \(A\) \(=\dfrac{h}{2}(a+b)\)
    \(=\dfrac{2}{2}\times (3 + 5)\)
    \(=8\ \text{m}^2\)

 

b.  

\(\text{Using Pythagoras,}\)

\(AD^2\) \(=2^2+2^2\)
\(AD^2\) \(=8\)
\(AD\) \(=\sqrt{8}\)
  \(=2.82\dots\ \text{m}\)

 

\(\therefore\ \text{Perimeter}\) \(=3+2+5+2.82\dots\)
  \(=12.8\ \text{m  (1 d.p.)}\)

Area, SM-Bank 014 MC

The top of a table is in the shape of a trapezium, as shown below.

The area of the tabletop, in square centimetres, is

  1. \(200\)
  2. \(4200\)
  3. \(4800\)
  4. \(288\ 000\)
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\(B\)

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\(\text{Area}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{60}{2}(80+60)\)
  \(=4200\)

 
\(\Rightarrow B\)

Area, SM-Bank 013

Lucy designs an outdoor table that is in the shape of a trapezium.

The dimensions of the table top are shown in the picture below.

What is the area of Lucy's table top?   (2 marks)

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\(2600\ \text{cm}^2\)

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\(\text{Method 1:  Composite}\)

\(\text{Area}\) \(=\text{Area of rectangle}+2\times \text{Area of triangle}\)
  \(=(50\times 40) + 2\times\Bigg(\dfrac{1}{2}\times 15\times 40\Bigg)\)
  \(=2000 + 600\)
  \(=2600\ \text{cm}^2\)

 

\(\text{Method 2:  Trapezium}\)

\(\text{Area}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{40}{2}(80+50)\)
  \(=2600\ \text{cm}^2\)

Area, SM-Bank 012

Luke designs a table that is in the shape of a trapezium.

The dimensions of the table top are shown in the picture below.
 

What is the area of Luke's table top?   (2 marks)

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\(880\ \text{cm}^2\)

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\(\text{Method 1:  Composite}\)

\(\text{Area}\) \(=\text{Area of rectangle}+2\times \text{Area of triangle}\)
  \(=(38\times 20) + 2\times\Bigg(\dfrac{1}{2}\times 6\times 20\Bigg)\)
  \(=760 + 120\)
  \(=880\ \text{cm}^2\)

 

\(\text{Method 2:  Trapezium}\)

\(\text{Area}\) \(=\dfrac{h}{2}(a+b)\)
  \(=\dfrac{20}{2}(38+50)\)
  \(=880\ \text{cm}^2\)