// Numbas version: finer_feedback_settings {"name": "Area of geometric shapes", "metadata": {"description": "
Apply formulas to calculate the areas of various shapes.
", "licence": "Creative Commons Attribution 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questions": [{"name": "Chris's copy of Use formulae for the area and volume of geometric shapes", "extensions": ["geogebra"], "custom_part_types": [], "resources": [["question-resources/icecrea_QqVaCIf.svg", "/srv/numbas/media/question-resources/icecrea_QqVaCIf.svg"], ["question-resources/frisbee_variable_TESZa4J.svg", "/srv/numbas/media/question-resources/frisbee_variable_TESZa4J.svg"], ["question-resources/tennis-ball_with_variable_MBOLQeM.svg", "/srv/numbas/media/question-resources/tennis-ball_with_variable_MBOLQeM.svg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Aiden McCall", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1592/"}], "tags": [], "metadata": {"description": "Substitute values into formulae for the area or volume of various geometric objects.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Answer the following questions by substituting the correct values into the given equations.
", "advice": "When inserting numbers into your calculator, make sure that you place brackets correctly.
\nWe can see from the diagram that the radius of the frisbee is $\\var{mccall[2]}$ $\\mathrm{cm}$.
Replacing the letter $r$ in the formula for the area of a circle with $\\var{mccall[2]}$ gives,
\\begin{align}
\\mathrm{Area} &= \\pi r^2 \\\\
&= \\pi\\times(\\var{mccall[2]})^2 \\\\
&= \\var{dpformat((mccall[2])^2, 2)}\\pi\\, \\mathrm{cm}^2 \\\\
&= \\var{dpformat(pi (mccall[2])^2, 2)}\\, \\mathrm{cm}^2 \\quad \\text{to 2 d.p.}
\\end{align}
We can see from the diagram that the triangle has two sides with lengths $\\var{length_cdp2}$ $\\mathrm{cm}$, $\\var{length_bdp2}$ $\\mathrm{cm}$ and an angle $\\var{c_thetadp2}\\mathrm{°}$ .
Replacing the letters $a$, $b$ and $C$ in the formula for the area of a triangle with $\\var{length_cdp2}$, $\\var{length_bdp2}$ and $\\var{c_thetadp2}$ respectively gives,
\\begin{align}
\\mathrm{Area} &= \\frac{1}{2}ab\\sin{C} \\\\
&= \\frac{1}{2} \\times \\var{length_cdp2} \\times \\var{length_bdp2} \\times \\sin(\\var{c_thetadp2}) \\\\
&= \\var{dpformat(0.5*(length_cdp2)*(length_bdp2)*sin(c_thetadp2* pi/180), 5)}\\, \\mathrm{cm}^2 \\\\
&= \\var{dpformat(0.5*(length_cdp2)*(length_bdp2)*sin(c_thetadp2 * pi/180), 2)}\\, \\mathrm{cm}^2 \\quad \\text{to 2 d.p.}
\\end{align}
We can see from the diagram that the radius of the cone is $\\var{r}$ $\\mathrm{cm}$ and the height is $\\var{h}$ $\\mathrm{cm}$.
Replacing the letters $r$ and $h$ in the formula for the volume of a cone with $\\var{r}$ $\\mathrm{cm}$ and $\\var{h}$ $\\mathrm{cm}$ respectively gives,
\\begin{align}
\\mathrm{Volume} &= \\frac{h}{3} \\pi r^2 \\\\
&= \\frac{(\\var{h})}{3} \\times \\pi \\times (\\var{r})^2 \\\\
&= \\var{dpformat((pi)*(h/3)*(r)^2 , 5)}\\, \\mathrm{cm}^3 \\\\
&=\\var{dpformat(h/3 * pi * (r)^2, 1)}\\, \\mathrm{cm}^3 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
\n
We can see from the diagram that the radius of the tennis ball is $\\var{mccall[1]}$ $\\mathrm{cm}$.
Replacing the letter $r$ in the formula for the volume of a sphere with $\\var{mccall[1]}$ gives,
\\begin{align}
\\mathrm{Volume} &= \\frac{4}{3} \\pi r^3 \\\\
&= \\frac{(4)}{(3)} \\times \\pi \\times (\\var{mccall[1]})^3 \\\\
&= \\var{dpformat((4/3)*pi*mccall[1]^3, 5)}\\, \\mathrm{cm}^3 \\\\
&= \\var{precround(((4/3)* pi) *(mccall[1])^3, 1)}\\, \\mathrm{cm}^3 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
We can see from the diagram that the trapezium has two parallel sides with length $\\var{trap_length_a}$ $\\mathrm{cm}$, $\\var{trap_length_b}$ $\\mathrm{cm}$ and height $\\var{trap_h}$ $\\mathrm{cm}$.
Replacing the letters $a$, $b$ and $h$ in the formula for the area of a trapezium with $\\var{trap_length_a}$, $\\var{trap_length_b}$ and $\\var{trap_h}$ respectively gives,
\\begin{align}
\\mathrm{Area} &= \\frac{1}{2} (a + b) h \\\\
&= \\frac{1}{2} \\times (\\var{trap_length_a} + \\var{trap_length_b}) \\times \\var{trap_h} \\\\
&= \\var{precround((0.5) (trap_length_a + trap_length_b) trap_h, 1)}\\, \\mathrm{cm}^2 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
\\begin{align}
\\mathrm{Area} &= \\frac{1}{2} (a + b) h \\\\
&= \\frac{1}{2} \\times (\\var{trap_length_a} + \\var{trap_length_b}) \\times \\var{trap_h} \\\\
&= \\var{precround((0.5)(trap_length_a +trap_length_b) trap_h, 2)}\\, \\mathrm{cm}^2 \\\\
&= \\var{precround((0.5) (trap_length_a + trap_length_b) trap_h, 1)}\\, \\mathrm{cm}^2 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
Rounded value for the length of c.
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\n\\[\\mathrm{Area} = \\pi r^2.\\]
\n\n$\\mathrm{Area}$ = [[0]] $\\mathrm{cm}^2$ Round your answer to 2 decimal places.
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\n\\[\\mathrm{Volume} = \\frac{h}{3} \\pi r^2.\\]
\n\n$\\mathrm{Volume}$ = [[0]] $\\mathrm{cm}^3$ Round your answer to 1 decimal place.
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\n\\[\\mathrm{Volume}= \\frac{4}{3} \\pi r^3.\\]
\n\n$\\mathrm{Volume}$ = [[0]] $\\mathrm{cm}^3$ Round your answer to 1 decimal place.
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\n\\[\\mathrm{Area} = \\frac{1}{2}(a+b) h .\\]
\n{geogebra_applet('https://www.geogebra.org/m/Gtjzajb6',trap_defs)}
\nAll lengths are given in metres.
\n$\\mathrm{Area}$ = [[0]] $\\mathrm{m}^2$ Round your answer to 1 decimal place.
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", "Underestimate and therefore we round each measurement down.
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\n[[0]]
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\n[[0]] m2
\n", "steps": [{"scripts": {}, "variableReplacements": [], "type": "information", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "showFeedbackIcon": true, "prompt": "
The room is {length}m long, {width}m wide, and {height}m high.
\nRound each measurement in the direction you decided on above.
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", "minValue": "ceil(width)", "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "variableReplacements": [], "marks": "0.1", "showCorrectAnswer": true}, {"correctAnswerFraction": false, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "maxValue": "ceil(height)", "showFeedbackIcon": true, "prompt": "Round the height to the nearest metre.
", "minValue": "ceil(height)", "correctAnswerStyle": "plain", "mustBeReducedPC": 0, "mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "variableReplacements": [], "marks": "0.1", "showCorrectAnswer": true}], "marks": 0}, {"scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "gaps": [{"correctAnswerFraction": false, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "mustBeReducedPC": 0, "maxValue": "buckets", "showFeedbackIcon": true, "minValue": "buckets", "correctAnswerStyle": "plain", "allowFractions": false, "mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "variableReplacements": [], "marks": 1, "showCorrectAnswer": true}], "showFeedbackIcon": true, "prompt": "One bucket of {colour} paint is enough to paint an area of 15m2. How many buckets should {person['name']} buy to ensure {pronouns['they']} {if(person['gender']='neutral','have','has')} enough paint?
\n[[0]]
", "marks": 0}], "ungrouped_variables": [], "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Estimate the number of buckets of paint to buy, by rounding measurements of a room up to the nearest metre and estimating the total area.
"}, "preamble": {"css": "", "js": ""}, "advice": "It is much better to have spare paint than not to have enough of it. So it is better to overestimate the area.
\nTherefore, we round each measurement up.
\nWe round each of our measurements up to the nearest whole metre:
\nLength: $\\var{length}\\,\\mathrm{m} \\approx \\var{l}\\,\\mathrm{m}$.
\nWidth: $\\var{width}\\,\\mathrm{m} \\approx \\var{w}\\,\\mathrm{m}$.
\nHeight: $\\var{height}\\,\\mathrm{m} \\approx \\var{h}\\,\\mathrm{m}$.
\nThe total area consists of five areas: two walls of $\\var{l}\\,\\text{m} \\times \\var{h}\\,\\text{m}$ (length by height); two walls of $\\var{w}\\,\\text{m} \\times \\var{h}\\,\\text{m}$ (width by height); and a ceiling of $\\var{l}\\,\\text{m} \\times \\var{w}\\,\\text{m}$ (length by width).
\n\\[ \\begin{align}
\\var{l}\\,\\text{m} \\times \\var{h}\\,\\text{m} &= \\var{l*h}\\,\\text{m}^2
\\\\ \\var{w}\\,\\text{m} \\times \\var{h}\\,\\text{m} &= \\var{w*h}\\,\\text{m}^2
\\\\ \\var{l}\\,\\text{m} \\times \\var{w}\\,\\text{m} &= \\var{l*w}\\,\\text{m}^2
\\end{align}\\]
Therefore, the total area {person['name']} needs to paint is
\n\\[ \\var{2*l*h} + \\var{2*w*h} + \\var{l*w} \\,\\mathrm{m}^2 = \\var{rall}\\,\\mathrm{m}^2 \\text{.} \\]
\nThe exact number of buckets needed is
\n\\[\\var{rall}\\,\\text{m}^2 \\div 15\\,\\text{m}^2 = \\var{rall/15} \\text{.}\\]
\n{person['name']} can only buy a whole number of buckets, so {pronouns['they']} need{verbs} to decide between {buckets-1} and {buckets} paint buckets. As it is better to buy more paint than not buy enough, {pronouns['they']} should buy {buckets} buckets of {colour} paint.
"}, {"name": "Calculate the areas of polygons", "extensions": [], "custom_part_types": [], "resources": [["question-resources/trapezium.svg", "/srv/numbas/media/question-resources/trapezium.svg"], ["question-resources/trangle.svg", "/srv/numbas/media/question-resources/trangle.svg"], ["question-resources/parallelogram.svg", "/srv/numbas/media/question-resources/parallelogram.svg"], ["question-resources/Parallelogram_area_animated.gif", "/srv/numbas/media/question-resources/Parallelogram_area_animated.gif"], ["question-resources/rectangle_zISmvoz.svg", "/srv/numbas/media/question-resources/rectangle_zISmvoz.svg"], ["question-resources/hardertrapezium_8GqMwOo.svg", "/srv/numbas/media/question-resources/hardertrapezium_8GqMwOo.svg"], ["question-resources/Trap_advice.svg", "/srv/numbas/media/question-resources/Trap_advice.svg"], ["question-resources/Triangle_advice_lD6eKvD.svg", "/srv/numbas/media/question-resources/Triangle_advice_lD6eKvD.svg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Vicky Hall", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/659/"}, {"name": "Aiden McCall", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1592/"}], "rulesets": {}, "functions": {}, "ungrouped_variables": [], "metadata": {"description": "This question tests the students ability to calculate the area of different 2D shapes given the units and measurements required. The formulae for the areas are available if required but students are encouraged to try to remember them themselves.
\nThe shapes are: a rectangle, a parallelogram, a right-angled triangle, and a trapezium.
\nAuthor of gif: Picknick
https://commons.wikimedia.org/wiki/File:Parallelogram_area_animated.gif
This file is licensed under the Creative Commons Attribution-Share Alike 4.0 International license.
a)
\nThe area of a rectangle is calculated using the formula
\n\\[\\mathrm{Area} = \\mathrm{base} \\times \\mathrm{height}\\text{.}\\]
\nWe have a base of $\\var{w0}$m and a height $\\var{h0}$m, therefore
\n\\begin{align}
\\mathrm{Area} &= \\mathrm{base} \\times \\mathrm{height} \\\\
&= \\var{w0} \\times \\var{h0} \\\\ &= \\var{w0*h0} \\\\
&= \\var{dpformat(w0*h0,1)}\\, \\mathrm{m}^2 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
\\begin{align}
\\mathrm{Area} &= \\mathrm{base} \\times \\mathrm{height} \\\\
&= \\var{w0} \\times \\var{h0} \\\\
&= \\var{dpformat(w0*h0,1)}\\, \\mathrm{m}^2 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
b)
\nThe parallelogram is just a slanted rectangle:
\n\n\nTherefore, the area of a parallelogram is calculated using the formula
\n\\[\\mathrm{Area} = \\mathrm{base} \\times \\mathrm{height}.\\]
\nWe have a base $\\var{w1}$m and perpendicular height $\\var{h1}$m.
\n\\begin{align}
\\mathrm{Area} &= \\mathrm{base} \\times \\mathrm{height} \\\\
&= \\var{w1} \\times \\var{h1} \\\\ &= \\var{{w1}{h1}}\\, \\mathrm{m}^2 \\\\
&= \\var{dpformat({w1}{h1},1)}\\, \\mathrm{m}^2 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
\\begin{align}
\\mathrm{Area} &= \\mathrm{base} \\times \\mathrm{height} \\\\
&= \\var{w1} \\times \\var{h1} \\\\
&= \\var{dpformat({w1}{h1},1)}\\, \\mathrm{m}^2 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
The area of a triangle is calculated using the formula
\n\\[\\mathrm{Area} = \\frac{\\mathrm{base} \\times \\mathrm{height}}{2}.\\]
\nNote that the triangle is half of a rectangle:
\n\nOur triangle has a base $\\var{w2}$m and a height $\\var{h2}$m, therefore
\n\\begin{align} \\mathrm{Area} &= \\frac{1}{2} \\times \\mathrm{base} \\times \\mathrm{height} \\\\
&= \\frac{1}{2} \\times \\var{w2} \\times \\var{h2} \\\\
&= \\var{0.5*w2*h2}\\, \\mathrm{m}^2 \\\\
&= \\var{dpformat(0.5*w2*h2, 1)}\\, \\mathrm{m}^2 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
\\begin{align} \\mathrm{Area} &= \\frac{1}{2} \\times \\mathrm{base} \\times \\mathrm{height} \\\\
&= \\frac{1}{2} \\times \\var{w2} \\times \\var{h2} \\\\
&= \\var{dpformat(0.5*w2*h2, 1)}\\, \\mathrm{m}^2 \\quad \\text{to 1 d.p.} \\\\
\\end{align}
d)
\n\nA trapezium can be interpreted as half of a parallelogram, this is shown below:
\n\nAs we only want the area of one half of this shape, the area is half of
\n\\[\\mathrm{area} = (a+b) \\times \\mathrm{height}\\text{,}\\]
\nwith ${a} = \\var{w5a}$m, ${b} = \\var{w5b}$m, and height $\\var{h5}$m.
\n\\begin{align}
\\mathrm{Area} &= \\frac{(a+b)}{2} \\times \\mathrm{height} \\\\
&= \\frac{(\\var{w5a}+\\var{w5b})}{2} \\times \\var{h5} \\\\
&= \\var{(w5a+w5b)*0.5} \\times \\var{h5} \\\\
&= \\var{(w5a+w5b)*(h5)/2}\\, \\mathrm{m}^2 \\\\
&= \\var{dpformat((w5a+w5b)*(h5)/2, 1)}\\, \\mathrm{m}^2 \\quad \\text{to 1 d.p.}
\\end{align}
\\begin{align}
\\mathrm{Area} &= \\frac{(a+b)}{2} \\times \\mathrm{height} \\\\
&= \\frac{(\\var{w5a}+\\var{w5b})}{2} \\times \\var{h5} \\\\
&= \\var{(w5a+w5b)*0.5} \\times \\var{h5} \\\\
&= \\var{dpformat((w5a+w5b)*(h5)/2, 1)}\\, \\mathrm{m}^2 \\quad \\text{to 1 d.p.}
\\end{align}
Calculate the area of the following shapes.
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