// Numbas version: finer_feedback_settings {"name": "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}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Use formulae for the area and volume of geometric shapes", "tags": ["Area", "area", "area of a circle", "area of a trapezium", "area of a triangle", "Area of a triangle", "Circle", "circle", "cone", "Cone", "geometry", "taxonomy", "trapezium", "triangle", "Triangle", "volume", "Volume", "volume of a cone", "volume of a sphere"], "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{geogebra_applet('https://www.geogebra.org/m/jcUJu6F4',defs)}
\nAll lengths are in centimetres.
\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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