Calamvale Community College Year 10 Formative Exam Notes

Detailed calculations for determining the volume of a composite structure involving a triangular prism situated atop a rectangular prism:
- The total volume is the summation of the individual volumes of the two primary components: the triangular prism and the rectangular prism.
Volume of the Triangular Prism:
- The formula for the volume of a triangular prism is defined as the area of the triangular base multiplied by the length: $\text{Volume} = (\frac{1}{2} \times \text{base} \times \text{height}) \times \text{length}$ .
- The dimensions provided are a base of $13\,cm$ , a vertical height of $7\,cm$ , and a length (depth) of $6\,cm$ .
- Calculation: $\frac{1}{2} \times 13\,cm \times 7\,cm \times 6\,cm = 273\,cm^3$ .
Volume of the Rectangular Prism:
- The formula for the volume of a rectangular prism is $\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}$ .
- The dimensions provided for the base structure are a length of $13\,cm$ , a width of $6\,cm$ , and a height of $8\,cm$ .
- Calculation: $13\,cm \times 6\,cm \times 8\,cm = 624\,cm^3$ .
Total Volume Integration:
- $\text{Total Volume} = 273\,cm^3 + 624\,cm^3 = 897\,cm^3$ .

Calculations for the surface area of a composite shape consisting of a sphere and a cylinder, with specified dimensions: a radius ( $r$ ) of $8\,m$ and a height ( $h$ ) for the cylinder of $11\,m$ .
Surface Area of the Sphere:
- The formula used is $SA_{\text{sphere}} = 4 \times \pi \times r^2$ .
- Calculation: $4 \times \pi \times 8^2 = 804.3\,m^2$ .
Surface Area of the Cylinder:
- The relevant component of the cylinder surface area is calculated as $2 \times \pi \times r \times h$ .
- Calculation: $2 \times \pi \times 8\,m \times 11\,m = 553\,m^2$ .
Total Surface Area (TSA):
- The sum of the sphere surface area and the specific cylinder area results in the total surface area.
- Calculation: $804.3\,m^2 + 553\,m^2 = 1357.2\,m^3$ .

Application of the Pythagorean theorem to solve a real-world navigational problem involving distance on a plane.
Scenario and Triangle Construction:
- A canoeist paddles $2.1\,km$ North and then $3.8\,km$ West.
- To find the shortest path back to the starting point, the distance is treated as the hypotenuse ( $c$ ) of a right-angled triangle where the legs are $a = 3.8\,km$ and $b = 2.1\,km$ .
Mathematical Execution:
- Formula: $c^2 = a^2 + b^2$ .
- Substitution: $c^2 = 3.8^2 + 2.1^2$ .
- Intermediate Step: $c^2 = 14.44 + 4.41 = 18.85$ .
- Solving for $c$ : $c = \sqrt{18.85}$ .
- Final Result: The distance is $4.3\,km$ (noted as the calculated hypotenuse).

Determination of an aeroplane's altitude using trigonometry and the angle of depression.
Problem Parameters:
- Distance from the runway: $2\,km$ .
- Angle of Depression: $10^{\circ}$ .
- Target: Altitude ( $h$ ) in metres.
Calculations:
- The relationship is defined by the tangent of the angle: $\tan(10^{\circ}) = \frac{\text{altitude}}{\text{distance}}$ .
- Rearranging to solve for altitude: $h = \tan(10^{\circ}) \times 2$ .
- Altitude in kilometres: $h = 0.35\,km$ .
- Conversion to metres: $0.35\,km \times 1000 = 352.7\,m$ .
- Rounded altitude to the nearest metre: $353\,m$ .

Analysis of a road sign based on an equilateral triangle to find height and total area.
Determining Height ( $h$ ):
- The sign has a side length of $84\,cm$ .
- To find the height, the triangle is divided into two right-angled triangles with a hypotenuse of $84\,cm$ and a base of $42\,cm$ ( $84$ \div $2$ ).
- Using Pythagoras: $a^2 = c^2 - b^2$ .
- Calculation: $a = \sqrt{84^2 - 42^2} = \sqrt{5292}$ .
- Resulting height ( $a$ ): $72.75\,cm$ .
Calculating Total Area:
- Formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ .
- Substitution: $\text{Area} = \frac{1}{2} \times 84\,cm \times 72.75\,cm$ .
- Final Area: $3055.50\,cm^2$ .