Math Exam: Trapeziums, Cones, Triangles

• This is a practice exam with 36 questions covering various math topics.

Question 1: Area of a Trapezium

• The area of a trapezium is calculated using A = 1/2 (a+b)h
• Where a and b are the parallel sides, and h is the perpendicular height.
• In this case the area of the trapezium is 621cm²

Question 2: Volume Between Two Ice Cream Cones

• The volume of ice cream fitting between two cones is determined by V = V2 - V1
• V1 = (1/3)πr1^2h is the inner cone with radius 1.5cm and height 10cm.
• V2 = (1/3)πr2^2h is the outer cone with radius 5cm and height 10cm.
• The volume of ice-cream fitting between the two cones, rounded to the nearest cm³, is 238cm³.

Question 3: Length of a Triangle Side

• Using the Pythagorean theorem, a² = c² - b²
• The length of the unknown side (a) is approximately 14.32 cm.

Question 4: Length of the Hypotenuse

• The length of the hypotenuse (c) is determined by c = √(a² + b²)
• The length of the hypotenuse is approximately 33.30 cm.

Question 5: Value of y in a Triangle

• Applying the Pythagorean theorem: y = √(15² - 8²)
• The value of y, rounded to two decimal places, is 12.69 cm.

Question 6: Area of a Triangle

• Using Heron's formula, the area A = √[s(s-a)(s-b)(s-c)]
• s = (a+b+c)/2 where a, b, and c are the sides of the triangle.
• The area of the triangle, rounded to one decimal place, is 280.4 cm².

Question 7: Distance Between Locations in a Right-Angled Triangle

• The distance between locations A and C (hypotenuse) is determined by c = √(a² + b²)
• The distance between locations A and C, rounded to the nearest kilometre, is 12 km.

Question 8: Area of Earth Visible During a Solar Eclipse

• The area of the Earth visible is calculated by subtracting the Moon's area from the Earth's area: A = E - M
• AM = π * 1737^2 = 9478715.965 km²
• AE = π * 6371^2 = 127516118 km²
• The area of Earth visible ≈ 118037402 km².

Question 9: Perimeter of a Triangle

• The perimeter of the triangle is calculated by adding all sides.
• The perimeter of the triangle is 78 cm

Question 10: Volume of a Truncated Pyramid

• The volume of the truncated pyramid is the difference in volume between the whole and the top pyramids.
• V1 = (1/3) * (20*30) * 30 = 6000 m³
• V2 = (1/3) * (3*4.5) * 5 = 22.5 m³
• The volume of the truncated pyramid is 5977.5 m³.

Question 11: Volume of a Cylinder

• The volume of a cylinder is V = πr²h
• The volume of the cylinder is approximately 904.78 cm³.

Question 12: Length of PS in a 3D Rectangular Prism

• First, find the length of PR using the Pythagorean theorem: PR = √(19² + 11²) ≈ 21.95cm
• Then, find the length of PS using the Pythagorean theorem: PS = √(21.95² + 10²) ≈ 24.12cm
• The length of PS, rounded to two decimal places, is 24.12 cm.

Question 13: Similarity of Triangles

• Two triangles are similar if two pairs of corresponding sides are in the same ratio and the included angles are equal.
• Here, 580 x 2 / 87 = 20/3
• Here, 940 /141 = 20/3

Question 14: Finding the Value of y in Similar Triangles

• Determine the value of y using similar triangles and ratios of their corresponding sides
• SF (Scale Factor) = 9/15 y / 10 = (3/5) * 10
• The value of y is 6 cm.

Question 15: Volume of a Sphere After Scaling

• The volume is calculated as V = 300 * 125
• k³ = 5 *3
• The volume of the new sphere is 37,500 cm³.

Question 16: Identifying Similar Triangles

• Not similar

Question 17: Area of a Scaled Rectangular Poster

• New area is equal to A= 1500 x 2.25
• The area of a new poster is 3375 cm².

Question 18: Radius of a Smaller Cylinder

• Here, D = 12 x 3
• The radius of the smaller cylinder is 4cm.

Question 19: Determining the Value of z

• If the remaining area is mathematically similar to the swimming pool, then z = 10m

Question 20: Height of a Pine Tree

• Here, Height = (3 x 2)
• The pine tree is 6 m tall.

Question 21: Weight of a Scaled Chocolate Bar

• New weight = 250 x k3
• The increased weight would be 16,000 g.

Question 22: Value of x in Similar Triangles

• Using similar triangles and the ratios of corresponding sides, x = 5 cm.

Question 23: Scale Diagram Dimensions

• A scale diagram of the side view should be drawn with 6.25 cm by 5 cm dimensions

Question 24: Volume of Scaled Container

• The increased volume would be 64 L.

Question 25: Transposing the Formula

• The formula W = 4P + 2R transposed to make P the subject is P = (W - 2R) / 4.

Question 26: Solving for z in an Equation

• z = 8

Question 27: Value of C on a Segmented Graph

• The value of C when t = 5 equals 206

Question 28: Cost For An Electrician

• C = 100(4) + 200 = 600

Question 29: Rearranging Equation

• The expression 3x + 4y = 15 rearranged for y is y

Question 30: Solving for x

• x = c-b / 2a

Question 31: Value of b

• b = 15

Question 32: Cost of an Electrician

• C = 200(0) + 100 = 100.0

Question 33: Finding Value of a

• v = 22

Question 34: Expression Rearranged for y

• y = 11-22 / 4

Question 35: The Area of an Equilateral Triangle

• Finn wants to calculate the area for diamonds of bases 7 cm to 12 cm.

Question 36:

• Finn wants to calculate the area for diamonds of bases 5 cm to 10 cm.