NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0

Area of triangles and parallelograms

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A triangular display stand in a library in Melbourne is constructed using two identical right-angled triangles joined along their hypotenuses to form a parallelogram. One of the right-angled triangles has a base of 27/4 metres and a perpendicular height of 68/5 metres.

Using the formula for the area of a triangle, calculate the exact area of one triangular face in square metres. Express your answer as a simplified fraction.

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Base and perpendicular height. Using base 4 and perpendicular height 3, area = ½ × 4 × 3 = 6 square units.345
Using base 4 and perpendicular height 3, area = ½ × 4 × 3 = 6 square units.

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Explanation

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Worked solution

  1. The problem asks for the area of one triangular face of a display stand in Melbourne. The triangle is right-angled, with base = 27/4 metres and perpendicular height = 68/5 metres.
  2. The formula for the area of a triangle is: Area = (1/2) × base × height. This applies because the triangle is right-angled and the base and height are perpendicular.
  3. Substitute the given values into the formula: Area = (1/2) × (27/4) × (68/5).
  4. Multiply the numerators: 1 × 27 × 68 = 1836.
  5. Multiply the denominators: 2 × 4 × 5 = 40.
  6. The area is now 1836 / 40 square metres.
  7. Simplify the fraction 1836 / 40 by dividing both numerator and denominator by their greatest common divisor (GCD). First, divide both by 4: 1836 ÷ 4 = 459, 40 ÷ 4 = 10.
  8. Now the fraction is 459 / 10. Check if 459 and 10 have any common factors. 10 = 2 × 5; 459 is odd (not divisible by 2), and 459 ÷ 5 = 91.8 (not integer), so no further simplification is possible.
  9. The simplified area is 459/10 square metres.
  10. Check the calculation: (1/2) × (27/4) × (68/5) = (27 × 68) / (2 × 4 × 5) = 1836 / 40 = 459 / 10. Correct.
  11. The area of one triangular face is 459/10 square metres.
  12. Final answer: 459/10 square metres.

Reference answer: 459/10 square metres

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