Question 1: Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.

Solution:

The diagonal AC divides the quadrilateral ABCD into two triangles: ΔABC and ΔADC.

1. For ΔABC: Sides are a = 3 cm, b = 4 cm, c = 5 cm.
Since 3² + 4² = 9 + 16 = 25 = 5², ΔABC is a right-angled triangle at B.
Area(ΔABC) = 1/2 × Base × Height = 1/2 × 3 × 4 = 6 cm².

2. For ΔADC: Sides are a = 5 cm, b = 4 cm, c = 5 cm.
Semi-perimeter, s = (5 + 4 + 5) / 2 = 14 / 2 = 7 cm.
Area(ΔADC) = √(s(s - a)(s - b)(s - c)) = √(7(7 - 5)(7 - 4)(7 - 5)) = √(7 × 2 × 3 × 2) = 2√21 cm² ≈ 9.16 cm².

Total Area of Quadrilateral ABCD: Area(ΔABC) + Area(ΔADC) = 6 + 9.16 = 15.16 cm².

Question 2: A park in the shape of a quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m, and AD = 8 m. How much area does it occupy?

Solution:

Join diagonal BD to divide the quadrilateral into ΔBCD and ΔABD.

1. In right-angled ΔBCD:
By Pythagoras Theorem, BD² = BC² + CD² = 12² + 5² = 144 + 25 = 169 ⇒ BD = 13 m.
Area(ΔBCD) = 1/2 × Base × Height = 1/2 × 12 × 5 = 30 m².

2. In ΔABD: Sides are a = 9 m, b = 8 m, c = 13 m.
Semi-perimeter, s = (9 + 8 + 13) / 2 = 30 / 2 = 15 m.
Area(ΔABD) = √(15(15 - 9)(15 - 8)(15 - 13)) = √(15 × 6 × 7 × 2) = 6√35 m² ≈ 35.5 m².

Total Area of the Park: Area(ΔBCD) + Area(ΔABD) = 30 + 35.5 = 65.5 m².