In the given figure, PQR is a triangle and quadrilateral ABCD is inscribed in it. QD = 2 cm, QC = 5 cm, CR = 3 cm. BR = 4 cm. PB = 6 cm. PA = 5 cm and AD = 3 cm. What is the area (in cm²) of the quadrilateral ABCD?

Solution

We draw a Ʇ PT bisect QR PT = 2√21 BY PGT   ∆ PQR = ½ × 8 × 2√21 = 8√21 Area of ∆ QDC = ½ × 2 × 5 × sin θ                                 = ½ × 2 × 5 × (2√21/10)                                 = √21 Area of ∆ BCR = ½ × 3 × 4 × sin θ = ½ × 3 × 4 × (2√21/10) =6√21 / 5 Area of ∆ PAB = ½ × 5 × 6 × sin (180 – 200) = ½ × 5 × 6 × sin2 θ = ½ × 5 × 6 × 2 sin θ cos θ = ½ × 5 × 6 × 2 × 2√21/10 × 4/10 = 12√21 / 5 = 8√21 – (√21 + 6√21/5 + 12√21/5) = 8√21 – (√21 + 18√21/5) = 8√21 – (5√21 + 18√21/5) = 8√21 – √21 (23/5) = 40√21 – √21 (23) / 5 = 17√21 / 5