In the given figure, PQR is the 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 five times area (in cm²) of the quadrilateral ABCD?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e. 17√21.
Area ΔPQR = (QR/4) √(4PQ² - QR²)
⇒ (8/4) √(400 - 64)
⇒ 2 × 4√21 = 8√21 cm²
Area of ΔPAB in ratio to Area of ΔPQR
⇒ (5 × 6)/(10 × 30) = 3/10
Area of ΔQDC ratio to area of ΔPQR
⇒ (2 × 5)/(10 × 8) = 10/80 = 1/8
ΔBCR ratio to ΔPQR
⇒ (4 × 3)/(10 × 8) = 12/80 = 3/20
Area (ABCD) = Area of triangle PQR - (Ar Δ PAB + ΔDQC +ΔBCR)
⇒ 8√21 - [3/10 + 1/8 + 3/20] × 8√21
⇒ 8√21 - (46/80) 8√21
⇒ (17√21)/5 cm²
Five times of the area = 17√21