In PQRS, the areas of triangles PQS, QRS & PQR are in ratio 3:4:1. A line through Q cuts PR at A & RS at B such that PA:PR = RB:RS. Prove that A is the midpoint of PR & B is the midpoint of RS.
Answers
A is the midpoint of PR & B is the midpoint of RS.
Step-by-step explanation:
Let sat Area of PQS = 3x
then Area of QRS = 4x
Area of PQR = x
Area of PQRS = Area of PQS + Area of QRS = 3x + 4x = 7x
Area of PRS = Area of PQRS - Area of PQR = 7x - x = 6x
Let say Area of ΔPOS = a
Then Area of ΔPOQ = Area of PQS - Area of ΔPOS = 3x - a
Area of ΔROS = Area of PRS - Area of ΔPOS = 6x - a
Area of ΔQOR = Area of PQR - Area of ΔPOQ = x - (3x - a) = a - 2x
Let say
PA/PR = RB/RS = k/(k + 1)
Area of ΔPOS/ Area of ΔROS = Area of ΔPOQ/Area of ΔQOR
=> a / (6 x - a) = (3x - a)/(a - 2x)
=> a² - 2ax = 18x² + a² - 9ax
=> 18x² = 7ax
=> 18x = 7a
=> x = 7a/18
Area of Δ AOS = Area of Δ PAS - Area of ΔPOS
=> Area of Δ AOS = 6x * k/(k + 1) - a = 7a k/3(k + 1) - a
Finding all areas & Equating on solving we get
4k² - 3k - 1 =0
=> 4k² - 4k + k - 1 =0
=> 4k(k - 1) + 1 (k - 1) = 0
=> k = 1 or k = -1/4
k = 1
Hence PA/PR = RB/RS = 1/2
=> A is the midpoint of PR & B is the midpoint of RS.
