Math, asked by sreenu204308, 9 months ago

In a Convex quadrilateral PQRS the
areas of triangle PQS , QRS and PQR
are in the ratio 3:4 1 . A line
through Q Cuts PR at o and
RS at B Such that PA. RR = RB:
RS Prove that A is the
mid point of PR and B is midpoind of Rs.​

Answers

Answered by amitnrw
1

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.​

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