A Quadrilateral PQRS circumscribes a circle with center O and PQ||RS and PQ thrice of RS.length of QR,SP are equal but not parallel.Perimeter of quadrilateral iss 24...then area of quadrilateral=?

• from diagram,
  PQ=3RS => if RS=x, then PQ = 3x
  PS=QR = y(let)
  
  perimeter = 24
  => x+3x+y+y=24
  => 4x+2y=24
  
  now, PQ+RS=PS+QR
  => X+3X=Y+Y
  => 4x=2y
  
  solving eqns we get y=6,x=3
  sides of quadilateral are 3,9,6,6
  
  s=(a+b+c+d)/2 = (3+9+6+6)/2 = 12
  A=[(s-a)(s-b)(s-c)(s-d)]^1/2 = [(12-3)(12-9)(12-6)(12-6)]^1/2 = 18√3
• 10 years agoHelpfull: Yes(47) No(4)
• @rakesh- how PQ+RS=PS+QR ??
• 10 years agoHelpfull: Yes(14) No(0)
• @SHLAGHA SHARMA
  we know that,two tangents from a point outside a circle to the circle are equal,
  using this property it can be proved that PQ+RS=PS+QR
  If the lengths of the tangents from vertices P, Q, R, S of a quadrilateral
  circumscribed around a circle are denoted by p, q, r, s then
  
  PQ=p+q
  RS=r+s
  PS=p+s
  QR=q+r
  
  then PQ+RS=PS+QR=(p+q+r+s)
• 10 years agoHelpfull: Yes(14) No(1)
• @rakesh
  area of quadrilateral??
  From where have you applied this formulae??
• 10 years agoHelpfull: Yes(1) No(0)
• xplanation :
  
  
  As the area of square is 36cm2 its side will be 6cm
  
  So the perimeter of quadrilateral = perimeter of square = 6x4cm = 24cm
  
  Therefore (2a + 2b + 2c + 2d) = 24cm or (a + b + c + 2) = 12cm
  
  As RS = 3PQ , therefore (c + d) = 3(a + b)
  
  Therefore: (a +b + 3a + 3b) = 12cm or (a +b ) = 3cm
  
  therefore PQ = 3cm , RS = 9cm . Also PS = QR = 6cm
  
  now the height of quadrilateral can be calculated as: h2 = 62 - 32 = 27
  
  Therefore height of quadrilateral = √27 = 3√3
  
  As it is a trapezium ,so its area = 1/2 [PQ + RS] x height
  
  = 1/2 x (3 + 9) x 3√3
  
  = 1/2 x 12 x 3√3 = 18√3 cm2
• 7 years agoHelpfull: Yes(1) No(0)
• @RAKESH
  can u please tell me where u have got the area of quadrilateral = ((s-a)(s-b)(s-c)(s-d))^1/2
  
  i know that
  area of quadrilateral = ((s-a)(s-b)(s-c)(s-d) - abcd (cos^2(theta)/2))^1/2
  where theta = sum of two opposite angles
• 7 years agoHelpfull: Yes(0) No(0)
• @sir pls upload full question atleast......
• 7 years agoHelpfull: Yes(0) No(0)
• check out thi solution:-
  http://www.logicguns.com/q/a-quadrilateral-pqrs-circumscribes-a-circle-with-centre-oit-is-given-that-pq-is-parallel-to-rsalso-length-of-pq-is-thrice-that-of-rs-while-length-of-q/5603a2d82b78726726a80909/
• 7 years agoHelpfull: Yes(0) No(0)