Answer :

kvnmurty
we are to prove that  area of PQRS = area of ABRS.

see diagram.  Draw a perpendicular from B on to SR.  This is the perpendicular distance between PQ and SR. This is the perpendicular distance between AB and SR. 

Hence, area of parallelogram ABRS = RS * BG = area of parallelogram PQRS.

we have to prove that  area of triangle ANS = 1/2 area of PQRS.

N is on BR.  Draw a perpendicular NF from N on to SA.  Now the area of triangle ANS is = 1/2 * base * altitude = 1/2 * SA * NF

area of parallelogram ABRS = base * altitude = SA * NF

hence, area of triangle ANS = 1/2 area of parallelogram ABRS.

  
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kaynat87

Given : PQRS and ABRS are parallelograms and X is any point on side BR.

To prove : (i) ar (PQRS) = ar (ABRS) (ii) ar (AXS) = 1 2 ar (PQRS)

Proof : (i) In ∆ASP and BRQ, we have

∠SPA = ∠RQB [Corresponding angles] ...(1)

∠PAS = ∠QBR [Corresponding angles] ...(2)

∴ ∠PSA = ∠QRB [Angle sum property of a triangle] ...(3)

Also, PS = QR [Opposite sides of the parallelogram PQRS] ...(4)

So, ∆ASP ≅ ∆BRQ [ASA axiom, using (1), (3) and (4)]

Therefore, area of ∆PSA = area of ∆QRB [Congruent figures have equal areas] ...(5)

Now, ar (PQRS) = ar (PSA) + ar (ASRQ] = ar (QRB) + ar (ASRQ] = ar (ABRS)

So, ar (PQRS) = ar (ABRS) proved

(ii) Now, ∆AXS and ||gm ABRS are on the same base AS and between same parallels AS and BR

∴ area of ∆AXS = 1/ 2 area of ABRS

⇒ area of ∆AXS = 1/ 2 area of PQRS [ ar (PQRS) = ar (ABRS]

⇒ ar of (AXS) = 1/ 2 ar of (PQRS) proved

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