Q. 63.7( 22 Votes )

P is a point in t

Answer :


i) Construct segment GH parallel to AB and CD passing through point P as shown


Also construct perpendiculars PI and PJ on segments CD and AB respectively


Consider parallelogram DCGH


Base = CD …from figure


Height = PI …from figure


Area of parallelogram = base × height


Area of parallelogram DCGH = CD × PI …(i)


Consider parallelogram ABGH


Base = AB …from figure


Height = PJ …from figure


Area of parallelogram = base × height


Area of parallelogram ABGH = AB × PJ …(ii)


Area of triangle = × base × height


For ΔPCD


Base = CD


Height = PI


Area of ΔPCD = × CD × PI


Using (i)


Area of ΔPCD = × Area of parallelogram DCGH …(iii)


For ΔAPB


Base = AB


Height = PJ


Area of ΔAPB = × AP × PJ


Using (ii)


Area of ΔAPB = × Area of parallelogram ABGH …(iv)


Add (iii) and (iv)


Area of ΔPCD + Area of ΔAPB = × Area of parallelogram DCGH + × Area of parallelogram ABGH


Area of ΔPCD + Area of ΔAPB = × (Area of parallelogram DCGH + Area of parallelogram ABGH)


From figure Area of parallelogram DCGH + Area of parallelogram ABGH = area of parallelogram ABCD


Therefore, Area of ΔPCD + Area of ΔAPB = × area of parallelogram ABCD


area of parallelogram ABCD = 2 × Area of ΔPCD + 2 × Area of ΔAPB …(*)


ii) from figure


area(ΔDPC) = area(DCGH) – area(ΔDHP) – area(CPG) …(i)


area(ΔAPB) = area(ABGH) – area(ΔAPH) – area(BPG) …(ii)


add equation (i) and (ii)


area(ΔDPC) + area(ΔAPB) = [area(DCGH) + area(ABGH)] – [area(ΔDHP) + area(ΔAPH)] – [area(CPG) + area(BPG)]


area(ΔDPC) + area(ΔAPB) = area(ABCD) – area(APD) – area(BPC)


Using equation (*) from first part of question


area(ΔDPC) + area(ΔAPB) = 2 × Area of ΔPCD + 2 × Area of ΔAPB – area(APD) – area(BPC)


Rearranging the terms we get


area(APD) + area(BPC) = 2 × Area of ΔPCD + 2 × Area of ΔAPB - area(ΔDPC) - area(ΔAPB)


therefore, area(APD) + area(BPC) = area(ΔPCD) + area(ΔAPB)


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