Given : A trapezium PBCQ with its PARALLEL sides QC and PB in the ratio of 7:5 is cut off from a rectangle ABCD if the area of trapezium is4/7 part of area of rectangle

समलम्ब PBCQ जिसकी समांतर भुजाओं QC तथा PB में  7:5 का अनुपात है , एक आयत ABCD में से काटी गई

समलम्ब का क्षेत्रफल, आयत के क्षेत्रफल का  4/7 भाग है,

To  find : the LENGTH of QC and PB  

QC तथा PB की लम्बाइयाँ

Solution :

Area of Rectangle  आयत के क्षेत्रफल  ABCD =  AB  * BC

Area of Trapezium  PBCQ  समलम्ब  क्षेत्रफल   = (1/2) ( PB + QC)  * BC

QC : PB = 7 : 5

Let say QC = 7x  then PB = 5x

=> Area of Trapezium ,  समलम्ब  क्षेत्रफल   = (1/2)(5x + 7x) * BC

=> Area of Trapezium ,  समलम्ब  क्षेत्रफल   = 6x * BC

Area of Trapezium  = (4/7) Area of Rectangle

 समलम्ब  क्षेत्रफल = (4/7)  आयत के क्षेत्रफल

=> 6x * BC = (4/7)  *  AB * BC

=> X = 2AB/21

QC = 7x =  14AB/21

PB = 5x = 10AB/21

 AB = 21 cm as PER attached fig आकृति में दिखाया गया

=> QC = 14 cm

   PB = 10 cm

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A trapezium PBCQ with its parallel sides QC and PB in the ratio of 7:5

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