Question

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 find the length of QC and PB

Answer 1

Answer:

Step-by-step explanation:

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 find the length of QC and PB

Area of Rectangle =  AB  * BC

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

Let say QC = 7x  then PB = 5x

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

=> Area of Trapezium = 6x * BC

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

=> 6x * BC = (4/7)  *  AB * BC

=> x = 2AB/21

QC = 7x =  2AB/3

PB = 5x = 10AB/21

if we say AB = 21 cm as per attached fig

then QC = 14cm   & PB  = 10 cm

Answer 2

Ans 14 and 10

Step-by-step explanation:

AREA OF RECTANGLE ABCD=21*5=105

AREA OF TRAPEZIUM IS 4/7AREA OF RECTANGLE SO AREA OF TRAPEZIUM IS 4/7*105

AREA OF TRAPEZIUM IS 60

RATIO OF ll sides of trapezium is 7:5

so sides are 7x and 5x

height is 5

so 60=1/2*(7x+5x)*5

x=2

so ll sides are 7*2=14

and 5*2=10

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