Question

The dimensions of a rectangle ABCD are 51cmx25cm.A trapezium with its

parallel sides QC and PD in the ratio 9:8 is cut off from the rectangle. If the area of

the trapezium PQCD is 5/6 th part of the rectangle .Find the length of QC and PD.​

Answer 1

FINAL ANSWER:-

$QC=$ $45 cm$

$PD=40cm$

GIVEN:-

• dimensions of a rectangle ABCD are  

⇒ $length\:AB,DC\:\:(l) =51cm$  

⇒ $breadth\:AD,BC\:(b)=25cm$  

• dimensions of a trapezium PQCB are

⇒ $parallel \:side\:QC\:(p1)=9x$  

⇒ $parallel \:side\:PB\:(p1)=8x$

⇒ $height\:BC \:(h)=25cm$

• area of the trapezium PQCD is  $\frac{5}{6} ^{th}$part of the rectangle

TO FIND:-

Find the length of QC(9x) and PD(8x)

FORMULAS USED:-

• $area\:of\:trapezium =\frac{1}{2} *(p1+p2)*h$

• $area\:of\:rectangle=l*b$

SOLUTION:-

according to question it is given that area of the trapezium PQCD is 5/6 th part of the rectangle

so

↔ $area\:of\:trapezium=\frac{5}{6} *area\:of\:rectangle$ ↔

$\frac{1}{2} *(p1+p2)*h=\frac{5}{6} *l*b$  

$\frac{1}{2} *(PB+QC)*BC=\frac{5}{6} *AB*BC$

∵ put values in equation

$\frac{1}{2} *(9x+8x)*25=\frac{5}{6} *51*25$

$\frac{1}{2} *17x*25=\frac{5}{6} *51*25$

$\frac{ 42 5x }{2} =\frac{6375}{6}$

$x=\frac{6375*2}{6*425}$ [∵ by cross multiplication]

$x=5$

∴ length of QC and PD

• QC ⇒ 9x ⇒ 9×5 ⇒ $45 cm$

• PD ⇒ 8x ⇒  8×5 ⇒ $40cm$

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Answer 2

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