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The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle as shown in the Figure. If the area of the trapezium PQCD is $\frac{5}{6}th$ part of the area of the rectangle, find the lengths QC and PD.

Answers (1)

$[QC = 45\; cm, PD = 40\; cm]$

$Given, BC = 51\; cm \; and \; CD = 25\; cm$

$Area \; of\; rectangle\; ABCD = (51 \times 25) cm^{2}$

$In\; trapezium\; PQCD\; , parallel\; sides\; QC\; and\; PD\; are\; in\; the\; ratio\, 9:8$

Let length of QC = 9x and PD = 8x

$We\; have\: , Area\: of\: trapezium\: PQCD = \frac{5}{6}th\: part\; of\; Area(ABCD)$

$\frac{1}{2} \times (Sum\; of\; ||\; sides) \times height =\frac{5}{6} \times length \times breadth$

$\frac{1}{2}\times (PD + QC) \times CD = \frac{5}{6}\times BC \times CD$

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

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

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

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

$x = 5$

$PD = 8x$

$= 8\times 5$

$= 40\; cm$

$QC = 9x$

= 9 x 5 = 45

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infoexpert21

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