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In the given figure, ABCD is a rectangle...

In the given figure, ABCD is a rectangle of length 51 cm and bredth 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 `(5)/(6)`th part of the area of the rectangle, find the lengths QC and PD.

Text Solution

Verified by Experts

Length of rectangle ABCD = 51 cm.
Breadth of rectangle ABCD = 25 cm.
Area of rectangle ABCD = `(51 xx 25)cm^(2) = 1275 cm^(2)`.
Let QC = 9k and PD = 8k. Then,
area(trap. PQCD)
`=((1)/(2) xx "sum of parallel sides"xx"distance between them")`
`={(1)/(2) xx (9k + 8k)xx 25}cm^(2)=((17 xx 25)/(2))"k cm"^(2)`.
Now, area (trap. PQCD) = `(5)/(6) xx` (area of rectangle ABCD)
`rArr" "((17 xx 25)/(2))k=((5)/(6)xx1275)`
`rArr" "k=((5)/(6)xx1275xx(2)/(17 xx 25)) rArr k = 5`.
`therefore" "QC = (9k)cm =(9 xx 5)`cm = 45 cm
and PD = (8k) cm = `(8 xx 5)`cm = 40 cm.
Hence, QC = 45 cm and PD = 40 cm.
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