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प्रश्न
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 following 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.
उत्तर
Given: ABCD is a rectangle, where AB = 51 cm and BC = 25 cm.
The parallel sides QC and PD of the trapezium PQCD are in the ratio of 9 : 8. Let QC = 9x and PD = 8x.
Now, the area of trapezium PQCD:
= `1/2` × (Sum of parallel sides) × (Distance between parallel sides)
= `1/2 xx (9x + 8x) xx 25 cm^2`
= `1/2 xx 17x xx 25`
Again, area of rectangle ABCD = BC × CD = 51 × 25
Now, according to the question,
Area of trapezium PQCD = `5/6` × Area of rectangle ABCD
= `1/2 xx 17x xx 25`
= `5/6 xx 51 xx 25`
x = `5/6 xx 51 xx 25 xx 2xx 1/(17 xx 25)`
x = 5
Therefore, the length of the trapezium PQCD, QC = 9x = 9 × 5 = 45 cm and PD = 8x = 8 × 5 = 40 cm.
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