# How do you solve the following question in which a trapezium is cut-off from a rectangle?The dimensions of a rectangle ABCD are 51 cm x 25 cm. A trapezium PQCD with its parallel sides QC and PD in...

How do you solve the following question in which a trapezium is cut-off from a rectangle?

The dimensions of a rectangle ABCD are 51 cm x 25 cm. A trapezium PQCD 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 5/6th of the area of the rectangle, find the length of QC and PD.

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For this answer, assume ABCD a rectangle with AD=51 and AB=25; also we assume `bar(PD)` lies on `bar(AD)` and `bar(QC)` lies on `bar(CB)` :

The area of the rectangle is 51x25=1275`cm^2` . Since the area of the trapezium (trapezoid) is 5/6 that of the rectangle, the area of the trapezium is `5/6(1275)=1062.5` .

Another way to find the area of a trapezium is the formula `A=1/2(b_1+b_2)h` where A is the area, `b_1,b_2` the lengths of the parallel sides, and h the perpendicular distance between the parallel sides. With our assumption theheight is 25; we let the lengths of theparallel sides be 8x and 9x (since they are in a ratio of 8:9), so:

`1062.5=1/2(8x+9x)25`

`2125=17x(25)`

`17x=85`

`x=5`

With QC:PD = 9:8, we have QC=9*5=45 and PD=8*5=40.

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**The lengths of the parallel sides are QC=45,PD=40**

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Checking we get the area of the trapezium is `1/2(45+40)(25)=1062.5` as required.

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This is not the only possible solution. There are an infinite number of solutions to this problem. For instance, if AD=25 and AB=51 then we have PD=`1000/51` and ` ` QC=`1125/51` with a height of the trapezium being 51.

Also we need not have assumed that QC and PD lie on the sides of the rectangle, unless there is some definition of "cut-off" that dictates this. We would only need the endpoints D and C to coincide with the vertices of the rectangle; P and Q could then be in the interior.