Question

Dimensions of a rectangle abcd are 51cm×25cm. A trapezium pbcq with its parallel sides qc and pb in the ratio 9:8is cut off from the rectangle. If the area of the trapezium is 5/6 th partof the area of the rectangle, find the length of qc and pb

Answer 1

One way to calculate this is by assuming that the height of the trapezium is the same as that of the rectangle.

Area of the rectangle:   51cm x 25 cm = 1275cm²
Area of the trapezium= 5/6th of that of the rectangle, = 5/6 x 1275 = 1062.5cm²

Now that we have the area of the trapezium, we can use the formula for calculating the area of a trapezium to find length of qc and pb.

Formula for area of trapezium... =average of the 2 parallel sides x the height= 1/2(a+b)xh

1/2(qc + pb) x h ....height in this case = 25(height of the original rectangle)

=  1/2(qc+pb) x 25 = 1062.5 ...multiple both sides of the equation by 2
= (qc+pb)25=2125.......divide both sides by 25
= qc+pb= 85

Ratio of qc:pb  = 9:8    that means qc/pb = 9/8  and hence 9pb=8qc

pb= 8/9qc (substitute in the above equation)

qc + 8/9qc= 85   .....multiply both sides by 9
9qc + 8qc= 765
17qc= 765  ......divide both sides by 17
qc= 45

substitute qc to find pb ..... pb +45 = 85
                                           pb= 85-45 = 40

Therefore qc = 45 cm and,
                pb=  40 cm

Answer 2

Answer:

Answer:

Qc=45

Pd=40

Step-by-step explanation:

qc/pd=9/8 or pd=8*qc/9

area pqcd=5*25*51/6

qc*25+25*(pd-qc)/2=5*25*51/6

25qc+25/2*(8qc/9-qc)=5*25*17/2

by solving above  

qc=45

pd=8*45/9=40