Question

The adjacent sides of a parallelogram are 36 and 27 in length . If the perpendicular distance between the shorter sides is 12 cm find the distance between the longer sides

Answer 1

Let the parallelogram be ABCD with AB=CD=36 cm and AD=CB=27 cm. Let DE be the altitude from D on BC and AF be the altitude from A on CD. The shortest distance between the shorter sides is 12 cm. We know that the perpendicular distance is the shortest distance between two points. So we can say that altitude DE is the shortest distance between the two short sides AD & BC. Therefore, DE=12 cm. Now, we know that Area of Parallelogram = ½ X Base X Height. So Area of Parallelogram ABCD = ½ X AD X DE = ½ X AF X DC     Area of Parallelogram ABCD = ½ X 27 X 12 = ½ X AF X 36     Area of Parallelogram ABCD = ½ X 324 = ½ X AF X 36                                                      =  324 =  AF X 36                                                  =  AF   = 324/36                                                =  AF   =  9 cm         Therefore the shortest distance between the two loger sides is 9 cm

Answer 2

Area of parallelogram = base × height
First taking shorter side as base
Area = 27 × altitude
= 27 × 12
= 324 cm^2
Now taking longer side
Area = 36 × altitude
324 = 36 × altitude
Altitude = 324 / 36
= 9 cm