Question

Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm

from each other. Calculate this area as

(i) the sum of the areas of two triangles and one rectangle.

(ii) the difference of the area of a rectangle and the sum of the areas of two triangles.​

Answer 1

Answer:

Length of the parallel sides of a trapezium are 10 cm and 15 cm.

The distance between them is 6 cm

Let us extend the smaller side and then draw perpendiculars from the ends of both sides.

(i) Area of trapezium ABCD= (Area of rectangle EFCD)+(Area of triangle AED+Area of triangle BFC) =(10×6)+[(

2

1

×AE×ED)+(

2

1

×BF×FC)]

=60+[(

2

1

×AE×6)+(

2

1

×BF×6)]

=60+[3AE+3BF]

=60+3×(AE+BF)

Here, AE+EF+FB=15 cm

And EF=10 cm

∴AE+10+BF=15

Or, AE+BF=15−10=5 cm

Putting this value in the above formula:

Area of the trapezium =60+3×(5)=60+15=75 cm

2

(ii) In this case,the figure will look as follows:

Area of trapezium ABCD= (Area of rectangle ABGH)-[(Area of triangle AHD) + (Area of triangle BGC)] =(15×6)−[(

2

1

×DH×6)+(

2

1

×GC×6)]

=90−[3×DH+3×GC]

=90−3[DH+GC]

Here, HD+DC+CG=15 cm

Here, DC=10 cm

So, HD+10+CG=15

⇒HD+GC=15−10=5 cm

Putting this value in the above equation:

Area of the trapezium =90−3(5)=90−15=75 cm

2