Our top 5% students will be awarded a special scholarship to Lido.

Rd sharma solutions

A rectangular field is 50 m by 40 m. It has two roads through its center, running parallel to its sides. The widths of the longer and shorter roads are 1.8 m and 2.5 m respectively. Find the area of the roads and the area of the remaining portion of the field.

Let ABCD be the rectangular field and KLMN and PQRS the two rectangular roads with width 1.8 m and 2.5 m, respectively.

Given that length of rectangular field = 50 m

Breadth of rectangular field = 40m

Area of rectangular field ABCD = 50 m x 40 m

= 2000 m^2

Area of the road KLMN = 40 m x 2.5 m

= 100 m^2

Area of the road PQRS = 50 m x 1.8 m

= 90 m^2

Clearly area of EFGH is common to the two roads.

Thus, Area of EFGH = 2.5 m x 1.8 m = 4.5 m^2

Hence, Area of the roads = Area of KLMN + Area of PQRS – Area of EFGH

= (100 m^2 + 90 m^2) – 4.5 m^2

= 185.5 m^2

Area of the remaining portion of the field = Area of the rectangular field ABCD – Area of the roads

= (2000 – 185.5) m^2

= 1814.5 m^2

Let ABCD be the rectangular field and KLMN and PQRS the two rectangular roads with width 1.8 m and 2.5 m, respectively.

Given that length of rectangular field = 50 m

Breadth of rectangular field = 40m

Area of rectangular field ABCD = 50 m x 40 m

= 2000 m^2

Area of the road KLMN = 40 m x 2.5 m

= 100 m^2

Area of the road PQRS = 50 m x 1.8 m

= 90 m^2

Clearly area of EFGH is common to the two roads.

Thus, Area of EFGH = 2.5 m x 1.8 m = 4.5 m^2

Hence, Area of the roads = Area of KLMN + Area of PQRS – Area of EFGH

= (100 m^2 + 90 m^2) – 4.5 m^2

= 185.5 m^2

Area of the remaining portion of the field = Area of the rectangular field ABCD – Area of the roads

= (2000 – 185.5) m^2

= 1814.5 m^2

Lido

Courses

Quick Links

Terms & Policies

Terms & Policies

2021 © Quality Tutorials Pvt Ltd All rights reserved