Question

An oblong piece of ground measures 19m 2.5 dm by 12m5dm.fom centre of each side of the ground,a path 2 m wide goes across to the center of the opposite side.what is the area of the path?

a. 59.5 m2

b. 54 m2

c. 43 m2

d. 34 m2

Answer 1

1 dm = 0.1 m

Now ,

Dimensions of the ground :

Length = AB = 19m 2.5 DM

AB = 19.25 m

Breadth = BC = 12m 5dm

BC = 12.5 m

If 2 m wide path goes across to the

center of the opposite sides .

width of the path = w = 2 m

From the figure ;

area of the path( A ) = area of EFGH + area of

PQRS - area of Square KLMN

A = EF × w + QR × w - w²

= AB × w + BC × w - w²

= 19.25 × 2 + 12.5 × 2 - 2²

= 38.5 + 25 - 4

= 63.5 - 4

A = 59.5 m²

Therefore ,

Area of the path = A = 59.5 m²

Option ( a ) is correct.

I hope this helps you.

: )

Answer 2

Answer:

59.5m A

Step-by-step explanation:

1dm=0.1m

Thus 19m2.5dm=19.25 & 12m5dm=12.5m

Area of path with overlapping = (19.25m X 2m) + (12.50m X 2m) = 63.5m^2

Since area of path is overlapped. So, subtract 2m X 2m =4m^2

Hence, the actual area of path = (63.5m^2 – 4m^2) = 59.5m^2(Answer)