[tex]{\huge{\red{\colorbox{lavender}{Question: -}}}}[/tex]
A field is 70m long and 40m broad. a path of uniform width runs all around inside the field and covers an area of 1000m²
A field is 70m long and 40m broad. a path of uniform width runs all around inside the field and covers an area of 1000m²
Answers & Comments
Given:
Length of the field
=
70
m
.
Breadth of the field
=
40
m
.
Area of the path
=
1000
m
2
.
To find:
We have to find the width of the path.
Solution:Let the width of the field be
w
.
In the above diagram A'B'C'D' is the rectangular field, Shaded portion is the path inside the field.
The area of the path is the difference in the area of A'B'C'D' and the area of ABCD.
Now,
Area of ABCD = Length(AB)
×
Breadth(BC)
Area of ABCD = (70-2w)
×
(
40
−
2
w
)
m2
Area of ABCD $=2800-220w+4w^2\ m2
And,
Area of A'B'C'D' = Length(A'B')
×
Breadth(B'C')
Area of A'B'C'D' = 70
×
40 m2
Area of A'B'C'D' = 2800 m2
So,
Area of path = Area of A'B'C'D'
−
Area of ABCD
1000
=
2800
−
(
2800
−
220
w
+
4
w
2
)
m
2
1000
=
220
w
−
4
w
2
m
2
250
=
55
w
−
w
2
w
2
−
55
w
+
250
=
0
w
2
−
50
w
−
5
w
+
250
=
0
w
(
w
−
50
)
−
5
(
w
−
50
)
=
0
(
w
−
50
)
(
w
−
5
)
=
0
w
=
50
or
w
=
5
w
=
5
(
w
cannot be greater than 40 m)
Verified answer
[tex]\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\orange{E}\orange{}\blue{R}}}}[/tex]
Area of an internal field = (70-2x)(40-2x)
Width of the path = Area of a field - Area of an
internal field.
→1000=2800-(70-2x)(40-2x)
→x255x + 250 = 0
→(x-50)(x-5) = 0
x = 50 is not acceptable.
x = 5
Therefore the width of the path is 5 m..
[tex]\scriptsize{\tt{Hope \: it \: helps \: you!!!}}[/tex]