1. Find the area of a rectangle of dimensions (i) 30 cm x 16 cm
(ii) 20.5 cm x 10 cm (iii) 12 m by 6 m.
2. If the length of a rectangle is
35.5 m and its breadth is 5.5 m less than its length, find its area.
3. Find the area of a square of side (i) 10 m (u) 21 cm (ii) 0.5 m
4.(i) Find the length and perimeter of a rectangle of area 900 m² and breadth 20 m.
(ii) Find the breadth and perimeter of a rectangle of area 962 cm² and length 26 cm.
5. Find the side of a square of area (i) 1936 m² (1) 900 cm²
(ii) 62 500 m² (iv) 7.29 m².
6.(i) The perimeter of a rectangle is 480 cm and its length is 140 cm. Find its area in square metres.
(ii) The perimeter of a rectangle is 12.4 m and its breadth is 2.2 m. Find its area.
7. Find the area of a square if its perimeter is (i) 36 cm (ii) 720 m.
8. If the area of a rectangle is 54 m² and its width is 75 cm, find its length in metres.
9. Two plots of land have the same perimeter. One is a square of side 40 m, while the other is
a rectangle of breadth 25 m. Which plot is larger and by how much?
10. A path 2 m wide surrounds a garden 20 m by 10 m. Find
(i) the area of the path, and
(ii) the cost of paving the path at
10 per m².
11. A rectangular field has a path 5 m wide running round it. If the rectangle formed by the
path and the field is 200 m x 60 m, find
(i) the area of the path, and
(ii)the cost of constructing the path at ₹15per m².
12. A path of width 2m runs around a square garden of side 30m .Find the area of the path.
Answers
Find the area of a rectangle of dimensions (i) 30 cm x 16 cm
(ii) 20.5 cm x 10 cm (iii) 12 m by 6 m.
answer
Answer
Error in product of quantities: Suppose x=a×b
Let Δa=absolute error in measurement of a,
Δb=absolute error in measurement of b,
Δx=absolute error in calculation of x, i.e. product of a and b.
The maximum fractional error in x is
x
Δx
=±(
a
Δa
+
b
Δb
)
Percentage error in the value of x=(Percentage error in value of a)+(Percentage error in value of b)
According to the problem, length l=(16.2±0.1)cm
Breadth b=(10.1±0.1)cm
Area A=l×b=(16.2cm)×(10.1cm)=163.62cm
2
As per the rule area will have only three significant figures and error will have only one significant figure.Rounding off we get,area A=164cm
2
If ΔA is error in the area, then relative error is calculated as
A
δ4
.
A
Δ4
=
l
Δl
+
b
Δb
=
16.2cm
0.1cm
+
10.1cm
0.1cm
=
16.2×10.1
1.01+1.62
=
163.62
2.63
⇒ΔA=A×
163.62
2.63
cm
2
=162.62×
163.62
2.63
=2.63cm
2
ΔA=3cm
2
(By rounding off to one significant figure)
Area, A=A±ΔA(164±3)cm
hii,
refer above two pictures for your answer.
I hope you get your answer
thnx for asking
