Exercise 14A
Question 1
Calculate the perimeter of each shape
(a) (4×8)cm= 32 cm
(b) (1+2+3+2+3+1+2+2)cm=16 cm
(c) (15+20+20) cm =55 cm
(d) 2(15+25) cm
=(2×40)=80 cm
(e) (5×14) cm
=70 cm
(f) (11+8+5+3+7) cm
=34 cm
(g) 3rd side $=\sqrt{(8)^2+(6)^2}$
=√64+36
=√100
=10
Perimeter=(6+8+10) cm
=24 cm
(h) (20+8+5+15+5+8+20+8+5+15+5+8) cm
=122 cm
Question 2
Find the perimeter of a square with side length
(a) (4×16) cm
=64 cm
(b) (4×3.12) cm
=12.48 cm
(c) (4×4.5) m
=18 cm
(d) (4×17) mm
Question 3
Find the perimeter of each rectangle given in the table below
(a) 2(8+5) cm
=2×13=26 cm
(b) 2(26+15) mm
=2×41=82 mm
(c) 2(8.3+4.9)cm
=2×13.2=26.4 cm
(d) 2(28.9+19.5) km
Question 4
What is the cost of following
(a) A rectangular field 80 m by 45 m, if fencing cost 85.40 per metre ?
Perimeter of field=2(80+45)m
=2×125=250 m
∴Fencing cost=(250×25.40)=21.350
(b) A square field with sides 57.5 m , when fencing costs 76.25 per metre
Perimeter of square field=(4×57.5)m
=230 m
∴Cost of fencing=(230×76.25)
=17537.50
(c) A triangular garden with sides 20.4 m, 63.8 m and 48.7 m, when fencing costs 39.80 per meter?
Perimeter of triangular garden=(20.4+63.8+48.7) m
=132.9 m
∴Cost of fencing=(132.9×39.80)
=5289.42
Question 5
(a) The perimeter of a square is 80 m. What is the length of each side of the sqaure?
∴Length of each side$=\frac{80}{4}$m
=20 m
(b) The perimeter of a rectangular card is 400 mm and its length is 6cm. What is its breadth ?
Let, the breadth be x
400 mm=40 cm
∴2(x+6)=40
$x+6=\frac{40}{2}$
x+6=20
x=20-6=14
Ans : 14
Question 6
This rectangular block of land is to be enclosed with fencing panels 2.6m long. How many panels will be needed if one panel is left out to put in a gate ?
Figure to be added
Perimeter of the rectangular block=2(39+208)=119.6 m
∴Number of panels$=\frac{119.6}{2.6}$
=46
∴One pannel is left out then total number of pannels
=46-1=45
Question 7
A 1.8 m wide path s built around a rectangular pool 20 m long and 8.7 m wide . What is the distance around the outside of
Figure to be added
(a) The pool
(b) The path
Sol :
(a) Perimeter of the pool=2(87+20) =574 m
Ans : The distance around the outside of pool is 57.4 m
(b) Length of the path=(20+1.8+1.8) =23.6 m
Breadth of the path=(8.7+1.8+1.8)=12.3 m
∴Perimeter of the path=2(23.6+12.3)=71.8 m
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