Class 7 Maths Chapter 9 Question Answers - Perimeter and Area

Q1. Find the area of a square park whose perimeter is 480m.
Ans:  Given, Perimeter(P)= 480m
s = Side of square
P = 4×s
480= 4×s
s = 480/4
s = 120m
Area of Square=s²
= (120⁾²
= 14,400sq.m

Q2. If the perimeter of a rectangle is 390cm and the length is 30cm. Find its breadth in the area.
Ans:  Given, Perimeter = 390cm , length = 30cm
B = Breadth
A = Area
P = 2(l + b)
390 = 2(30 + b)
390 / 2 = (30 + b)
195=(30 + b)
b = 195 − 30
b = 165cm
A = l × b
A = 30 × 165
A = 4,950 sq.cm

Q3. A wire bent in the shape of a rectangle. Its length is 30cm and breadth is 15cm and if the same wire is rebent in the shape of a square. What will be the measure of slides and which encloses more area.
Ans:  We know, perimeter of rectangle
P = 2(l + b)
Where, l=length and b = breadth
= 2(30 + 15)
= 2(45)
= 90cm
Perimeter of Square= Perimeter of rectangle
90 = 4s s = side
s = 904
s = 22.5 cm
Area of rectangle = l × b
= 30 × 15
= 450 sq. cm
Area of Square = s² = (22.5)²
= 506.25 sq.cm
Hence, Square  encloses a large area.

Q4. Find the area of    

Ans: Given
height = 3 cm  
breadth = 6 cm
Area of parallelogram = b × h
= 6 × 3
= 18cm²

Q5. Find the area of

Ans: Given base = 5.5cm , height = 2.2cm
Area of triangle = 12×b×h
= 1 / 2 × 5.5 × 2.2
= 6.05sq. cm

Q6. ΔABC  is isosceles with AB = AC = 5.5cm and BC = 8cm. What will be the height from C to AB i.e., CE? If the height AD from A to BC is 4.5cm. Find the area of ΔABC?

Ans: Given BC = base = 8 cm , AD=heigth=4.5cm
Area of ΔABC
= 1/2 × b × h
= 1/2 × 8 × 4.5
= 18sq.cm    
Area of ΔABC
= 12 ×  b × h
= 12 × AB×CE
18  = 1/2 × 5.5 × CE
CE = 36 / 5.5
CE = 6.55cm

Q7. If the circumference of a circular sheet is 132cm. Find its radius and area.
Ans:  Given, Circumference = 132cm
C = 2πr
r = radius
132 = 2 × 22 / 7×r
r = 132 × 7 / 2 × 22
r = 21cm
Area = πr²
Area = 22 / 7 × (21)²
Area = 66×21
Area = 1386sq. cm

Q8. Find the circumference and area of the circle if radius is 7cm.
Ans: Given, radius = 7cm
C = 2πr
C = Circumference
C = 2 × 22  / 7 × 7
C = 44cm
Area = πr²
Area = 22/7 × (7)²
Area = 22 × 7
Area = 154sq. cm

Q9. A garden is 50m long and 42m broad. A path of 2m wide is built outside and around it. Find the area of the path in hectares?
Ans: Given,
Length of garden = 50m
Breadth of garden = 42m
Area of garden = l × b
= 50 × 42
= 2100sq. m
Area of garden where path is included
Area of garden = l × b
= 54 × 46
= 2484sq. m
Area of path = Area of garden including path – Area of garden
= 2484 − 2100
= 384sq.m
1 hectare = 10000m²
Area of garden= 384 / 10000
= 0.0384hectare

Q10. Find the area and perimeter of the square whose side is 4cm.
Ans: We know,
Area of square = s²
s = side
= 4²
= 16sq. cm
Perimeter = 4s
= 4 × 4
= 16cm

Q11. The length and breadth of a rectangular piece of land are 350m and 150m respectively.
Find
(a) The area
Ans: We know,
Area of garden = l × b
l = length
b =  Breadth
= 350×150
=52,500sq. m

(b) The cost of the land, if 1m²  of the land costs Rs. 10,000
Ans: Given, Cost of 1m² of  land = Rs. 10,000
Cost of 52,500m² of land = Rs. 52,500×10,000
= Rs. 52,50,00,000

Q12. A rectangular field of length 50m and breadth 45m need to be fenced. Find the cost of fencing if the changes are Rs. 4 per meter.
Ans: We know, perimeter of rectangle
P=2(l + b)
l = length
b = Breadth
= 2(50 + 45)
= 2(95)
= 190m
Cost of fencing= 190 × 4
= Rs. 760

Q13. ΔABC is right angles at A, AD BC ⊥ .If AB = 8cm, BC = 17cm and AC = 15cm. Find the area of ABC and also length of AD.

Ans: Here,
Area of ΔABC = 1/2 × b × h
b= base
h=height

=1/2 × AC × AB
= 1/2 × 8 × 15
= 60sq.cm    
Area of ΔABC = 1/2 × b × h
= 1/2 × BC × AD
60 = 1/2 × 17 × AD
AD = 7.03cm

Q14. Find the area and circumference of the circle whose radius is
(a) 2cm
Ans: Given, radius = 2cm
C = 2πr
C = Circumference
C = 2 × 22/7 × 2
C = 88/7cm
Area = πr²
Area = 22/7×(2)²
Area = 12 x 4 / 7sq.cm

(b) 21cm
Ans: Given, radius = 21cm
C = 2πr
C = 2 × 22 /7 × 21
C = 132cm
Area = πr²
Area = 22 / 7×(21)²
Area = 66×21
Area = 1386sq. cm

Q15. Find the area of quadrilateral ABCD here AC = 20cm, BM = 4cm, DN = 4cm and BM ⊥AC and DN ⊥ AC .

Ans: Using formula
Area of ΔABC
= 12 × b × h
B = Base
H = Height
= 12 × 4 × 20
= 40sq.cm    
Area of ΔADC
= 12 × b × h
=12 × 20 × 4
= 40 sq.cm
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC
= 40 + 40
= 80sq. cm