Important Questions: Perimeter and Area - Class 7 MCQ

πd²/4 = 2464
22/7xd²=2464x4
d²=2464x4x7
            22
⇒ d =  = 56m

r = 35 m; R = 35+7=42 m
Area of circular path 
= π(R+r)(R−r)
= 22/7(42+35)(42−35) = 1694m²

2πr − r = 37 

r = 37/2π−1 = = 7m

Circumference = 2πr = 2 × 22/7 × 7 = 44m

Circumference = 30 cm Area 

= C²/4π = =71.64cm²

Number of plants = A/4 = 71.64/4 

Area of circle = Circumference of circle
πr² = 2πr = 2 ⇒ d = 2 + 2 = 4units

2πr₁/2πr₂ =   4/9
r₁/r₂ = 4/9
πr₁²/πr₂² = r₁²/r₂² = 16/81

The perimeter of a rectangular sheet = 100 cm

Length of the rectangular sheet (l) = 35 cm

Let the breadth be 'b' cm

Perimeter of a rectangular sheet = 2 (Length + Breadth)

100 = 2l + 2b

100 = 2 × 35 + 2b

100 = 70 + 2b

2b =100 - 70

b = 30/2

b = 15 cm

The perimeter of the wall hanging is given by the sum of circumferences of the 4 semicircles −4 x diameter. Clearly, the diameter of each semicircle is 14 cm. The required perimeter 

= 2 × circumference of circle of radius 7 cm. 

= 2×2 × 22/7 × 7 = 88cm

Length of wire = 2π × 42 = 84πcm
Let x be the side of the square. We have,  4x = 84π ⇒ x = 21π
Area of the circle: Area of the square 
= π(42)²:(21π)² 
= 4:π = 4:22/7 = 14:11

∴ Area = 400 × 250m²
= 100000 m²
Cost of the land per square metre 
= Rs.1000 
∴  Cost of total land 
100000 × Rs.1000 = Rs.10 crores

Here we equate the areas, i.e., AB×DL = BC×DM 

⇒ 18×DL = 12×10 
⇒ DL = (12×10)/18 = 20/3 =

We have, area =156 cm² 
⇒ b×h = 156 
AB×DL=156 
DL = 156/AB =156/13 = 12cm
In right ΔADL, AD²=DL²+LA²
AL²=AD²−DL²=13²−12² = 25
∴ AL = =5cm

Distance between two opposite corners 

= 
 = 50cm

The area of the verandah is 63 m².

• Outer dimensions: 10 m × 8.5 m = 85 m²
• Room area: 5.5 m × 4 m = 22 m²
• Verandah area: 85 m² - 22 m² = 63 m².