Area & Perimeter - 2 - Class 5 MCQ

There are 6×4=24 squares and each square has an area of = 1cm². So, the area of rectangle =6×4×1=24cm².

Given figure is a rectangle. Also, given I = length = 8m

Let b be the breadth of the rectangle.

Perimeter of rectangle = 2(1 + b) = 28

⇒   l + b = 28 / 2

⇒   l + b = 14

⇒ b = 14 - 8 = 6 m

Solution:

• Base of the triangle: 12 cm
• Height of the triangle: 2 × 12 = 24 cm
• Area of the triangle: (1/2) × Base × Height = (1/2) × 12 × 24 = 144 sq. cm

The perimeter of a rectangle is given by the formula: P=2×(l+b) Where:

• l = length of the rectangle
• b = breadth of the rectangle

For the new rectangle, the new length is increased by 2 units, i.e., l′=l+2, while the breadth remains the same.

The perimeter of the new rectangle is calculated as: P′=2×(l′+b)
Substitute l′=l+2l' into the equation: P′=2×((l+2)+b)
Simplifying: P′=2l+2b+4
Thus, the new perimeter becomes: P′=P+4
This shows that the perimeter of the rectangle increases by 4 units when the length is increased by 2 units, while the breadth remains the same.

The area of a square is calculated by multiplying one side by itself. This is often expressed as side × side.

Given each block is of 2m², so there should be 12 blocks in a figure to get an area of 24m². In option (c), there are 2×6=12 blocks in figure.

Total Boxes = 88
No. of shaded boxes = 28,
No. of unshaded boxes = 88 -28 = 60
Area of 1 unshaded box =3m²
So, area of unshaded region = 60×3=180m².

Distance covered in one round = Perimeter of the park
= 2 × (Length + Width)
= 2 × (150 m + 100 m)
= 2 × 250 m
= 500 m
Distance covered in 3 rounds = 3 × 500 m
= 1500 m

Tiling the hall means tiling the area of the hall.
Area (A) = Length × Width
= 40 m × 20 m
= 800 sq. m
Cost of tiling 1 sq. m = ₹ 10
Cost of tiling 800 sq. m = 800 × ₹ 10 = ₹ 8000
Thus, the cost of tiling the hall is ₹ 8000.