Perimeter and Area — NCERT Solutions, AI Tutor & Practice

The perimeter of a rectangle is the total length of its boundary.

Formula: $\text{Perimeter} = 2 \times (\text{length} + \text{breadth})$

Substitute the values: $\text{Perimeter} = 2 \times (45 + 30)$ $= 2 \times 75$ $= 150 \text{ m}$

So, the perimeter of the field is $150$ m.

A square has 4 equal sides.

Formula: $\text{Perimeter} = 4 \times \text{side}$

Substitute the value: $\text{Perimeter} = 4 \times 60 = 240 \text{ m}$

So, the perimeter of the park is $240$ m.

We know: $\text{Perimeter} = 2 \times (\text{length} + \text{breadth})$

Substitute the known values: $100 = 2 \times (30 + \text{breadth})$

Divide both sides by 2: $50 = 30 + \text{breadth}$

Subtract 30: $\text{breadth} = 50 - 30 = 20 \text{ cm}$

So, the breadth is $20$ cm.

Area is the amount of surface enclosed by a shape.

Formula: $\text{Area} = \text{length} \times \text{breadth}$

Substitute the values: $\text{Area} = 12 \times 8 = 96 \text{ cm}^2$

So, the area of the rectangle is $96$ square cm.

For a square, all sides are equal, so length = breadth = side.

Formula: $\text{Area} = \text{side} \times \text{side}$

Substitute the value: $\text{Area} = 15 \times 15 = 225 \text{ cm}^2$

So, the area of the tile is $225$ square cm.

We know: $\text{Area} = \text{length} \times \text{breadth}$

So: $\text{length} = \dfrac{\text{Area}}{\text{breadth}}$

Substitute: $\text{length} = \frac{240}{12} = 20 \text{ m}$

So, the length of the hall is $20$ m.

The length of the wire equals the perimeter of the square.

So, $\text{Perimeter} = 48$ cm.

Since $\text{Perimeter} = 4 \times \text{side}$: $\text{side} = \frac{\text{Perimeter}}{4} = \frac{48}{4} = 12 \text{ cm}$

So, each side of the square is $12$ cm.

Fencing is done along the boundary, so we need the perimeter.

$\text{Perimeter} = 2 \times (25 + 15) = 2 \times 40 = 80 \text{ m}$

Cost of fencing $= \text{Perimeter} \times \text{rate per metre}$ $= 80 \times 20 = ₹1600$

So, the cost of fencing the garden is $₹1600$.

Tiling covers the whole floor, so we need the area.

$\text{Area} = \text{length} \times \text{breadth} = 6 \times 4 = 24 \text{ m}^2$

Cost of tiling $= \text{Area} \times \text{rate per square metre}$ $= 24 \times 150 = ₹3600$

So, the cost of tiling the floor is $₹3600$.

First check the perimeters:

• Sheet 1: $2 \times (6 + 4) = 2 \times 10 = 20$ cm ✓
• Sheet 2: $2 \times (7 + 3) = 2 \times 10 = 20$ cm ✓

Both have the same perimeter. Now compare areas:

• Sheet 1: $\text{Area} = 6 \times 4 = 24 \text{ cm}^2$
• Sheet 2: $\text{Area} = 7 \times 3 = 21 \text{ cm}^2$

Since $24 > 21$, Sheet 1 (6 cm by 4 cm) has the greater area.

This shows that equal perimeters do not mean equal areas.

Area: Split the L-shape into the two given rectangles.

• Rectangle 1: $8 \times 3 = 24 \text{ cm}^2$
• Rectangle 2: $4 \times 3 = 12 \text{ cm}^2$

Total area $= 24 + 12 = 36 \text{ cm}^2$

Perimeter: Add all the outer boundary edges of the combined L-shape.

Going around the boundary the sides are: $8, 3, 4, 3, 4, 6$ cm (the two rectangles joined along a 3 cm edge give an outer boundary of these lengths).

$\text{Perimeter} = 8 + 3 + 4 + 3 + 4 + 6 = 28 \text{ cm}$

So, area $= 36 \text{ cm}^2$ and perimeter $= 28$ cm.

Area of the inner plot: $40 \times 25 = 1000 \text{ m}^2$

The path of width 2 m is on all sides, so the outer dimensions increase by $2 + 2 = 4$ m on both length and breadth.

• Outer length $= 40 + 4 = 44$ m
• Outer breadth $= 25 + 4 = 29$ m

Area of the outer rectangle (plot + path): $44 \times 29 = 1276 \text{ m}^2$

Area of the path $=$ outer area $-$ plot area: $1276 - 1000 = 276 \text{ m}^2$

So, the area of the path is $276 \text{ m}^2$.