Mensuration — NCERT Solutions, AI Tutor & Practice

For a rectangle:

• Area $= \text{length} \times \text{breadth} = 60 \times 40 = 2400 \text{ m}^2$
• Perimeter $= 2(\text{length} + \text{breadth}) = 2(60 + 40) = 2 \times 100 = 200 \text{ m}$

So the area is $2400 \text{ m}^2$ and the boundary length is $200 \text{ m}$.

Area of a trapezium $= \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

Substituting the values:

$\text{Area} = \frac{1}{2} \times (12 + 8) \times 5 = \frac{1}{2} \times 20 \times 5 = 50 \text{ cm}^2$

So the area of the trapezium is $50 \text{ cm}^2$.

Area of a rhombus $= \dfrac{1}{2} \times d_1 \times d_2$

Given area $= 240 \text{ cm}^2$ and $d_1 = 16 \text{ cm}$.

$240 = \frac{1}{2} \times 16 \times d_2$

$240 = 8 \times d_2$

$d_2 = \frac{240}{8} = 30 \text{ cm}$

So the other diagonal is $30 \text{ cm}$.

The hexagon is divided into two congruent trapeziums, each with parallel sides $5 \text{ cm}$ and $11 \text{ cm}$ and height $4 \text{ cm}$.

Area of one trapezium:

$= \frac{1}{2} \times (5 + 11) \times 4 = \frac{1}{2} \times 16 \times 4 = 32 \text{ cm}^2$

Total area of hexagon $= 2 \times 32 = 64 \text{ cm}^2$.

Total surface area of a cuboid $= 2(lb + bh + hl)$

Here $l = 15$, $b = 10$, $h = 8$.

$= 2(15 \times 10 + 10 \times 8 + 8 \times 15)$

$= 2(150 + 80 + 120)$

$= 2 \times 350 = 700 \text{ cm}^2$

So the total surface area is $700 \text{ cm}^2$.

Lateral surface area of a cube $= 4a^2$, where $a$ is the edge.

$4a^2 = 144 \implies a^2 = 36 \implies a = 6 \text{ cm}$

Total surface area of a cube $= 6a^2 = 6 \times 36 = 216 \text{ cm}^2$.

So the edge is $6 \text{ cm}$ and total surface area is $216 \text{ cm}^2$.

Curved surface area (CSA) $= 2\pi r h$

$= 2 \times \frac{22}{7} \times 7 \times 10 = 2 \times 22 \times 10 = 440 \text{ cm}^2$

Total surface area (TSA) $= 2\pi r(r + h)$

$= 2 \times \frac{22}{7} \times 7 \times (7 + 10) = 44 \times 17 = 748 \text{ cm}^2$

So CSA $= 440 \text{ cm}^2$ and TSA $= 748 \text{ cm}^2$.

Volume of cuboid $= l \times b \times h = 12 \times 9 \times 5 = 540 \text{ cm}^3$

Since $1000 \text{ cm}^3 = 1 \text{ litre}$:

$540 \text{ cm}^3 = \frac{540}{1000} = 0.54 \text{ litres}$

So the volume is $540 \text{ cm}^3$ or $0.54$ litres.

Volume of cylinder $= \pi r^2 h$

$= \frac{22}{7} \times (3.5)^2 \times 4$

$= \frac{22}{7} \times 12.25 \times 4$

$= \frac{22 \times 12.25 \times 4}{7} = \frac{1078}{7} = 154 \text{ m}^3$

So the tank can hold $154 \text{ m}^3$ of water.

When two cubes of edge $4 \text{ cm}$ are joined, the resulting cuboid has:

• Length $= 4 + 4 = 8 \text{ cm}$
• Breadth $= 4 \text{ cm}$
• Height $= 4 \text{ cm}$

Total surface area $= 2(lb + bh + hl)$

$= 2(8 \times 4 + 4 \times 4 + 4 \times 8)$

$= 2(32 + 16 + 32) = 2 \times 80 = 160 \text{ cm}^2$

So the total surface area of the cuboid is $160 \text{ cm}^2$.

Volume of metal $=$ (volume of outer cylinder) $-$ (volume of inner cylinder)

$= \pi R^2 h - \pi r^2 h = \pi h (R^2 - r^2)$

Here $R = 5$, $r = 4$, $h = 21$.

$= \frac{22}{7} \times 21 \times (5^2 - 4^2)$

$= \frac{22}{7} \times 21 \times (25 - 16)$

$= 22 \times 3 \times 9 = 594 \text{ cm}^3$

So the volume of metal used is $594 \text{ cm}^3$.

When the sheet is rolled along its length (44 cm), that side becomes the circumference of the base and the other side (20 cm) becomes the height.

Step 1: Find the radius from the circumference.

$2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44$

$\frac{44}{7} r = 44 \implies r = 7 \text{ cm}$

Step 2: Height $h = 20 \text{ cm}$.

Step 3: Volume $= \pi r^2 h$

$= \frac{22}{7} \times 7^2 \times 20 = \frac{22}{7} \times 49 \times 20 = 22 \times 7 \times 20 = 3080 \text{ cm}^3$

So the volume of the cylinder is $3080 \text{ cm}^3$.