Skip to content

Mensuration.md

Latest commit

 

History

History
184 lines (155 loc) · 8.72 KB

Mensuration.md

File metadata and controls

184 lines (155 loc) · 8.72 KB
Raw

Quantitative Aptitude: Mensuration

1. 2D Shapes (Area & Perimeter)

Shape Area Perimeter / Circumference Key Variables
Rectangle $L \times B$ $2(L + B)$ $L = \text{Length}, B = \text{Breadth}$
Square $a^2$ or $\frac{d^2}{2}$ $4a$ $a = \text{Side}, d = \text{Diagonal } (a\sqrt{2})$
Triangle $\frac{1}{2} \times b \times h$ $a + b + c$ $b = \text{Base}, h = \text{Height}$
Scalene Triangle $\sqrt{s(s-a)(s-b)(s-c)}$ $2s = a+b+c$ $s = \text{Semi-perimeter}, a,b,c = \text{Sides}$
Equilateral Triangle $\frac{\sqrt{3}}{4} a^2$ $3a$ $a = \text{Side}, \text{Height } (h) = \frac{\sqrt{3}}{2}a$
Circle $\pi r^2$ $2\pi r$ $r = \text{Radius}$
Sector of Circle $\frac{\theta}{360} \times \pi r^2$ $2r + \left(\frac{\theta}{360} \times 2\pi r\right)$ $\theta = \text{Angle of Sector}$

2. 3D Shapes (Volume & Surface Area)

Shape Volume Lateral / Curved Surface Area (LSA/CSA) Total Surface Area (TSA) Key Variables
Cuboid $l \cdot w \cdot h$ $2h(l + w)$ $2(lw + wh + hl)$ $l, w, h = \text{dimensions}$, $\text{Diagonal} = \sqrt{l^2+w^2+h^2}$
Cube $a^3$ $4a^2$ $6a^2$ $a = \text{Side}$, $\text{Diagonal} = a\sqrt{3}$
Cylinder $\pi r^2 h$ $2\pi r h$ $2\pi r (h + r)$ $r = \text{Radius of base}, h = \text{Height}$
Cone $\frac{1}{3} \pi r^2 h$ $\pi r l$ $\pi r (l + r)$ $r = \text{Radius}, h = \text{Height}, l = \text{Slant height } (\sqrt{r^2+h^2})$
Sphere $\frac{4}{3} \pi r^3$ $4\pi r^2$ $4\pi r^2$ $r = \text{Radius}$
Hemisphere $\frac{2}{3} \pi r^3$ $2\pi r^2$ $3\pi r^2$ $r = \text{Radius}$

3. Practice Problems

Problem 1

Find the cost of fencing a circular field of radius 28 meters at the rate of Rs. 15 per meter.

Problem 2

A copper sphere of radius 6 cm is melted and drawn into a wire of cylinder shape of radius 0.2 cm. Find the length of the wire.

Problem 3

If the radius of a cylinder is increased by 20% and its height is decreased by 10%, find the percentage change in its volume.

Problem 4

The length of a rectangle is increased by 60%. By what percentage must the width be decreased to maintain the same area?

Problem 5

Find the length of the longest pole that can be placed in a room 12 m long, 9 m broad, and 8 m high.

Problem 6

If the diagonal of a square is doubled, how many times does its area become?

Problem 7

A cylindrical tank of base radius 7 m and height 10 m is filled with water. If this water is poured into a cuboidal pond of length 11 m and breadth 10 m, what will be the height of the water level in the pond?

Problem 8

A sector of a circle of radius 12 cm has an angle of $120^\circ$. Find the length of the arc of this sector.

Problem 9

A rectangular plot measures 40 m by 30 m. A path of uniform width 2 m is built around the outside of the plot. Find the area of the path.

Problem 10

The radius of the base and the height of a right circular cone are in the ratio $5 : 12$. If its volume is $314 \text{ cm}^3$, find the slant height of the cone. (Use $\pi = 3.14$)

4. Step-by-Step Solutions

Solution 1

  1. Identify the parameter to calculate:
    • Fencing is done along the boundary (Circumference).
    • Circumference $C = 2\pi r$.
  2. Calculate the circumference:
    • Given radius $r = 28$ m, and $\pi \approx \frac{22}{7}$: $$C = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176 \text{ meters}$$
  3. Calculate the total cost: $$\text{Total Cost} = \text{Circumference} \times \text{Rate} = 176 \times 15 = \text{Rs. } 2,640$$
  4. Answer: The cost of fencing is Rs. 2,640.

Solution 2

  1. Formulate the volume conservation:
    • Volume of Sphere melted = Volume of Cylinder wire created.
  2. Set up the values:
    • Sphere radius $R = 6$ cm.
    • Cylinder radius $r = 0.2$ cm.
    • Cylinder length (height) $= h$.
  3. Equate the volumes: $$\text{Volume of Sphere} = \text{Volume of Cylinder}$$ $$\frac{4}{3} \pi R^3 = \pi r^2 h$$ $$\frac{4}{3} \times 6^3 = (0.2)^2 \times h$$ $$\frac{4}{3} \times 216 = 0.04 \times h$$ $$4 \times 72 = 0.04 \times h$$ $$288 = 0.04 \times h \implies h = \frac{288}{0.04} = 7200 \text{ cm}$$
  4. Convert to meters:
    • $7200 \text{ cm} = 72$ meters.
  5. Answer: The length of the wire is 72 meters.

Solution 3

  1. Understand volume dependence:
    • Volume of a cylinder $V = \pi r^2 h$.
    • The volume depends on $r \times r \times h$.
  2. Use successive percentage formula:
    • First, calculate percentage change in $r^2 = r \times r$ (both increased by 20%): $$\text{Change in } r^2 = 20 + 20 + \frac{20 \times 20}{100} = 40 + 4 = 44% \text{ increase}$$
    • Next, combine this $44%$ increase with the $10%$ height decrease ($h$): $$\text{Net Change in } V = 44 - 10 + \frac{44 \times (-10)}{100} = 34 - 4.4 = 29.6%$$
  3. Answer: The volume increases by 29.6%.

Solution 4

  1. Analyze area dependency:
    • Area $A = L \times W$.
    • Let initial length $L_1 = 100$ and width $W_1 = 100$. Initial Area $A_1 = 10000$.
    • New length $L_2 = 160$ (increased by 60%).
  2. Find new width $W_2$:
    • $L_2 \times W_2 = A_1$
    • $160 \times W_2 = 10000 \implies W_2 = \frac{10000}{160} = 62.5$.
  3. Calculate percentage decrease:
    • Decrease in width $= 100 - 62.5 = 37.5%$.
  4. Answer: The width must be decreased by 37.5%.

Solution 5

  1. Identify the shape and required parameter:
    • The room is a cuboid with dimensions $l = 12$ m, $w = 9$ m, $h = 8$ m.
    • The longest pole corresponds to the space diagonal of the cuboid.
  2. Apply the diagonal formula:
    • $\text{Diagonal} = \sqrt{l^2 + w^2 + h^2}$.
    • $\text{Diagonal} = \sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17$ m.
  3. Answer: The length of the longest pole is 17 meters.

Solution 6

  1. Understand area-diagonal relationship of a square:
    • Area of a square in terms of diagonal $d$ is: $$A = \frac{d^2}{2}$$
  2. Determine the change:
    • Let original diagonal be $d_1$ and area $A_1 = \frac{d_1^2}{2}$.
    • New diagonal $d_2 = 2d_1$.
    • New Area $A_2 = \frac{d_2^2}{2} = \frac{(2d_1)^2}{2} = 4 \times \frac{d_1^2}{2} = 4A_1$.
  3. Answer: The area becomes 4 times the original area.

Solution 7

  1. Formulate the volume conservation:
    • Volume of water in Cylinder = Volume of water in Cuboid.
  2. Calculate the volume of the cylindrical tank:
    • Base radius $r = 7$ m, height $h_c = 10$ m.
    • Volume $V = \pi r^2 h_c = \frac{22}{7} \times 7^2 \times 10 = 22 \times 7 \times 10 = 1540 \text{ m}^3$.
  3. Set up the cuboid volume equation:
    • Let the height of the water level in the pond be $h_p$.
    • Dimensions of pond: $l = 11$ m, $w = 10$ m.
    • Volume in pond $= l \times w \times h_p = 11 \times 10 \times h_p = 110 h_p$.
  4. Equate the volumes and solve:
    • $110 h_p = 1540 \implies h_p = \frac{1540}{110} = 14$ m.
  5. Answer: The height of the water level in the pond will be 14 meters.

Solution 8

  1. Identify parameters:
    • Radius $r = 12$ cm.
    • Angle $\theta = 120^\circ$.
  2. Apply the arc length formula:
    • $\text{Arc Length } L = \frac{\theta}{360} \times 2\pi r$.
    • $L = \frac{120}{360} \times 2 \times \pi \times 12 = \frac{1}{3} \times 24\pi = 8\pi$ cm.
    • Using $\pi \approx 3.1416$: $L \approx 8 \times 3.1416 = 25.13$ cm.
  3. Answer: The length of the arc is $8\pi$ cm (approx. 25.13 cm).

Solution 9

  1. Find the area of the rectangular plot:
    • Area of plot $= 40 \times 30 = 1200 \text{ m}^2$.
  2. Find the dimensions and area of the plot including the path:
    • Length including path $= 40 + 2 \times 2 = 44$ m.
    • Width including path $= 30 + 2 \times 2 = 34$ m.
    • Area including path $= 44 \times 34 = 1496 \text{ m}^2$.
  3. Calculate the area of the path:
    • Area of path $= \text{Area including path} - \text{Area of plot} = 1496 - 1200 = 296 \text{ m}^2$.
  4. Answer: The area of the path is 296 $\text{m}^2$.

Solution 10

  1. Express radius and height in terms of a variable:
    • Let base radius $r = 5x$ and height $h = 12x$.
  2. Formulate the volume equation:
    • Volume of cone $= \frac{1}{3}\pi r^2 h = 314$.
    • $\frac{1}{3} \times 3.14 \times (5x)^2 \times (12x) = 314$.
    • $\frac{1}{3} \times 3.14 \times 25x^2 \times 12x = 314$.
    • $3.14 \times 25x^2 \times 4x = 314$.
    • $314 x^3 = 314 \implies x^3 = 1 \implies x = 1$.
  3. Calculate base radius, height and slant height:
    • Radius $r = 5$ cm, height $h = 12$ cm.
    • Slant height $l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ cm.
  4. Answer: The slant height of the cone is 13 cm.