Mensuration Concepts

1. Basic Area and Perimeter of Common Shapes

• Formula:

$$\text{Rectangle: }A=lb,\quad P=2(l+b)$$ $$\text{Square: }A=a^2,\quad P=4a$$ $$\text{Triangle: }A=\frac12 bh$$ $$\text{Circle: }A=\pi r^2,\quad C=2\pi r$$

• Example: Find the area and perimeter of a rectangle of length 12 cm and breadth 8 cm. Solution:

$$A=12\times8=\boxed{96\text{ cm}^2}$$ $$P=2(12+8)=\boxed{40\text{ cm}}$$

---

2. Basic Volume and Surface Area of Solids

• Formula:

$$\text{Cuboid: }V=lbh,\quad TSA=2(lb+bh+hl)$$ $$\text{Cube: }V=a^3,\quad TSA=6a^2$$ $$\text{Cylinder: }V=\pi r^2h,\quad CSA=2\pi rh$$ $$\text{Sphere: }V=\frac43\pi r^3,\quad SA=4\pi r^2$$

• Example: Find the volume of a cylinder with radius 7 cm and height 10 cm. Solution:

$$V=\pi(7)^2(10)$$

Using (\pi=\frac{22}{7}):

$$V=\frac{22}{7}\times49\times10 =\boxed{1540\text{ cm}^3}$$

---

3. Melting & Recasting Shapes

• Formula: Volume remains constant during melting and recasting:

$$V_{\text{original}}=V_{\text{new}}$$

Number of new objects:

$$n=\frac{V_{\text{original}}}{V_{\text{one new object}}}$$

• Example: A solid cube of side 8 cm is melted and recast into cubes of side 2 cm. How many cubes are formed? Solution:

$$V_{\text{large}}=8^3=512\text{ cm}^3$$ $$V_{\text{small}}=2^3=8\text{ cm}^3$$ $$n=\frac{512}{8} =\boxed{64}$$

---

4. Melting & Recasting — Different Shapes

• Formula:

$$V_1=V_2$$

For example, cylinder into spheres:

$$\pi R^2h=n\left(\frac43\pi r^3\right)$$

• Example: A cylinder of radius 6 cm and height 8 cm is melted into spheres of radius 2 cm. How many spheres are formed? Solution:

$$\pi(6)^2(8) =n\left(\frac43\pi(2)^3\right)$$ $$288\pi=n\left(\frac{32}{3}\pi\right)$$ $$n=27$$ $$\boxed{27\text{ spheres}}$$

---

5. Path Inside a Rectangular Field

• Formula:

$$\text{Path area} =\text{Outer area}-\text{Inner area}$$

For a path of width (x) inside a rectangle (l\times b):

$$A_{\text{path}} =lb-(l-2x)(b-2x)$$

• Example: A rectangular park is 60 m × 40 m. A 5 m wide path runs inside all around. Find the path area. Solution: Inner dimensions:

$$60-10=50,\qquad40-10=30$$

Outer area:

$$60\times40=2400$$

Inner area:

$$50\times30=1500$$

Therefore:

$$\boxed{900\text{ m}^2}$$

---

6. Path Outside a Rectangular Field

• Formula: If a path of width (x) runs outside:

$$A_{\text{path}} =(l+2x)(b+2x)-lb$$

• Example: A 50 m × 30 m rectangular park has a 2 m wide path outside it. Find the area of the path. Solution: Outer dimensions:

$$54\times34$$

Outer area:

$$1836\text{ m}^2$$

Park area:

$$50\times30=1500\text{ m}^2$$

Path area:

$$1836-1500 =\boxed{336\text{ m}^2}$$

---

7. Circular Path / Ring

• Formula: For an annulus with outer radius (R) and inner radius (r):

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

If the path has width (x):

$$R=r+x$$

• Example: A circular garden has radius 14 m and an outside path of width 2 m. Find the area of the path. Solution:

$$R=14+2=16,\qquad r=14$$ $$A=\pi(16^2-14^2)$$ $$=\pi(256-196)=60\pi$$ $$\boxed{\frac{1320}{7}\text{ m}^2\approx188.57\text{ m}^2}$$

---

8. Composite Figures — Area

• Formula:

$$\text{Area of composite figure} =\sum\text{areas of component shapes}$$

Subtract areas of removed/cut-out regions.

• Example: A rectangle 20 cm × 14 cm has a semicircle of diameter 14 cm attached to one side. Find the total area. Solution: Rectangle:

$$20\times14=280$$

Semicircle:

$$\frac12\pi(7)^2=\frac{49\pi}{2}$$

Total:

$$280+\frac{49\pi}{2}$$

Using (\pi=\frac{22}{7}):

$$280+77 =\boxed{357\text{ cm}^2}$$

---

9. Hollow Cylinder / Pipe

• Formula:

$$V_{\text{material}} =\pi(R^2-r^2)h$$

where (R) = outer radius and (r) = inner radius.

• Example: A hollow cylindrical pipe has outer radius 7 cm, inner radius 5 cm and length 20 cm. Find the volume of material. Solution:

$$V=\pi(7^2-5^2)(20)$$ $$=\pi(49-25)(20) =480\pi$$ $$\boxed{\approx1508\text{ cm}^3}$$

---

10. Hollow Cylinder — Material Thickness

• Formula:

$$\text{Thickness}=R-r$$

and:

$$V_{\text{material}}=\pi(R^2-r^2)h$$

• Example: A pipe has outer diameter 14 cm and inner diameter 10 cm. Find its thickness. Solution:

$$R=7,\qquad r=5$$ $$\text{Thickness}=7-5 =\boxed{2\text{ cm}}$$

---

11. Sector — Area

• Formula:

$$A_{\text{sector}} =\frac{\theta}{360}\pi r^2$$

• Example: Find the area of a sector of radius 14 cm and angle (60^\circ). Solution:

$$A=\frac{60}{360}\times\frac{22}{7}\times14^2$$ $$=\frac16\times616 =\boxed{\frac{308}{3}\text{ cm}^2}$$ $$\approx\boxed{102.67\text{ cm}^2}$$

---

12. Arc Length

• Formula:

$$L=\frac{\theta}{360}\times2\pi r$$

• Example: Find the arc length of a (90^\circ) sector of radius 14 cm. Solution:

$$L=\frac{90}{360}\times2\times\frac{22}{7}\times14$$ $$=\frac14\times88 =\boxed{22\text{ cm}}$$

---

13. Perimeter of a Sector

• Formula:

$$\text{Perimeter}=2r+\text{Arc length}$$

• Example: Find the perimeter of a (90^\circ) sector of radius 7 cm. Solution: Arc length:

$$\frac{90}{360}\times2\pi(7)=11\text{ cm}$$

Therefore:

$$P=14+11 =\boxed{25\text{ cm}}$$

---

14. Segment of a Circle

• Formula:

$$\text{Area of segment} =\text{Area of sector}-\text{Area of triangle}$$

• Example: A sector of radius 7 cm has angle (90^\circ). Find the area of the minor segment. Solution: Sector area:

$$\frac14\pi(7)^2=\frac{49\pi}{4}$$

Triangle area:

$$\frac12(7)(7)=\frac{49}{2}$$

Therefore:

$$A_{\text{segment}} =\frac{49\pi}{4}-\frac{49}{2}$$

Using (\pi=\frac{22}{7}):

$$=38.5-24.5 =\boxed{14\text{ cm}^2}$$

---

15. Surface Area of Combined Solids

• Formula: Add only the exposed surfaces:

$$SA_{\text{composite}} =\text{sum of exposed component surfaces}$$

Do not count surfaces where two solids are joined.

• Example: A hemisphere of radius 7 cm is attached to the top of a cylinder of radius 7 cm and height 10 cm. Find the exposed surface area. Solution: Exposed area = cylinder CSA + cylinder bottom + hemisphere CSA:

$$2\pi rh+\pi r^2+2\pi r^2$$ $$=2\pi(7)(10)+3\pi(49)$$ $$=140\pi+147\pi=287\pi$$

Using (\pi=\frac{22}{7}):

$$\boxed{902\text{ cm}^2}$$

---

Advanced Variants

16. Recasting with Loss of Material

• Formula: If a percentage (p%) of material is lost:

$$V_{\text{new}}=V_{\text{original}}\left(1-\frac p{100}\right)$$

• Example: A metal cube of volume 1,000 cm³ is recast after a 10% loss of material. Find the volume of the final product. Solution:

$$V_{\text{new}}=1000(1-0.10)$$ $$=\boxed{900\text{ cm}^3}$$

---

17. Recasting into Equal-Capacity Containers

• Formula:

$$n=\frac{V_{\text{original}}}{V_{\text{one container}}}$$

Equivalently, for similar cubes:

$$n=\left(\frac{a}{b}\right)^3$$

• Example: A cube of side 12 cm is recast into cubes of side 3 cm. Find the number of smaller cubes. Solution:

$$n=\left(\frac{12}{3}\right)^3 =4^3 =\boxed{64}$$

---

18. Path with Corner Squares

• Formula: For an inner path around a rectangle:

$$A_{\text{path}}=lb-(l-2x)(b-2x)$$

This automatically includes the four corner regions.

• Example: A 40 m × 30 m rectangular field has a 2 m wide path inside. Find the path area. Solution:

$$A=40(30)-(36)(26)$$ $$=1200-936 =\boxed{264\text{ m}^2}$$

---

19. Frustum of a Cone

• Formula:

$$V=\frac13\pi h(R^2+r^2+Rr)$$

Curved surface area:

$$CSA=\pi(R+r)l$$

• Example: A frustum has radii 6 cm and 3 cm and height 4 cm. Find its volume. Solution:

$$V=\frac13\pi(4)(36+9+18)$$ $$=\frac43\pi(63) =\boxed{84\pi\text{ cm}^3}$$

---

20. Similar Solids — Volume Ratio

• Formula: For similar solids:

$$\frac{V_1}{V_2} =\left(\frac{l_1}{l_2}\right)^3$$

Surface-area ratio:

$$\frac{SA_1}{SA_2} =\left(\frac{l_1}{l_2}\right)^2$$

• Example: Two similar cubes have sides in the ratio (2:3). Find their volume ratio. Solution:

$$V_1:V_2=2^3:3^3$$ $$=\boxed{8:27}$$