Topic: Area and Perimeter of Basic 2D Shapes (Right-Angled Triangle) Exams: SSC MTS, RRB Group D Find the perimeter and the area of a right-angled tr...

1. Identify the missing side (height): In a right-angled triangle, the relation between the base ($b$), height ($h$), and hypotenuse ($c$) is given by Pythagoras' theorem: $$c^2 = b^2 + h^2$$ Substitute the given values ($c = 13\text{ cm}$, $b = 12\text{ cm}$): $$13^2 = 12^2 + h^2$$ $$169 = 144 + h^2$$ $$h^2 = 169 - 144 = 25$$ $$h = \sqrt{25} = 5\text{ cm}$$ 2. Calculate the perimeter ($P$): The perimeter of a triangle is the sum of all three sides: $$P = a + b + c = 5 + 12 + 13 = 30\text{ cm}$$ 3. Calculate the area ($A$): The area of a right-angled triangle is: $$A = \frac{1}{2} \times \text{base} \times \text{height}$$ $$A = \frac{1}{2} \times 12 \times 5 = 30\text{ cm}^2$$ Final Answer: Perimeter $= 30\text{ cm}$, Area $= 30\text{ cm}^2$.

Topic: Frustum Volume and Melting/Recasting into Hollow Cylinder Exams: SSC CGL Tier-II, NDA A solid metallic right circular cone of height $20\text{...

1. Understand the dimensions of the frustum: - The total height of the cone is $20\text{ cm}$. - The vertical angle is $60^\circ$. Therefore, the semi-vertical angle ($\theta$) of the cone is: $$\theta = \frac{60^\circ}{2} = 30^\circ$$ - The plane cuts the cone at the middle of its height (at height $10\text{ cm}$ from the vertex). - Find the upper radius ($r$) and lower radius ($R$) of the resulting frustum using trigonometry: $$r = 10 \tan(30^\circ) = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}}\text{ cm}$$ $$R = 20 \tan(30^\circ) = 20 \times \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}}\text{ cm}$$ - The height of the frustum ($h$) is: $$h = 20 - 10 = 10\text{ cm}$$ 2. Calculate the volume of the frustum ($V_{\text{frustum}}$): $$V_{\text{frustum}} = \frac{1}{3}\pi h \left( R^2 + Rr + r^2 \right)$$ Substitute $h = 10$, $R = \frac{20}{\sqrt{3}}$, and $r = \frac{10}{\sqrt{3}}$: $$V_{\text{frustum}} = \frac{1}{3}\pi (10) \left[ \left(\frac{20}{\sqrt{3}}\right)^2 + \left(\frac{20}{\sqrt{3}}\right)\left(\frac{10}{\sqrt{3}}\right) + \left(\frac{10}{\sqrt{3}}\right)^2 \right]$$ $$V_{\text{frustum}} = \frac{10}{3}\pi \left[ \frac{400}{3} + \frac{200}{3} + \frac{100}{3} \right]$$ $$V_{\text{frustum}} = \frac{10}{3}\pi \left[ \frac{700}{3} \right] = \frac{7000}{9}\pi\text{ cm}^3$$ 3. Establish the properties of the recast wire: - The wire is a long solid cylinder. - Diameter of the wire $= \frac{1}{16}\text{ cm} \implies$ Radius ($r_{\text{wire}}$) $= \frac{1}{32}\text{ cm}$. - Let the length of the wire be $L$ (in cm). - Volume of the wire ($V_{\text{wire}}$): $$V_{\text{wire}} = \pi (r_{\text{wire}})^2 L = \pi \left(\frac{1}{32}\right)^2 L = \frac{\pi L}{1024}\text{ cm}^3$$ 4. Equate the volumes to solve for $L$: $$\text{Volume of wire} = \text{Volume of frustum}$$ $$\frac{\pi L}{1024} = \frac{7000}{9}\pi$$ $$L = \frac{7000 \times 1024}{9}\text{ cm}$$ $$L = \frac{7,168,000}{9}\text{ cm} \approx 796,444.44\text{ cm}$$ 5. Convert the length into meters: $$\text{Length in meters} = \frac{L}{100} = \frac{71,680}{9}\text{ m} \approx 7964.44\text{ m}$$ Final Answer: The length of the wire is approximately $7964.44\text{ meters}$ (or $\frac{71680}{9}\text{ m}$).

Quantitative Aptitude

Mensuration Guide & Practice

Q: Topic: Volume of Composite 3D Solids (Cylinder, Cone, and Hemisphere) Exams: SSC CHSL, Railway NTPC A solid cylindrical container of radius $6\text{ ...

1. Calculate the volume of the cylindrical container ($V_{\text{cylinder}}$): Given: Radius of cylinder $R = 6\text{ cm}$, Height $H = 15\text{ cm}$. $$V_{\text{cylinder}} = \pi R^2 H$$ $$V_{\text{cylinder}} = \pi \times 6^2 \times 15 = 540\pi\text{ cm}^3$$ 2. Calculate the volume of a single ice cream cone set ($V_{\text{ice cream}}$): Each child receives a cone surmounted by a hemisphere. Both have radius $r = 3\text{ cm}$, and the height of the cone is $h = 12\text{ cm}$. $$\text{Total volume } V_{\text{ice cream}} = V_{\text{cone}} + V_{\text{hemisphere}}$$ $$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 3^2 \times 12 = 36\pi\text{ cm}^3$$ $$V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi \times 3^3 = 18\pi\text{ cm}^3$$ $$V_{\text{ice cream}} = 36\pi + 18\pi = 54\pi\text{ cm}^3$$ 3. Determine the number of cones ($N$): Equating the total volume of the container to $N$ individual ice cream servings: $$N \times V_{\text{ice cream}} = V_{\text{cylinder}}$$ $$N \times 54\pi = 540\pi$$ $$N = \frac{540\pi}{54\pi} = 10$$ Final Answer: The number of cones that can be filled is $10$.

Master area, perimeter, surface area, and volume formulas for 2D and 3D shapes with solved examples and free SSC/Railways mock tests. Explore dynamic solver blueprints, master fundamental equations, examine step-by-step solved examples, and practice with real exam-grade mock test sets.

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Practice Question Papers

15 min

Easy

1. Fundamentals & Definitions

Key Definitions

Mensuration: The branch of mathematics that deals with the geometric measurement of various parameters of 1D, 2D, and 3D figures, such as length, area, perimeter, and volume.
2D Shapes: Flat, two-dimensional geometric figures bounded by a closed path on a single plane. They possess length and width but have no depth or height. Key parameters measured are perimeter and area.
3D Shapes (Solid Shapes): Three-dimensional figures that occupy space. They possess length, width, and depth/height. Key parameters measured are volume, curved/lateral surface area, and total surface area.
Perimeter (P) / Circumference (C): The total length of the continuous boundary enclosing a 2D shape. Measured in linear units (e.g., $m$ , $cm$ ).
Area (A): The region or space enclosed by a closed 2D shape. Measured in square units (e.g., $m^2$ , $cm^2$ ).
Volume (V): The total 3D space occupied or enclosed by a solid shape. Measured in cubic units (e.g., $m^3$ , $cm^3$ ).
Curved Surface Area (CSA): The area of only the curved surfaces of a 3D solid (such as cylinders, cones, and spheres), excluding the flat ends.
Lateral Surface Area (LSA): The total area of all the lateral (vertical side) faces of a solid (such as prisms, pyramids, cubes, and cuboids), excluding the areas of the bases (top and bottom).
Total Surface Area (TSA): The sum of the areas of all surfaces (both flat and curved) of a 3D solid.
Inradius ( $r$ ): The radius of the circle inscribed inside a polygon (such as a triangle or regular polygon) that touches all its sides.
Circumradius ( $R$ ): The radius of the circle circumscribed around a polygon that passes through all its vertices.
Apothem ( $r_a$ ): The line segment from the center of a regular polygon to the midpoint of one of its sides (corresponds to the inradius of the polygon).
Slant Height ( $l$ ): The distance from the apex of a cone or regular pyramid along the surface to the boundary of its base.

2. Core Concepts & Formulas

2.1 Key Geometric Rules and Properties

Triangles

Semi-perimeter ( $s$ ): Half of the perimeter of the triangle: $s = \frac{a+b+c}{2}$
Trigonometric Area: If two sides $a$ and $b$ and their included angle $\theta$ are known: $\text{Area} = \frac{1}{2}ab\sin\theta$
Medians Formula: If $d, e, f$ are the lengths of the medians of a triangle, and $u = \frac{d+e+f}{2}$ : $\text{Area} = \frac{4}{3}\sqrt{u(u-d)(u-e)(u-f)}$
Inradius and Circumradius Relations: $\text{Area} = r \times s$ $\text{Area} = \frac{abc}{4R}$

Quadrilaterals

Brahmagupta's Formula: For a cyclic quadrilateral (vertices lie on a circle) with sides a, b, c, d and semi-perimeter $s = \frac{a+b+c+d}{2}$ : $\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}$
Bretschneider's Formula: For any general quadrilateral with sides a, b, c, d, semi-perimeter $s$ , and opposite angles $\alpha$ and $\beta$ : $\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d) - abcd\cos^2\left(\frac{\alpha+\beta}{2}\right)}$
Ptolemy's Theorem: A quadrilateral is cyclic if and only if the product of its diagonals equals the sum of the products of its opposite sides: $ac + bd = pq \quad (\text{where } p, q \text{ are diagonals})$
Trapezium Intersecting Diagonals Property: In a trapezium $ABCD$ where parallel sides are $AB$ and $CD$ , and diagonals intersect at $O$ : $\text{Area}(\Delta AOD) = \text{Area}(\Delta BOC)$ $\text{Area}(\Delta AOB) \times \text{Area}(\Delta COD) = \text{Area}(\Delta BOC) \times \text{Area}(\Delta AOD)$

Regular Polygons

Interior Angles: $\text{Sum of interior angles} = (n-2) \times 180^\circ$ $\text{Each interior angle (regular)} = \frac{(n-2) \times 180^\circ}{n}$
Exterior Angles: $\text{Each exterior angle (regular)} = \frac{360^\circ}{n}$
Diagonals: $\text{Number of diagonals} = \frac{n(n-3)}{2}$

Solids and 3D Principles

Scaling and Percentage Changes:
- If all linear dimensions of a 2D shape are scaled by a factor $k$ , the perimeter changes by a factor of $k$ and the area changes by a factor of $k^2$ .
- If all linear dimensions of a 3D solid are scaled by a factor $k$ , the surface area changes by a factor of $k^2$ and the volume changes by a factor of $k^3$ .
- If only the radius $r$ of a cylinder/cone changes by a factor $x$ and height $h$ changes by a factor $y$ : $\text{Volume change ratio} = x^2y$ $\text{Curved surface area change ratio (for cylinder)} = xy$
Melting and Recasting Rule: When a solid is melted and recast into another shape (or multiple smaller shapes), the total volume remains constant (assuming no wastage). If there is a wastage percentage $w$ : $\text{Volume of newly formed solid(s)} = \left(1 - \frac{w}{100}\right) \times \text{Volume of original solid(s)}$

2.2 Comprehensive Formula Reference Table

Dimension Type	Shape Name	Key Parameters	Perimeter / LSA / CSA	Area / TSA	Volume / Diagonal / Special Properties
2D	Square	Side $a$	$P = 4a$	$A = a^2$ or $A = \frac{1}{2}d^2$	Diagonal $d = a\sqrt{2}$
2D	Rectangle	Length $l$ , width $w$	$P = 2(l+w)$	$A = l \times w$	Diagonal $d = \sqrt{l^2 + w^2}$
2D	General Triangle	Sides $a, b, c$ ; semi-perimeter $s = \frac{a+b+c}{2}$	$P = a+b+c$	$A = \sqrt{s(s-a)(s-b)(s-c)}$ (Heron's) $A = \frac{1}{2} \times \text{base} \times \text{height}$	Inradius $r = \frac{A}{s}$ Circumradius $R = \frac{abc}{4A}$
2D	Equilateral Triangle	Side $a$	$P = 3a$	$A = \frac{\sqrt{3}}{4}a^2$	Height $h = \frac{\sqrt{3}}{2}a$ Inradius $r = \frac{a}{2\sqrt{3}}$ , Circumradius $R = \frac{a}{\sqrt{3}}$
2D	Isosceles Triangle	Equal sides $a$ , base $b$	$P = 2a + b$	$A = \frac{b}{4}\sqrt{4a^2-b^2}$	Height $h = \frac{1}{2}\sqrt{4a^2-b^2}$
2D	Right-Angled Triangle	Base $b$ , height $h$ , hypotenuse $c$	$P = b + h + c$	$A = \frac{1}{2}bh$	Hypotenuse $c = \sqrt{b^2 + h^2}$
2D	Circle	Radius $r$	Circumference $C = 2\pi r$	$A = \pi r^2$	Diameter $d = 2r$
2D	Circular Sector	Radius $r$ , arc angle $\theta$	Arc Length $L = \frac{\theta}{360} \times 2\pi r$ $P = L + 2r$	$A = \frac{\theta}{360} \times \pi r^2$ or $A = \frac{1}{2}Lr$	$\theta$ is in degrees
2D	Circular Segment	Radius $r$ , angle $\theta$	Boundary length = $L + 2r\sin\left(\frac{\theta}{2}\right)$	$A = r^2 \left( \frac{\pi\theta}{360} - \frac{\sin\theta}{2} \right)$	Area = Area of Sector $-$ Area of Triangle
2D	Parallelogram	Adjacent sides $a, b$ , height $h$	$P = 2(a+b)$	$A = b \times h$ or $A = ab\sin\theta$	Opposite angles are equal; diagonals bisect
2D	Rhombus	Side $a$ , diagonals $d_1, d_2$	$P = 4a$	$A = \frac{1}{2}d_1 d_2$	Diagonals bisect at $90^\circ$ Side relation: $4a^2 = d_1^2 + d_2^2$
2D	Trapezium	Parallel sides $a, b$ , height $h$	$P = a+b+c+d$	$A = \frac{1}{2}(a+b)h$	$h$ is distance between parallel sides
2D	Regular Polygon	Number of sides $n$ , side $s$	$P = ns$	$A = \frac{1}{2} \times r_a \times P$ (where $r_a$ is apothem)	Interior angle = $\frac{(n-2)180^\circ}{n}$ Exterior angle = $\frac{360^\circ}{n}$
3D	Cube	Edge $a$	$\text{LSA} = 4a^2$	$\text{TSA} = 6a^2$	$V = a^3$ Diagonal $d = a\sqrt{3}$
3D	Cuboid	Length $l$ , width $w$ , height $h$	$\text{LSA} = 2h(l+w)$	$\text{TSA} = 2(lw + wh + hl)$	$V = lwh$ Diagonal $d = \sqrt{l^2 + w^2 + h^2}$
3D	Solid Cylinder	Radius $r$ , height $h$	$\text{CSA} = 2\pi rh$	$\text{TSA} = 2\pi r(r+h)$	$V = \pi r^2 h$
3D	Hollow Cylinder	Outer radius $R$ , inner $r$ , height $h$	$\text{CSA} = 2\pi h(R+r)$ (Total curved)	$\text{TSA} = 2\pi h(R+r) + 2\pi(R^2-r^2)$	Volume of material $V = \pi h(R^2 - r^2)$
3D	Right Circular Cone	Radius $r$ , height $h$	$\text{CSA} = \pi rl$	$\text{TSA} = \pi r(r+l)$	$V = \frac{1}{3}\pi r^2 h$ Slant height $l = \sqrt{r^2 + h^2}$
3D	Frustum of a Cone	End radii $R, r$ , height $h$	$\text{CSA} = \pi l(R+r)$	$\text{TSA} = \pi l(R+r) + \pi R^2 + \pi r^2$	$V = \frac{1}{3}\pi h(R^2 + Rr + r^2)$ Slant height $l = \sqrt{h^2 + (R-r)^2}$
3D	Sphere	Radius $r$	$\text{CSA} = 4\pi r^2$	$\text{TSA} = 4\pi r^2$	$V = \frac{4}{3}\pi r^3$
3D	Hemisphere	Radius $r$	$\text{CSA} = 2\pi r^2$	$\text{TSA} = 3\pi r^2$	$V = \frac{2}{3}\pi r^3$
3D	Quarter Sphere	Radius $r$	$\text{CSA} = \pi r^2$	$\text{TSA} = 2\pi r^2$	$V = \frac{1}{3}\pi r^3$
3D	Right Prism	Base Area $A_b$ , Base Perimeter $P_b$ , height $h$	$\text{LSA} = P_b \times h$	$\text{TSA} = \text{LSA} + 2A_b$	$V = A_b \times h$
3D	Right Pyramid	Base Area $A_b$ , Base Perimeter $P_b$ , height $h$	$\text{LSA} = \frac{1}{2} P_b \times l$ (where $l$ is slant height)	$\text{TSA} = \text{LSA} + A_b$	$V = \frac{1}{3} A_b \times h$
3D	Regular Tetrahedron	Side $a$	$\text{LSA} = \frac{3\sqrt{3}}{4}a^2$	$\text{TSA} = a^2\sqrt{3}$	$V = \frac{a^3}{6\sqrt{2}}$ or $V = \frac{\sqrt{2}}{12}a^3$ Height $h = a\sqrt{\frac{2}{3}}$

Typical Exam Weightage

Exam	Typical Questions
SSC (CGL / CHSL / MTS)	2–4 questions
Railways (RRB)	1–2 questions
Defense (NDA / CDS)	1–2 questions

3D mensuration (cone, cylinder, sphere combinations) is a recurring SSC CGL Tier 2 favorite.

Figures are typical ranges based on recent-year patterns, not a guarantee for any specific upcoming paper — always cross-check against the latest official syllabus and previous-year papers for Mensuration.

Solved Examples

1Example 1 (Easy)

Topic: Area and Perimeter of Basic 2D Shapes (Right-Angled Triangle) Exams: SSC MTS, RRB Group D

Question: Find the perimeter and the area of a right-angled triangle whose base is $12\text{ cm}$ and hypotenuse is $13\text{ cm}$ .

View Detailed Solution & Explanation

Step-by-Step Explanation

Identify the missing side (height): In a right-angled triangle, the relation between the base ( $b$ ), height ( $h$ ), and hypotenuse ( $c$ ) is given by Pythagoras' theorem: $c^2 = b^2 + h^2$ Substitute the given values ( $c = 13\text{ cm}$ , $b = 12\text{ cm}$ ): $13^2 = 12^2 + h^2$ $169 = 144 + h^2$ $h^2 = 169 - 144 = 25$ $h = \sqrt{25} = 5\text{ cm}$
Calculate the perimeter ( $P$ ): The perimeter of a triangle is the sum of all three sides: $P = a + b + c = 5 + 12 + 13 = 30\text{ cm}$
Calculate the area ( $A$ ): The area of a right-angled triangle is: $A = \frac{1}{2} \times \text{base} \times \text{height}$ $A = \frac{1}{2} \times 12 \times 5 = 30\text{ cm}^2$

Final Answer: Perimeter $= 30\text{ cm}$ , Area $= 30\text{ cm}^2$ .

2Example 2 (Moderate)

Topic: Volume of Composite 3D Solids (Cylinder, Cone, and Hemisphere) Exams: SSC CHSL, Railway NTPC

Question: A solid cylindrical container of radius $6\text{ cm}$ and height $15\text{ cm}$ is completely filled with ice cream. This ice cream is to be distributed among children in cones of height $12\text{ cm}$ and radius $3\text{ cm}$ , having a hemispherical shape on the top. Find the number of such cones that can be completely filled.

View Detailed Solution & Explanation

Step-by-Step Explanation

Calculate the volume of the cylindrical container ( $V_{\text{cylinder}}$ ): Given: Radius of cylinder $R = 6\text{ cm}$ , Height $H = 15\text{ cm}$ . $V_{\text{cylinder}} = \pi R^2 H$ $V_{\text{cylinder}} = \pi \times 6^2 \times 15 = 540\pi\text{ cm}^3$
Calculate the volume of a single ice cream cone set ( $V_{\text{ice cream}}$ ): Each child receives a cone surmounted by a hemisphere. Both have radius $r = 3\text{ cm}$ , and the height of the cone is $h = 12\text{ cm}$ . $\text{Total volume } V_{\text{ice cream}} = V_{\text{cone}} + V_{\text{hemisphere}}$ $V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 3^2 \times 12 = 36\pi\text{ cm}^3$ $V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi \times 3^3 = 18\pi\text{ cm}^3$ $V_{\text{ice cream}} = 36\pi + 18\pi = 54\pi\text{ cm}^3$
Determine the number of cones ( $N$ ): Equating the total volume of the container to $N$ individual ice cream servings: $N \times V_{\text{ice cream}} = V_{\text{cylinder}}$ $N \times 54\pi = 540\pi$ $N = \frac{540\pi}{54\pi} = 10$

Final Answer: The number of cones that can be filled is $10$ .

3Example 3 (Hard)

Topic: Frustum Volume and Melting/Recasting into Hollow Cylinder Exams: SSC CGL Tier-II, NDA

Question: A solid metallic right circular cone of height $20\text{ cm}$ with a vertical angle of $60^\circ$ is cut into two parts at the middle of its height by a plane parallel to its base. The bottom frustum portion is melted and drawn into a long cylindrical wire of uniform diameter $\frac{1}{16}\text{ cm}$ . Find the length of the wire in meters.

View Detailed Solution & Explanation

Step-by-Step Explanation

Understand the dimensions of the frustum:
- The total height of the cone is $20\text{ cm}$ .
- The vertical angle is $60^\circ$ . Therefore, the semi-vertical angle ( $\theta$ ) of the cone is: $\theta = \frac{60^\circ}{2} = 30^\circ$
- The plane cuts the cone at the middle of its height (at height $10\text{ cm}$ from the vertex).
- Find the upper radius ( $r$ ) and lower radius ( $R$ ) of the resulting frustum using trigonometry: $r = 10 \tan(30^\circ) = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}}\text{ cm}$ $R = 20 \tan(30^\circ) = 20 \times \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}}\text{ cm}$
- The height of the frustum ( $h$ ) is: $h = 20 - 10 = 10\text{ cm}$
Calculate the volume of the frustum ( $V_{\text{frustum}}$ ): $V_{\text{frustum}} = \frac{1}{3}\pi h \left( R^2 + Rr + r^2 \right)$ Substitute $h = 10$ , $R = \frac{20}{\sqrt{3}}$ , and $r = \frac{10}{\sqrt{3}}$ : $V_{\text{frustum}} = \frac{1}{3}\pi (10) \left[ \left(\frac{20}{\sqrt{3}}\right)^2 + \left(\frac{20}{\sqrt{3}}\right)\left(\frac{10}{\sqrt{3}}\right) + \left(\frac{10}{\sqrt{3}}\right)^2 \right]$ $V_{\text{frustum}} = \frac{10}{3}\pi \left[ \frac{400}{3} + \frac{200}{3} + \frac{100}{3} \right]$ $V_{\text{frustum}} = \frac{10}{3}\pi \left[ \frac{700}{3} \right] = \frac{7000}{9}\pi\text{ cm}^3$
Establish the properties of the recast wire:
- The wire is a long solid cylinder.
- Diameter of the wire $= \frac{1}{16}\text{ cm} \implies$ Radius ( $r_{\text{wire}}$ ) $= \frac{1}{32}\text{ cm}$ .
- Let the length of the wire be $L$ (in cm).
- Volume of the wire ( $V_{\text{wire}}$ ): $V_{\text{wire}} = \pi (r_{\text{wire}})^2 L = \pi \left(\frac{1}{32}\right)^2 L = \frac{\pi L}{1024}\text{ cm}^3$
Equate the volumes to solve for $L$ : $\text{Volume of wire} = \text{Volume of frustum}$ $\frac{\pi L}{1024} = \frac{7000}{9}\pi$ $L = \frac{7000 \times 1024}{9}\text{ cm}$ $L = \frac{7,168,000}{9}\text{ cm} \approx 796,444.44\text{ cm}$
Convert the length into meters: $\text{Length in meters} = \frac{L}{100} = \frac{71,680}{9}\text{ m} \approx 7964.44\text{ m}$

Final Answer: The length of the wire is approximately $7964.44\text{ meters}$ (or $\frac{71680}{9}\text{ m}$ ).