Calculation of Area and Volume MCQ Quiz - Objective Question with Answer for Calculation of Area and Volume - Download Free PDF

Last updated on Jun 11, 2025

Latest Calculation of Area and Volume MCQ Objective Questions

Calculation of Area and Volume Question 1:

To the sum of the first and the last ordinates, add twice the sum of the intermediate ordinates. The total sum thus obtained is multiplied by the common distance between the ordinates. One-half of this product gives the required area. This rule of finding the area is called

  1. Average ordinate rule
  2. Simpson's rule
  3. Trapezoidal rule
  4. Mid-ordinate rule

Answer (Detailed Solution Below)

Option 3 : Trapezoidal rule

Calculation of Area and Volume Question 1 Detailed Solution

Concept:

The rule described is used for approximating the area under a curve using straight-line segments (trapezoids) between data points. This is known as the Trapezoidal Rule.

Formula:

\( A = \frac{h}{2} \left[ y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1}) \right] \)

Where:

  • \( h \) is the spacing between ordinates
  • \( y_0 \) and \( y_n \) are the first and last ordinates
  • All others are intermediate ordinates

Calculation of Area and Volume Question 2:

If the ordinates are provided as 25 m, 44 m, and 20 m, then the value of d is 2 m. Determine the area using the mid-ordinate.

  1. 168 sq. m
  2. 198 sq. m
  3. 178 sq. m
  4. 188 sq. m
  5. 180 sq. m

Answer (Detailed Solution Below)

Option 3 : 178 sq. m

Calculation of Area and Volume Question 2 Detailed Solution

Concept:

The formula for the area of the mid-ordinate can be given as

Area = common distance × sum of mid-ordinates

Area of plot = h1 × d + h2 × d + … + hn × d = d (h1 + h2 + … hn)

A = d × ∑O.

Calculation:

Given that,

d = 2 m

Ordinate = 25 m, 44 m, and 20 m

On substitution,

A = 2 × (25 + 44 + 20)

A = 178 sq. m.

Calculation of Area and Volume Question 3:

The sum of first and last ordinates, add twice the sum of the remaining odd ordinates and four times the sum of all the even ordinates. The total sum thus obtained is multiplied by one-third of the common distance between the ordinates and the result gives the required area. This rule of finding the area is called:

  1. Mid-ordinate rule
  2. Trapezoidal rule
  3. Average ordinate rule
  4. Simpson's rule
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : Simpson's rule

Calculation of Area and Volume Question 3 Detailed Solution

Explanation:

Simpson's rule:

This rule is based on the assumption that the figures are trapezoids.

In order to apply Simpson's rule, the area must be divided in even number i.e., the number of offsets must be odd i.e., n term in the last offset 'On' should be odd. 

The area is given by Simpson's rule:

\(Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + ........ + {O_{n - 1}}) + 2({O_3} + {O_5} + ......{O_{n - 2}})} \right]\)

where O1, O2, O3, .........Ois the offset

Important Points 

  • In case of an even number of cross-sections, the end strip is treated separately and the area of the remaining strip is calculated by Simpson's rule. The area of the last strip can be calculated by either trapezoidal or Simpson's rule.

Calculation of Area and Volume Question 4:

Estimate the quantity of earthwork for a portion of road using the trapezoidal formula, in which distance between the sections of the road D=30 m and cross-sectional areas \(A_0 = 5.5 \, \text{m}^2, \quad A_1 = 11.5 \, \text{m}^2, \quad A_2 = 17 \, \text{m}^2, \quad A_3 = 24.5 \, \text{m}^2 \)

  1. 1478 m3
  2. 1256 m3
  3. 1189 m3
  4. 1305 m3

Answer (Detailed Solution Below)

Option 4 : 1305 m3

Calculation of Area and Volume Question 4 Detailed Solution

Concept:

To estimate the earthwork quantity between multiple cross-sections, we use the Trapezoidal Formula:

\( V = D \left[ \frac{A_0 + A_n}{2} + A_1 + A_2 + \dots + A_{n-1} \right] \)

Given:

Distance between sections, \( D = 30 \, \text{m} \)

Cross-sectional areas: \( A_0 = 5.5 \, \text{m}^2, \, A_1 = 11.5 \, \text{m}^2, \, A_2 = 17 \, \text{m}^2, \, A_3 = 24.5 \, \text{m}^2 \)

Calculation:

Apply the trapezoidal formula:

\(V = 30 \left[ \frac{5.5 + 24.5}{2} + 11.5 + 17 \right]\)

\(V = 30 \left[ \frac{30}{2} + 11.5 + 17 \right] = 30 \left[15 + 11.5 + 17 \right]\)

\(V = 30 \times 43.5 = 1305 \, \text{m}^3\)

Calculation of Area and Volume Question 5:

The trapezoidal formula is used to calculate ______.

  1.  Volume of excavation 
  2. Area of irregular plots 
  3.  Quantity of bricks
  4. Reinforcement steel weight

Answer (Detailed Solution Below)

Option 2 : Area of irregular plots 

Calculation of Area and Volume Question 5 Detailed Solution

Explanation:

Trapezoidal Formula Application

The trapezoidal formula is used to calculate the area of irregular plots. It is particularly useful in scenarios where the plot does not have a regular shape, and the boundaries are not straight lines. By dividing the irregular plot into several trapezoids, we can calculate the area of each trapezoid and then sum them to get the total area. This method is widely used in land surveying and civil engineering to accurately determine the area of land plots.

Analyzing the Given Options

  1. "Volume of excavation." (Incorrect)

    • The volume of excavation is usually calculated using other methods such as the cross-sectional area method or the prismoidal formula, not the trapezoidal formula.

    • The trapezoidal formula focuses on area calculation, not volume.

  2. "Area of irregular plots." (Correct)

    • The trapezoidal formula is specifically designed to calculate the area of irregularly shaped plots by dividing them into a series of trapezoids.

    • Summing the areas of these trapezoids gives the total area of the plot, making it the correct application of the trapezoidal formula.

  3. "Quantity of bricks." (Incorrect)

    • The quantity of bricks is generally calculated using the dimensions of the wall or structure and the size of individual bricks, not using the trapezoidal formula.

    • The trapezoidal formula does not apply to the calculation of brick quantities as it is intended for area calculations.

  4. "Reinforcement steel weight." (Incorrect)

    • The weight of reinforcement steel is typically calculated by determining the volume and density of the steel, not by using the trapezoidal formula.

    • While the trapezoidal formula is useful for area calculations, it does not apply to the weight calculations of reinforcement steel.

Top Calculation of Area and Volume MCQ Objective Questions

Prismoidal correction, while surveying is always?

  1. Exponentially subtractive
  2. Exponentially additive
  3. Subtractive
  4. Additive

Answer (Detailed Solution Below)

Option 3 : Subtractive

Calculation of Area and Volume Question 6 Detailed Solution

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Explanation:

The volume of earthwork by trapezoidal method  = V1

V1 =  \(common\: distance\left \{ \frac{First\: area + Last\: area}{2}+the \:sum \:of\: remaining \:area \right \}\)

The volume of earthwork by prismoidal formula = V2

V2 = \(=\frac{Common\: distance}{3}\left \{ First\: area+ Last\: area + 2(Sum\: of odd\: area) + 4(Sum\: of even\: area)\right \}\)

Prismoidal correction:

  • The volume by the prismoidal formula is more accurate than any other method
  • But the trapezoidal method is more often used for calculating the volume of earthwork in the field.
  • The difference between the volume computed by the trapezoidal formula and the prismoidal formula is known as a prismoidal correction.
  • Since the trapezoidal formula always overestimates the volume, the prismoidal correction is always subtractive in nature is usually more than calculated by the prismoidal formula, therefore the prismoidal correction is generally subtractive.
  • Volume by prismoidal formula = volume by the trapezoidal formula - prismoidal correction

Prismoidal correction (CP)

\(C_{P}=\frac{DS}{6}\left \{ d- d_{1}\right \}^{2}\)

Where, D = Distance between the sections, S (Horizontal) : 1 (Vertical) = Side slope, d and d1 are the depth of earthwork at the centerline

In earthwork computations on a longitudinal profile, the diagram prepared to work out the quantity of earthwork is:

  1. double mass curve
  2. mass haul diagram
  3. Mollier diagram
  4. flow net

Answer (Detailed Solution Below)

Option 2 : mass haul diagram

Calculation of Area and Volume Question 7 Detailed Solution

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Explanation:

Mass Haul Curve:

  • This is a curve representing the cumulative volume of earthwork at any point on the curve, the manner in which earth to be removed.
  • It is necessary to plan the movement of excavated soil of worksite from cuts to fill so that haul distance is minimum to reduce the cost of earthwork.
  • The mass haul diagram helps to determine the economy in a better way.
  • The mass haul diagram is a curve plotted on a distance base with the ordinate at any point on the curve representing the algebraic sum of the volume of earthwork up that point.

A haul refers to the transportation of your project’s excavated materials. The haul includes the movement of material from the position where you excavated it to the disposal area or a specified location. A haul is also sometimes referred to as an authorized haul.

Haul = Σ Volume of earthwork × Distance moved.

What is the volume of earthwork for constructing a tank that is excavated in the level ground to a depth of 4 m ? The top of the tank is rectangular in shape having an area of 50 m × 40 m and the side slope of the tank is 2: 1 (horizontal: vertical).

  1. 5461 m3
  2. 6688 m3
  3. 8866 m3
  4. 5632 m3

Answer (Detailed Solution Below)

Option 4 : 5632 m3

Calculation of Area and Volume Question 8 Detailed Solution

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Concept:

a) Trapezoidal Formula:

Volume (v) of earthwork between a number of sections having areas A1, A2,…, An spaced at a constant distance d.

\({\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1{\rm{\;}}}}} \right]\)

Also can be written as 

\(V = \frac{D}{2}\left[ {({A_1} + A_n +2 \times (A2 +..+ A_{n-1} )} \right]\)

Calculation:

Given, A1 = 50 × 40 = 2000 m2, As the side slope is given 2:1 i.e. H:V

So for a depth of 1 m, there is a change of 2 m in a Horizontal Direction.

So at 4 m vertical depth

Bottom Dimension is ( 50 - 16 ) = 34 m & (40 - 16) = 24 m

∴ The bottom area is 34 m × 24 m = 816 m2

Mean area (Am) = (2000 + 816) / 2 = 1408 m2

F1 Raam Shraddha 29.01.2021 D1

According to simple Trapezoidal rule for volume,

\(V = \frac{D}{2}\left[ {({A_1} + 2Am + {A_2})} \right]\)

\(V = \frac{2}{2}\left[ {(2000 + 2\times (1408) + 816)} \right]\) = 5632 m3

The Simpson’s rule for determination of areas is used when the number of offsets are:

  1. 2
  2. even
  3. 3
  4. odd

Answer (Detailed Solution Below)

Option 4 : odd

Calculation of Area and Volume Question 9 Detailed Solution

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Explanation:

Simpson's rule:

This rule is based on the assumption that the figures are trapezoids.

In order to apply Simpson's rule, the area must be divided in even number i.e., the number of offsets must be odd i.e., n term in the last offset 'On' should be odd. 

The area is given by Simpson's rule:

\(Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + ........ + {O_{n - 1}}) + 2({O_3} + {O_5} + ......{O_{n - 2}})} \right]\)

where O1, O2, O3, .........On is the offset

Important Points

  • In case of an even number of cross-sections, the end strip is treated separately and the area of the remaining strip is calculated by Simpson's rule. The area of the last strip can be calculated by either trapezoidal or Simpson's rule.

To the sum of the first and last ordinates, add twice the sum of the intermediate ordinates. The total sum thus obtained is multiplied by the common distance between the ordinates. One-half of this product gives the required area. This rule of finding the area is called

  1. mid-ordinate rule
  2. trapezoidal rule
  3. average ordinate rule
  4. Simpson's rule

Answer (Detailed Solution Below)

Option 2 : trapezoidal rule

Calculation of Area and Volume Question 10 Detailed Solution

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Concept:

i) Mid-Ordinate Rule

Area of plot = h1 × d + h2 × d + … + hn × d = d (h1 + h2 + … hn)

∴ Area = common distance × sum of mid-ordinates

ii) Average-Ordinate Rule

\(\text{Area}=\frac{{{\text{O}}_{1}}+{{\text{O}}_{2}}+\ldots +{{\text{O}}_{\text{n}}}}{{{\text{O}}_{\text{n}+1}}}\times \text{l}\)

\(\text{i}.\text{e}.\text{ }\!\!~\!\!\text{ Area}=\frac{\text{Sum }\;\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ ordinates}}{\text{No}.\;\text{ }\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ ordinates}}\times \text{length }\;\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ base }\;\!\!~\!\!\text{ line}\)

(iii) Trapezoidal Rule

\(\text{Total }{area}=\frac{\text{d}}{2}\left\{ {{\text{O}}_{1}}+2{{\text{O}}_{1}}+2{{\text{O}}_{2}}+\ldots +2{{\text{O}}_{\text{n}-1}}+{{\text{O}}_{\text{n}}} \right\}\)

\(\text{Total Area}=\frac{\text{Common distance}}{2}\times \{\left( 1\text{st ordinate}+\text{last ordinate} \right)+2\left( \text{sum of other ordinate} \right))\}\)

(iv) Simpson’s Rule

\(\text{Total }{ area}=\frac{\text{d}}{3}\left( {{\text{O}}_{1}}+4{{\text{O}}_{2}}+2{{\text{O}}_{3}}+4{{\text{O}}_{4}}+\ldots +{{\text{O}}_{\text{n}}} \right)\)

\(\text{Total Area}=\frac{\text{Common Distance}}{3}\times \left\{ \left( 1\text{st ordinate}+\text{last ordinate} \right)+4\left( \text{sum of even ordinates} \right)+2\left( \text{sum of remaining odd ordinates} \right) \right\}\)

Which of the following formulas explains the calculation of area of earthwork using its mean depth?

Area = BD + SD2

Where, B = Breadth of section

D = Mean depth of section

SD = Areas of sides 

  1. Average area formula
  2.  Prismoidal formula
  3. Trapezoidal formula
  4. Mid-section formula

Answer (Detailed Solution Below)

Option 4 : Mid-section formula

Calculation of Area and Volume Question 11 Detailed Solution

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Explanation:

Methods for measurement of earthwork:

1. Mid Section method  2. Trapezoidal Method  3. Prismoidal Method  4. Simpson's 3/8th rule

Mid Section Method:

In this method the quantity of earthwork is computed with the help of size of mid section.

Area of earthwork \(= \frac{1}{2}\left\{B+\left ( B+2SD \right ) \right\}\times D = BD+SD^{2}\) 

Volume of earthwork \(= Area\times Length=\left\{BD+SD^{2} \right\} \times L \)

F1 SSC  Priyas 14 5 24 D3

Determine the approximate quantity of earthwork for a road in embankment having a length of 120 m on a uniform level ground. The width of formation is 10 m and side slopes are 3 ∶ 1. The heights of the bank at the ends are 1 m and 1.5 m, respectively. Use trapezoidal method considering average of areas at the two ends.

  1. 1785 m3
  2. 1485 m3
  3. 1885 m3
  4. 2085 m3

Answer (Detailed Solution Below)

Option 4 : 2085 m3

Calculation of Area and Volume Question 12 Detailed Solution

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Concept:

Volume of Sloped earthwork (V) = (bd + sd2) × L   

Where, B   = Width, d  = Depth and S = Side slope of the cross-section

Calculation:

Given,

slope = s: 1 = 3: 1

b = 10 m

L = 120 m

d1 = 1.5 m and d2 = 1 m

Volume of Sloped earthwork (V) = (bd + sd2) × L   

\(V = \;\frac{1}{2}\left[ {(b{d_1}\; + {\rm{ }}s{d_1}^2) + (b{d_2}\; + {\rm{ }}s{d_2}^2)} \right] \times 120\)

\(V = \;\frac{1}{2}\left[ {(10 \times 1.5\; + {\rm{ }}3 \times {1.5^2}) + (10 \times 1\; + {\rm{ }}3 \times {{1}^2})} \right] \times 120 = 2085{m^3}\)

The approximate quantity of earthwork = 2085 m3

A road embankment 10 m wide at the formation level with side slopes 2:1 and with an average height of 5 m is constructed with an average gradient of 1:40 from the contour 220 m to 280 m. Find the volume of earthwork.

  1. 6,40,000 m3
  2. 1,40,000 m3
  3. 3,40,000 m3
  4. 2,40,000 m3

Answer (Detailed Solution Below)

Option 4 : 2,40,000 m3

Calculation of Area and Volume Question 13 Detailed Solution

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Concept:

Gradient: 

A gradient is the rate of rise or falls along the length of the road with respect to horizontal. It is expressed as ‘1' vertical unit to 'N' horizontal units.

F1 Abhishek M 1.3.21 Pallavi D1

\(Tan\ \theta \ =\frac{{\bf{h}}}{{\bf{l}}}\)

\(\frac{{\bf{h}}}{{\bf{l}}} = \frac{1}{{N}}\)

Area of trapezoidal:

According to the trapezoid area formula, the area of a trapezoid is equal to half the product of the height and the sum of the two bases.

Area = ½ x (Sum of parallel sides) x (perpendicular distance between the parallel sides).

Calculation:

F1 Abhishek M 1.3.21 Pallavi D2

Road embankment = 10 m

Average height = 5 m

Difference in elevation(h) = 280 - 220 = 60 m

Average gradient = \( \frac{1}{{40}}\)

\(\frac{{\bf{h}}}{{\bf{l}}} = \frac{1}{{40}}\)

\(\frac{{\bf{60}}}{{\bf{l}}} = \frac{1}{{40}}\)

L = 2400 m

F1 Abhishek M 1.3.21 Pallavi D3

Average cross-sectional area(A) = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaaGymaaWdaeaapeGaaGOmaaaacqGHxdaTdaqa % daWdaeaapeGaaGymaiaaicdacqGHRaWkcaaIZaGaaGimaaGaayjkai % aawMcaaiabgEna0kaaiwdaaaa!423F! \frac{1}{2} \times \left( {10 + 30} \right) \times 5\)\(\frac{1}{2} \times \left( {10 + 30} \right) \times 5\)

A = 100 m2

Volume of earthwork = A × L

Volume of earthwork = 100 × 2400

∴ Volume of earthwork = 2,40,000 m3

What is the volume of a 6 m deep tank having rectangular shaped top 6 m x 4 m and bottom 4 m x 2m (computed through the use of prismoidal formula)?

  1. 92 m3
  2. 96 m3
  3. 90 m3
  4. 94 m3

Answer (Detailed Solution Below)

Option 1 : 92 m3

Calculation of Area and Volume Question 14 Detailed Solution

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Concept:

Calculation:

Given: L = 6m 

Top area = 6 × 4 = 24 m2 = A1

Bottom area = 4 × 2 = 8 m2 = A2

\({{\rm{A}}_{\rm{m}}} = \left( {\frac{{6 + 4}}{2}} \right) \times \left( {\frac{{4 + 2}}{2}} \right) = 15{\rm{\;}}{{\rm{m}}^2}\)

\({\rm{Volume\;}} = \frac{{\rm{L}}}{6}\left( {{{\rm{A}}_1} + 4{{\rm{A}}_{\rm{m}}} + {{\rm{A}}_2}} \right)\)

\({\rm{Volume}} = \frac{6}{6}\left( {24 + 4 \times 15 + 8} \right) = 92{\rm{\;}}{{\rm{m}}^3}\)

To apply Simpson’s rule for computation of irregular area, number of segments should be-

  1. Of equal width and even number
  2. May be of varying width also 
  3. Of equal width and odd number
  4. May be even or odd

Answer (Detailed Solution Below)

Option 1 : Of equal width and even number

Calculation of Area and Volume Question 15 Detailed Solution

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Explanation: Simpson's rule:

  1. Simpson's rule is also known as Parabolic Rule.
  2. For calculation of volume, this formula is known as the Prismoidal rule 
  3. This rule is used when the ends of ordinates or straight to form an arc of a parabola.
  4. This rule is applicable only when the number of divisions is even, i.e. the number of ordinates is odd.  
  5. This rule can be stated as follows: One-third of common distance multiplied by the sum of last and first ordinate, four times the sum of even ordinates, and twice the sum of the remaining odd ordinates are added.

Additional Information

Simson's Rule  Trapezoidal Rule 
The boundary between the ordinates is considered to be an arc of a parabola. The boundary between the ordinates is considered to be straight.
It gives a more accurate result.

It gives an approximate result 

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