# Mensuration Formula And Questions For 2D And 3D Shapes

## Mensuration Formula And Questions

Mensuration is the branch of mathematics that deals with measurements of different figures and shapes of geometry. It involves the calculation of areas, volumes etc of shapes. Various Mensuration formula are covered in Geometry useful for the exam point. In order to excel in this topic, you must know the formulas and concepts used to solve the questions. Check out the important basic maths mensuration formulas covered in this post with proper explanation.

### What are 2D and 3D shapes?

2D Shapes- In geometry, a two-dimensional shape is a flat plane figure or a shape that has only two dimensions i.e. length and width. Two-dimensional or 2-D shapes do not possess any thickness and can be measured in only two faces. Only the area and perimeter of 2D shapes can be calculated.

3D Shapes- A three-dimensional shape is where the figure has three dimensions namely length, width and thickness. We calculate volume, curved surface area, the total surface area of 3D shapes.

### Mensuration Formula of 2D Shapes

Check out the formula for area and perimeter of some of the 2D shapes:

Shape Area (Square units) Perimeter (units)
Square 4a
Rectangle l × b 2 ( l + b)
Circle πr² 2 π r
Scalene Triangle √[s(s−a)(s−b)(s−c)],

Where, s = (a+b+c)/2

a+b+c
Isosceles Triangle ½ × b × h 2a + b
Equilateral Triangle (√3/4) × a² 3a
Right Angle Triangle ½ × b × h b + hypotenuse + h
Rhombus ½ × d1 × d2 4 × side
Parallelogram b × h 2(l+b)
Trapezium ½ h(a+b) a+b+c+d

Area Of Trapezium: Definition, Properties, Formula And Examples

### Mensuration Formula of 3D Shapes

Check out the formula for some of the 3D shapes in geometry:

Shape Volume (Cubic units) Curved Surface Area (CSA) or
Lateral Surface Area (LSA) (Square units)
Total Surface Area (TSA) (Square units)
Cube 4 a² 6 a²
Cuboid l × b × h 2 h (l + b) 2 (lb +bh +hl)
Sphere (4/3) π r³ 4 π r² 4 π r²
Hemisphere (⅔) π r³ 2 π r² 3 π r²
Cylinder π r² h 2π r h 2πrh + 2πr²
Cone (⅓) π r² h π r l πr (r + l)

### Mensuration Formulas in Detail

Volume of a Sphere: Definition, Formula and Examples

### Mensuration Questions

Q1. The radius of a cylinder is 10 cm and the height is 4 cm. The number of centimetres that may be added either to the radius or to the height to get the same increase in the volume of the cylinder is :
एक बेलन की त्रिज्या 10 सेमी और ऊंचाई 4 सेमी है. सिलेंडर के आयतन में समान वृद्धि प्राप्त करने के लिए या तो त्रिज्या या ऊंचाई में कितने सेंटीमीटर जोड़े जाने चाहिए?
(a) 5
(b) 4
(c) 25
(d) 16

Ans. (a)
Sol.
Let ‘a’ cm is added in radius and height
π(10+a)²4 = π (10)² (4 +a)
(10+a)²4 = 10² (4 +a)
⇒ a = 5 cm

Q2. A solid sphere of radius 6 cm is melted to form a hollow right circular cylindrical tube of length 8 cm and external radius 10 cm. The thickness of the tube in m is
एक 6 सेमी त्रिज्या वाले ठोस गोले को एक 8 सेमी लम्बाई और 10 सेमी बाहरी त्रिज्या वाले एक खोखली लम्ब वृत्तीय बेलनाकार ट्यूब बनाने के लिए पिघलाया जाता है। ट्यूब की मोटाई मीटर में कितनी है?
(a) 1
(b) 0.01
(c) 2
(d) .02

Ans.(d)
Sol.
Volume of solid sphere
=4/3 π(6)³=288π cu.cm
⇒ Volume of the metal of tube
π(R² – r²)h cu.cm
where R = 10 cm, h = 8 cm
r = inner radius
∴ π(R²-r²)×h = 288π
⇒ (100 -r²) = 36
⇒ r = 8 cm
∴ thickness of cube = (10 – 8)cm
= 2 cm
= 0.02 m

Q3. PQRS is a rectangle. The ratio of the sides PQ and QR is 3 : 1. If the length of the diagonal PR is 10 cm, then what is the area (in cm²) of the rectangle?
PQRS एक आयत है। भुजाओं PQ और QR का अनुपात 3: 1 है. यदि विकर्ण PR की लंबाई 10 सेमी है, तो आयत का क्षेत्रफल (वर्ग सेमी में) क्या है?
(a) 15
(b) 30
(c) 45
(d) 20

Ans.(b)
Sol.

PQRS is a rectangle
PR = 10 given
PQ : QR = 3 : 1
In ∆PQR
9x² + x² = 100
10x² = 100
x = √10
Area of rectangle = 3x × 1x
= 3x²
= 3 × 10
= 30

Q4. The base of a prism is in the shape of an equilateral triangle. If the perimeter of the base is 18 cm and the height of the prism is 20 cm, then what is the volume (in cm³) of the prism?
एक प्रिज्म का आधार समबाहु त्रिभुज के आकार में है। यदि आधार की परिधि 18 सेमी है और प्रिज्म की ऊंचाई 20 सेमी है, तो प्रिज्म का आयतन (घन सेमी में) क्या है?
(a) 180√3
(b) 30√6
(c) 60√2
(d) 120√3

Ans. (A)
Sol.

Given:-
Perimeter = 18 &height=20
Hence ,Side = 6
Volume of prism = area of base × height
= √3/4×6 × 6 × 20
V = 180√3 cm3

Q5. The height of a cone is 24 cm and the area of the base is 154 cm². What is the curved surface area (in cm²) of the cone?
एक शंकु की ऊंचाई 24 सेमी है और आधार का क्षेत्रफल 154 वर्ग सेमी है। शंकु का वक्र पृष्ठीय क्षेत्रफल (वर्ग सेमी में) क्या है?
(a) 484
(b) 550
(c) 525
(d) 515

Ans.(B)
Sol. Area of base = 154
πr² = 154
22/7×r^2=154
r = 7

Height = 24
Slant height(ℓ) = √(h²+r² )
ℓ =√(24²+7² )
ℓ=25
C.S.A. = πrℓ
= 22/7×7×25
C.S.A. ⇒ 550 cm²

Q6. A right circular solid cylinder has radius of base 7 cm and height is 28 cm. It is melted to form a cuboid such that the ratio of its side is 2 : 3 : 6. What is the total surface area (in cm²) cuboid?
एक लंब वृत्तीय ठोस बेलन का आधार 7 सेमी और ऊंचाई 28 सेमी है। यह एक घनाभ बनाने के लिए पिघलाया जाता है, जिसकी भुजाओं का अनुपात 2: 3: 6. है। घनाभ का कुल पृष्ठीय क्षेत्रफल (वर्ग सेमी में) क्या है?
(a)72 ∛(( (1078)²)/(9²))
(b) ∛(2156/9)
(c) ∛(2148/3)
(d) ∛(2048/3)

Ans.(A)
Sol. according to question a cylinder is transforming into a cuboid as shown in figure below

Volume of cylinder = volume of cuboid
22/7×7×7×28=2x×3x×6x
x^3=1078/9 ⇒x=∛(1078/9)
T. Surface area of cuboid = 2 [ℓb + bh + hℓ]
=2×(1078/9)^(2/3) [2×3+3×6+6×2]
T.S.A.=72×(1078/9)^(2/3)

Q7. ABCDEF is a regular hexagon. What is the ratio of the area of triangle ACE and area of triangle AEF?
ABCDEF एक नियमित षट्भुज है। त्रिभुज ACE के क्षेत्रफल और त्रिभुज AEF के क्षेत्रफल का अनुपात क्या है? (a) 6 : 1
(b) 4 : 1
(c) 3 : 1
(d) 5 : 1

Ans.(c)
Sol.

Given is a regular hexagon
In a regular hexagon, there are 6 equilateral triangle as shown by dotted line
Area of ∆ACE = 1/2×[Area of 6 equilateral triangle]
= 3 equilateral triangle.
Area of ∆AEF = 1/2×[Area of 2 equilateral ∆]
= 1 equilateral triangle.
Hence ratio of
Area ACE : Area AEF = 3 : 1

Numerically proved
Let length of side of hexagon = 12
Interior angle ∠ABC = ((x–2)×180)/n = ((6–2)×180)/6 = 120
Area of ∆ABC = 1/2×AB×BC×sin⁡120
=1/2×12×12×cos⁡30
Area of ∆ABC = 36√3
Similarly
Area ∆CDE = Area ∆AEF = 36√3
Area of hexagon = 6×√3/4×12×12 = 216√3
Area of ACE = Area of hexagon – [Area of ∆ABC + ∆CDE + ∆AEF]
= 216√3 –108√3
= 108√3
∴ Area of ACE : Area of AEF = 108√3 ∶36√3
= 3 : 1

Q8. ABCD is a trapezium. Sides AB and CD are parallel to each other. AB = 6 cm, CD = 18 cm, BC = 8 cm and AD = 12 cm. A line parallel to AB divides the trapezium in two parts of equal perimeter. This line cuts BC at E and AD at F. If BE/EC = AF/FD, than what is the value of BE/EC?
ABCD एक ट्रेपेज़ियम है। भुजा AB और CD एक दूसरे के समानांतर हैं। AB = 6 सेमी, CD = 18 सेमी, BC = 8 सेमी और AD = 12 सेमी है। A रेखा AB के समांतर है, जो ट्रेपेज़ियम को समान परिधि के दो भागों में विभाजित करती है। यह रेखा BC को E और AD को F पर काटती है। यदि BE/EC = AF/FD, तो BE/EC का मान क्या है?
(a) 1/2
(b) 2
(c) 4
(d) 1/4
Ans.(c)
Sol.

All given values are shown in diagram
Let BE = x then EC = 8 – x
BE/EC=AF/FD (condition given)
Reverse the given condition & add 1 both side
EC/BE+1=FD/AF+1
Put values in eq. (i) → 8/x=12/AF AF=3x/2 & FD=12–3x/2
Now perimeter FABE = FECD
FA + AB + BE + FE = EC + CD + DF + FE
3x/2+6+x=8 –x+18+(12–3x/2)
5x = 32
x=32/5=BE, hence EC=8 –32/5=8/5
∴ BE/EC=(32/5)/(8/5)=4

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