Area – Aptitude Questions and Answers

Surface area topic is an integral part of various competitive examinations in India. In fact, they are an important part of Geometry which is usually asked in Quantitative Aptitude. Being able to understand and solve area questions quickly and accurately is essential to help you score well in exams. This article aims to provide you with information on how to calculate the surface area of various 2D and 3D shapes, some important concepts, shortcuts, and tricks related to Areas along with practice questions and their detailed solutions. 

What are shapes?

A shape in geometry is defined as the figure encompassed by the boundary. The boundary is formed by combining lines, points, and curves. In general, there are two different categories of geometric shapes, including:

•  2 Dimensional Shapes (2 D)
•  3 Dimensional Shapes (3 D)

The shapes that can be drawn on a piece of paper are known as 2-dimensional shapes, also called flat shapes. The shapes have two dimensions only i.e. length and breadth, no thickness. Examples of 2D shapes are circles, rectangles, squares, triangles, eclipses, parallelograms, hexagons, pentagons, and so on.

3D shapes are also known as solid shapes, they have 3 dimensions such as length, breadth, and thickness. 3-D shapes examples are cones, cylinders, spheres, cubes, cuboids, hemispheres, prism, and so on.  

Shape	Types	Dimensions
Square	2 D	Side
Rectangle	2 D	Length and Breadth
Circle	2 D	Radius
Cube	3 D	Length = Breadth = Height
Cuboid	3 D	Length, Breadth, and Height
Cylinder	3 D	The radius of the Base, Height
Sphere	3 D	Radius
Right Circular Cone	3 D	Slant height and Radius of the base

What is Area?

The area is a measure of the space occupied by a 2D flat surface or the space taken by the outer surface of 3 D object. It’s usually measured in square units like square meters, square feet, or acres.

Total Surface Area includes complete surface area i.e. curved area + area of the base, whereas Lateral surface area or Curved surface area includes only the curved area of the shape.

In the below table, we have listed down a few important shapes and important terms related to them.

Name of the Shape	Perimeter	Total Surface Area	Curved Surface Area/ Lateral Surface Area	Shape
Square	4b	b2	—-	Square
Rectangle	2(w+h)	w*h	—-	Rectangle
Parallelogram	2(a+b)	b*h		Parallelogram
Trapezoid	a+b+c+d	1/2(a+b).h	—-	Trapezoid
Circle	2πr	πr2	—-	Circle
Ellipse	2π√(a2 + b2)/2	π a.b	—-	Ellipse
Triangle	a+b+c	1/2 (b*h)	—-	Triangle
Cuboid	4(l+b+h)	2(lb+bh+hl)	2h(l+b)	Cuboid
Cube	6a	6a2	4a2	Cube
Cylinder	—-	2 π r(r+h)	2πrh	Cylinder
Cone	—-	π r(r+l)	π r l	Cone
Sphere	—-	4πr2	4πr2	Sphere
Hemisphere	—-	3πr2	πr2	Hemisphere

Tips to Prepare for Surface Area Questions 

1. Learn the basic concepts thoroughly: Before attempting any question, it is important to understand the basic concepts and the formulae related to them. Make sure that you understand the theorems and the formulae related to each shape in detail. 

2. Practice regularly: To score well, it is important to practice as many questions as possible. This will help you to understand how to apply the concepts and formulae in different situations. 

3. Learn the formulas: Mensuration questions involve a lot of calculations. Therefore, it is important to remember the formulas for each shape. 

4. Try to solve the questions in a step-by-step manner: When attempting a question, try to solve it in a step-by-step manner. This will help you to avoid any mistakes and will also help you to understand the entire process better.

5. Understand the diagrams: Mensuration questions involve diagrams as well. Therefore, make sure that you understand the diagrams given in the questions properly. 

6. Revise regularly: After you have learned the concepts and practiced the questions, make sure that you revise them regularly. This will help you to remember the formulae and concepts. 

Perimeter and Area of the Sector of the Circle 

Surface Area Questions and Answers

Q1. A rectangular plot has its length and width ratio 4:1. At 10km/hr speed, Ankit was cycling along the plot’s edge. In six minutes, he completes one round of the plot. Determine the plot’s area.

A. 52000 m²
B. 51500 m²
C. 53500 m²
D. 52500 m²

Solution: 

Distance covered by Rahul in 6 minutes = 10000/60 × 6 = 1000 m 
[10 km/hr = 10000/60  m/min]
Therefore, perimeter = 1000m
Length: Breadth = 4:1
Therefore, Length = 4x and breadth = x
Then, Perimeter of the Rectangle= 2 (l +b) = 1000
2 (4x + x) = 1000
2 (5x) = 1000
10x = 1000
x = 100
Length = 4x = 4* 100 = 400
Breadth = x = 100 = 100
Therefore, Area = l * b = 400*100 = 40000 m²
Correct option: D

Q2. The height of a trapezium-shaped wall is 8 meters. The trapezium’s parallel sides measure 4 and 6 meters. Find the cost of painting a whole wall if the price per square meter is Rs. 50.

A. Rs. 400
B. Rs. 420
C. Rs. 540
D. Rs. 450

Solution:  

Area of trapezium =1/2 * (Sum of parallel sides) * Distance between them
Area of trapezium = 1/2 * (4 + 6) + 8
Area of trapezium = 1/2 * (18)
Area of trapezium = 9 square meter
Rate of painting per square meter is Rs. 50
Therefore, to paint 9 square meter, total cost of painting = 9 * 50 = Rs. 450
Correct option: D

Q3. A rope makes 120 rounds of a cylinder with a base radius of 10 cm. How many times it can go around a cylinder with a base radius of 20 cm?

A. 70 cm
B. 60 cm
C. 45 cm
D. 50 cm

Solution:

Let the round be x
If radius is more, then rounds will be less as the length of the ropes remains the same
L = 2*π*10*120 —-(1)
Similarly,
L = 2 * π * 20 * x —-(2)
From (1) and (2)
10 * 120 = 20 * x
=> x = 60
Correct option: B

Q4. A 10 cm cube was divided into as n number of cubes of side lenght 1 cm cubes . Determine the total surface area ratio of the larger cube to the total surface areas of the smaller cubes.

A. 100:1
B. 10:1
C. 1:10
D. 1:100

Solution: 

Volume of the original cube = 10³ = 1000 cm³
Volume of each smaller cubes = 1 cm³. It means there are 1000 smaller cubes.
Surface area of the cube = 6a²
Surface area of the larger cube = 6a² = 6 * 10² = 6 * 100 = 600
Surface area of one smaller cubes = 6 (1²) = 6
Now, surface area of all 1000 cubes = 1000* 6 = 6000
Therefore,
Required ratio = Surface area of the larger cube: Surface area of smaller cubes
= 600: 6000
= 1:10
Correct option: C

Q5. After being submerged in water, a rectangular piece of cloth was found to have lost 20% of its length and 10% of its width. Calculate the overall percentage of the rectangular piece of cloth’s area reduction.

A. 75% Decrease
B. 28 % Increase
C. 28 % Decrease
D. 20% Decrease

Solution:  

Let the original length = l
Let the original breadth = b
Original Area = l * b
New length =  80/100 l
New breadth =90/100 b
Decrease in the area = lb – 80/100 l * 90/100 b
Decrease in the area = 7/25 lb
Decrease percentage =  7/25 lb * 7/lb * 100
Decrease percentage =  700/25 = 25%
Correct Option: C

Q6. ABCD is a square. AD is tangent to a circle with a radius r and OE = ED. What is the ratio of the area of a circle to the area of the square?

a) π/3
b) πr²/3
c) πr²/4
d) πr²/2

Solution:

OD² = OA²+ AD²
(2r)² = r² + AD²
Thus PQ, which is also the side of square, is equal to r????. The area of square becomes:  3r²
Hence, the ratio of the area of circle to square is:
(area of circle)/(area of square)=(πr²)/(3r² )=π/3  
Correct Option: A

Q7. A kite is in the shape of a square with a diagonal 48cm attached to a triangle of the base 4cm and height 6cm. How much paper has been used to make it? 

a) 200 cm²
b) 288 cm²
c) 300 cm²
d) 325 cm²

Solution:

Area of square = 1/2(diagonal)²
=1/2 * (24)²
=288 cm²
Area of triangle = 1/2 * base * height
= 1/2 * 4 * 6 = 12 cm²
Total area = 288 + 12 = 300 cm²
Correct Option: C

Q8. If G is the centroid and AD, BE, CF are three medians of triangle ABC with area 72 cm², then the area of triangle BDG is: 

a) 12 cm²
b) 16 cm²
c) 24 cm²
d) 8 cm² 

Solution: 

Given

G is the centroid and AD, BE, CF are three medians and the area of ????GE = 12 cm²
As, we know the median divides the triangle into 6 triangles of equal area
Hence, area of the quadrilateral BDGF= 2*????GE = 2*12 cm²
Area of the quadrilateral BDGF = 24 cm²
Correct Option: 24 cm²

Q9. If the perimeter of rhombus is 150 cm and length of one diagonal is 50 cm. Then find the length of second diagonal and area of rhombus.

(a) 425???? cm²
(b) 525???? cm²
(c) 625???? cm²
(d) 725???? cm²

Solution:

Perimeter = 4a=150 
a= 75 cm
4a²= d₁² + d²₂
75²=50² + d²₂
d₂ = 25???? 
Area = ½(d₁d₂) = 625???? cm²
Correct Option: C

Q 10. How many copper rods, each measuring 7 metres in length and 2 cm in diameter, can be produced using 0.88 cubic metres of copper?

(a)145
(b)345
(c)380
(d)400

Solution:

Volume of 1 rod =

Volume of 1 rod

Volume of copper = 0.88 m³

Number of rods =

Number of rods

Correct Option: D