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Best explanation to understand and solve the area problems.

In this section, we will learn about the aptitude topic AREA. This topic is well known by everyone because we have learned a lot about this in our school days.

This has a various set of topics to cover. The most important point to remember in area is that you need to know all the formulas of the shapes. This is a challenging task. 

To learn all the formulas and to know more about area aptitude,read the full article.

We have given detailed explanations of different terms in area along with its formulas.

It will help you to learn the formulas easily and you can score good marks in your exams.

Also we have a set of aptitude questions based on AREA. you can take up the test and practice for your exams.

What is area?

Area is nothing but the measurement of the surface of the given shape.

What are the topics under area aptitude?

There are three important topics in area aptitude. They are as follows.

  1. Quadrilateral
  2. Circles
  3. Triangles 

Quadrilateral:

These are nothing but shapes with four sides. For eg: square, rectangle, parallelogram,etc. 

Circles:

A circle is nothing but the locus of all points equidistant from a central point.

Triangles:

These are shapes with three angles. 

What are the formulas to calculate area of quadrilaterals?

i. Area of a rectangle = Length x Breadth

Length of rectangle = Area / Breadth , Breadth of rectangle = Area / Breadth

(Diagonal) 2 = (Length) 2 + (Breadth) 2

Perimeter of a rectangle = 2(Length + Breadth)

ii. Area of Parallelogram:

If a, b, c, and d are the lengths of a parallelogram,

S = 1/2 * (a + b + c + d)

Area = 2s(s – a)(s – b)(s – c)(s – d)

iii. Area of a square: (side)2 = 1/ 2 *(diagonal)2

Perimeter of a square = 4 x side

Area of Rhombus = 1/2 × (Product of diagonals)

When d1 and d2 are the two diagonals, then the side of rhombus = 1/2 * d12+d22

v. Area of 4 walls of a room = 2 x (Length + Breadth) x Height.

vi. Area of Parallelogram = (Base x Height).

vii. Area of trapezium = 1/2 (sum of parallel sides perpendicular distance between them)

= 1/2 * (a+b)h

Where a, and b are the parallel sides of the trapezium and h is the perpendicular distance between the sides ‘a’ and ‘b’.

Where k = (a-b), i.e., the difference between the parallel sides and c and d are the two non-parallel sides of the trapezium. Also,

S = (k + c + d) / 2

Area = 1/2 * (a + b), h = (a + b) / k

s(s – k)(s – c)(s – d)

How to calculate area of circles?

(i) Circumference of a circle = 2πr

(ii) Area of Circle = πr2

(iii) Arc AB = 2r / 3600 , where ∠AOB= θ and O is the center.                      

(iv) Area of sector AOB = r2 / 3600

(v) Area of Sector AOB = 1/2 × arc AB × r

How to calculate area of triangle?

(i) Area of triangle = (1/2) × Base × height

(ii) If a, b, c are the lengths of the sides of a triangle and s = 1/2 * (a+b+c):

Area of a triangle = s(s – a)(s – b)(s -c)

(iii) Area of an isosceles triangle = b / A * 4a2-b2

(iv) Area of equilateral triangle= 3 /4 ×(side)2 

Perimeter of an equilateral triangle = 3 × side

What are the quick methods to solve area aptitude?

1. Carpeting a Floor:

(i) Length of the carpet required = Area Aptitude

(ii) Amount Required = Rate per meter × Area Aptitude

2. Paving with tiles:

(i) Number of tiles required = Area Aptitude

(ii) The side of the largest possible square tile = HCF of length and breadth of the room.

3. Path around a garden, verandah around a room.

(i) When verandah is outside the room, surrounding it 

Area of verandah = 2 (width of verandah) x [length + breadth of room + 2(width of verandah)].

(ii) When the path is within the garden, surrounded by it

Area of path = 2(width of path) x [length + breadth of garden – 2(width of path)].

(iii) When the area of the path is given, to find the area of the garden enclosed (the garden is square in shape).

Area of square garden = Area Aptitude

4. Rhombus

(i) Area of a rhombus = 1/2 * (product of diagonals)

Area Aptitude

5. If the length and breadth of a rectangle is increased by x and y percent respectively, the area is increased by Area Aptitude.

6. If all the measuring sides of any two-dimensional figure is changed by x%, its area changes by Area Aptitude.

7. If all the measuring sides of any two-dimensional figure are changed (increased or decreased) by x%, its perimeter also changes by x%.

8. If all the sides of a quadrilateral are increased by x%, its corresponding diagonals also increased by x%.

9. If the area of a square is x cm2, the area of the circle formed by the same perimeter is given by Area Aptitude .

10. Area of a square inscribed in a circle of radius r is 2r2.

11. The area of the largest triangle inscribed in a semicircle of radius r is r2.

These are the very important formulas that you need to know before attending area aptitude.

This will be an easy task since many of you would have learned most of these topics in your higher secondary school.

The only hard part is that you need to memorise all these formulas which will be a little confusing. But when you practice a lot, you will get used to the formulas and yes you can easily score marks in your tests.

Don’t forget that we have a wonderful set of aptitude in our website. This will be very useful for you to practice for your exams.