FACTS  AND  FORMULAE  FOR  AREA  QUESTIONS

 

 

FUNDAMENTAL CONCEPTS :

I. Results on Triangles:

1. Sum of the angles of  a triangle is 180o

2. The sum of any two sides of a triangle is greater than the third side.

3. Pythagoras Theorem : In a right - angled triangle,

Hypotenuse2=Base2+Height2

4. The line joining the mid-point of a side of a triangle to the opposite vertex is called the median.

5. The point where the three medians of a triangle meet, is called Centroid. The centroid divides each of the medians in the ratio 2 : 1.

6. In an Isosceles triangle, the altitude from the vertex bisects the base.

7. The median of a triangle divides it into two triangles of the same area.

8. The area of the triangle formed by joining the mid-points of the sides of a given triangle is one-fourth of the area of the given triangle.

 

II.Results on Quadrilaterals :


1. The diagonals of a parallelogram bisect each other

2. Each diagonal of a parallelogram divides it into two triangles of the same area.

3. The diagonals of a rectangle are equal and bisect each other.

4. The diagonals of a square are equal and bisect each other at right angles

5. The diagonals of a rhombus are unequal and bisect each other at right angles

6. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.

7. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.

 

IMPORTANT FORMULAE

I. 

1. Area of a rectangle = (length x Breadth)

Length =AreaBreadth  and  Breadth=AreaLength

2. Perimeter of a rectangle = 2( length + Breadth)

 

 

II. Area of square = side2=12diagonal2 

 

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

 

 

IV.

1. Area of a triangle =12×base×height

2. Area of a triangle = s(s-a)(s-b)(s-c), where a, b, c are the sides of the triangle and s=12a+b+c

3. Area of an equilateral triangle =34×side2

4. Radius of incircle of an equilateral triangle of side a=a23

5. Radius of circumcircle of an equilateral triangle of side a=a3

6. Radius of incircle of a triangle of area  and semi-perimeter s=s

 

 

V.

1. Area of a parallelogram = (Base x Height)

2. Area of a rhombus = 12×Product of diagonals

3. Area of a trapezium = 12×(sum of parallel sides)×distance between them

    

 

VI.

1. Area of a cicle = πR2, where R is the radius.

2. Circumference of a circle = 2πR.

3. Length of an arc = 2πRθ360, where θ is the central angle.

4. Area of a sector = 12arc×R=πR2θ360 

 

VII.

1. Area of a semi-circle = πR22

2. Circumference of a semi - circle = πR

Q:

Square units 13 by 9 of an office area is

A) 97 B) 117
C) 107 D) 127
 
Answer & Explanation Answer: B) 117

Explanation:

Square units 13 by 9 of an office means office of length 13 units and breadth 9 units.

 

Now its area is 13x 9 = 117 square units or units square.

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24 14745
Q:

The perimeter of a square is equal to twice the perimeter of a rectangle of length 8cm and breadth 7cm. What is the circumference of a semicircle whose diameter is equal to the side of the square ?

A) 55.12 cm B) 22.54 cm
C) 42.51 cm D) 38.57 cm
 
Answer & Explanation Answer: D) 38.57 cm

Explanation:

We know that perimeter of rectangle = 2(l+b) = 2(8+7)= 30cm
Given perimeter of square is twice the perimeter of rectangle = 2(30) = 60cm
Therefore, side of the square is = 1/4 x 60 = 15cm
Circumference of the required semicircle = πr + 2r = 227× 152 + 2×152 = 23.57 + 15 = 38.57cm.

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6 14737
Q:

The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

A) 24 B) 48
C) 60 D) 72
 
Answer & Explanation Answer: B) 48

Explanation:

Let ABC be the isosceles triangle and AD be the altitude 

Let AB = AC = x. Then, BC = (32 - 2x). 

Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC,AC2=AD2+DC2 

x2=82+16-x2x=10 

BC = (32- 2x) = (32 - 20) cm = 12 cm. 

Hence, required area =  12*BC*AD=12*12*8=48cm2

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11 14516
Q:

Calculate the value of angle FED from the figure shown below.

 

 

A) 30 deg B) 40 deg
C) 60 deg D) 80 deg
 
Answer & Explanation Answer: B) 40 deg

Explanation:
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0 14450
Q:

select the letter missing from the following series.


U, O, I, ?,A

 

A) D B) Q
C) p D) E
 
Answer & Explanation Answer: D) E

Explanation:
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0 14293
Q:

Find the circumference (in cm) of a circle of radius 7 cm.

 

A) 56 B) 44
C) 16 D) 32
 
Answer & Explanation Answer: B) 44

Explanation:
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2 14286
Q:

For the figure given below, find the angle OAB (in degrees).

 

A) 25 B) 50
C) 75 D) 90
 
Answer & Explanation Answer: C) 75

Explanation:
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1 14083
Q:

A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?

A) 56m B) 65m
C) 34m D) 36m
 
Answer & Explanation Answer: A) 56m

Explanation:

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.

 

Hence number of poles required = 280 / 5 = 56

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19 14067