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:

2 metres broad pathway is to be constructed around a rectangular plot on the inside. The area of the plots is 96 sq.m. The rate of construction is Rs. 50 per square metre. Find the total cost of the construction?

A) 20 B) 30
C) 40 D) data is inadequate
 
Answer & Explanation Answer: D) data is inadequate

Explanation:

Given lb =96

Area of pathway = [(l-4)(b-4)-lb] = 16-4(l+b)

Which cannot be determined. so, data is inadequate.

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7 14259
Q:

A man walking at the speed of 4 kmph crosses a square field diagonally in 3 meters.The area of the field is

A) 10000 B) 20000
C) 30000 D) 40000
 
Answer & Explanation Answer: B) 20000

Explanation:

speed of the man = 4 x (5/18) m/sec = 10/9 m/sec

time taken = 3 x 60 sec = 180 sec

length of diagonal = speed x  time = (10/9) x 180 = 200m

Area of the field = (1/2) x (dioagonal)²

= (1/2) x 200 x 200 sq.m = 20000sq.m

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6 14189
Q:

Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively.Find the width of the ring.

A) 5 B) 4
C) 3 D) 2
 
Answer & Explanation Answer: B) 4

Explanation:

let the inner and outer radii be r and R meters
then, 2πr = 352/7

 

=> r = (352/7) * (7/22) * (1/2) = 8m
2πR = 528/7

 

=> R= (528/7) * (7/22) * (1/2)= 12m
width of the ring = R-r = 12-8 = 4m

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5 14132
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 13853
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 13725
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 13590
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 13503
Q:

The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec.What is the speed of motorcycle n km/hr?

A) 13.68kmph B) 31.68kmph
C) 41.68kmph D) 45.68kmph
 
Answer & Explanation Answer: B) 31.68kmph

Explanation:

In this type of question, we will first calculate the distance covered in given time.

 

Distance covered will be : Number of revolutions x Circumference 

 

So we will be having distance and time, from which we can calculate the speed. 

 

Radius of wheel = 70 / 2 = 35 cm

 

Distance covered in 40 revolutions will be 40*circumferene=40*2πr=40*2*227*35=8800cm = 88 m

 

Distance covered in 1 sec  = 8810=8.8m/s=8.8*185km/hr=31.68km/hr

 

 

 

 

 

 

 

 

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4 13468