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:

The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m

A) 53800 B) 43800
C) 83800 D) 73800
 
Answer & Explanation Answer: D) 73800

Explanation:

perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126 - 6 = 120m
area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m
cost of pavement = 246*6*50 = Rs. 73800

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30 30603
Q:

The circumference of a circle is 66 cm. Find its radius (in cm).

 

A) 21 B) 10.5
C) 4.5 D) 9
 
Answer & Explanation Answer: B) 10.5

Explanation:
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4 30259
Q:

Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is

A) 1.967 B) 1.867
C) 1.767 D) 1.567
 
Answer & Explanation Answer: A) 1.967

Explanation:

required area = (area of an equilateral triangle of side 7 cm)- (3 * area of sector with à = 60 degrees and r = 3.5cm)
34*7*7-3*227*3.5*3.5*60360 sq cm

 

=34*49-11*0.5*3.5 sq cm

 

= 1.967 sq cm

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34 30233
Q:

Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq meters.

A) 56m B) 16m
C) 14m D) 76m
 
Answer & Explanation Answer: A) 56m

Explanation:

clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope.


Let the length of the rope be r mts 

then, πr2=9856  r2=9856×722=3136 r =56m

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31 29526
Q:

Calculate the area (in sq.cm) of a circle whose diameter is 14 cm.

 

A) 154 B) 308
C) 224 D) 448
 
Answer & Explanation Answer: A) 154

Explanation:
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1 29470
Q:

The volume of a cube is 274.625 cm³. Find its side (in cm).

 

A) 7.5 B) 6.5
C) 5.5 D) 3.5
 
Answer & Explanation Answer: B) 6.5

Explanation:
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0 29463
Q:

The two lines 3x - 8y = 16 and 2x + 4y = 6 intersect at (a,b). Find the value of a2-4b2

 

 

A) 5   B)  10  
C) 15   D) 20
 
Answer & Explanation Answer: C) 15  

Explanation:
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0 29095
Q:

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

A) B=900;H=300 B) B=300;H=900
C) B=600;H=700 D) B=500;H=900
 
Answer & Explanation Answer: A) B=900;H=300

Explanation:

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
13.5*10000m2=135000m2 
Let altitude = x metres  and  base = 3x metres.
Then, 12*3x*x=135000x2=90000x=300 

Base = 900 m and Altitude = 300 m.

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62 28635