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:

If 4 circles of equal radius are drawn with vertices of a square as the centre , the side of the square being 7 cm, find the area of the circles outside the square ?

A) 119.21 sq cm B) 115.395 sq cm
C) 104.214 sq cm D) 111.241 sq cm
 
Answer & Explanation Answer: B) 115.395 sq cm

Explanation:

Because each vertice of the square is the center of a circle
(1/4) part of the total area of each circle inside of the square.
(3/4) part of the total area of each circle outside of the square.
Thus total area outside the square is 434×227×3.5×3.5 = 115.3955 sq cm.

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Q:

A class is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width. Find the least number of square tiles of equal size required to cover the entire floor of the class room ?

A) 115 B) 117
C) 116 D) 114
 
Answer & Explanation Answer: B) 117

Explanation:

Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 x 432)/(48 x 48) = 13 x 9 = 117.

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6 6828
Q:

If the radius of a circle is decreased by 50% , find the percentage decrease in its area.

A) 75% B) 65%
C) 35% D) 25%
 
Answer & Explanation Answer: A) 75%

Explanation:

let original radius = r and  new radius = (50/100) r = r/2

 


original area = πr2 and new area = πr/22

 


decrease in area =  3πr2/4*1πr2 *100 = 75%

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Q:

A circular piece of thin wire is converted into a rhombus of side 11 cm. Find the diameter of the circular piece?

A) 28 cm B) 3.5 cm
C) 7 cm D) 14 cm
 
Answer & Explanation Answer: D) 14 cm

Explanation:

Circular piece is 4 x 11 = 44 cm long,

Then Circumference of circle is given by,

44 = pi x D, where D is the diameter

D = 44 / pi

Take pi = 22 / 7, then

D = 44 / (22/7) = (44 x 7) / 22

D = 14 cm.

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Q:

A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

A) rs.458 B) rs,558
C) rs.658 D) rs.758
 
Answer & Explanation Answer: B) rs,558

Explanation:

Area to be plastered= [2(l + b) x h] + (l x b)

= [2(25 + 12) x 6] + (25 x 12)

= (444 + 300)

= 744 sq.m

Cost of plastering = Rs.744 x (75/100)

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Q:

Find the area of the rectangle or square with the given dimensions L = 5 inches, W= 3 inches?

A) 13 B) 14
C) 15 D) 16
 
Answer & Explanation Answer: C) 15

Explanation:

We know that,

Area of rectangle A = L x W = 5 x 3 = 15 sq.inches.

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Q:

Area of a square and a rectangle are equal. If side of the square is 40 cm and length of the rectangle is 64 cm, what is the perimeter of the rectangle ?

A) 187 cm B) 178 cm
C) 149 cm D) 194 cm
 
Answer & Explanation Answer: B) 178 cm

Explanation:

Area of square = 40 x 40

= 1600 sq.cm

Given that the areas of Square and Rectangle are equal

=> Area of rectangle = 1600 Sq.cm

We know that, Area of rectangle = L x B

Given L = 64 cm

Breadth of rectangle = 1600/64 = 25 cm

Perimeter of the rectangle = 2(L + B) = 2(64+25) = 178 cm.

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32 6167
Q:

If the area of a square with side a s equal to the area of a triangle with base a, then the altitude of the triangle is

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

Explanation:

Area of a square with side a = a² sq unts

Area of a triangle with base a = (1/2) * a * h sq.unts

a² =1/2 *a *h

=> h = 2a

altitude of the triangle is 2a

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