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:

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 6366
Q:

The sides of a right-angled triangle are 12 cm, 16 cm, 20 cm respectively. A new right angle Δ is made by joining the midpoints of all the sides. This process continues for infinite then calculate the sum of the areas of all the triangles so made.

A) 312 sq.cm B) 128 sq.cm
C) 412 sq.cm D) 246 sq.cm
 
Answer & Explanation Answer: B) 128 sq.cm

Explanation:

Area =   SSC Quiz : Quantitative Aptitude | 06 - 01 - 18 = 96

 

Sum of Area =  SSC Quiz : Quantitative Aptitude | 06 - 01 - 18

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21 6301
Q:

The length of a rope in meter by which a cow must be tethered in order that she may be able to graze an area of 2826 sq.m is ?

A) 15 mts B) 30 mts
C) 24 mts D) 36 mts
 
Answer & Explanation Answer: B) 30 mts

Explanation:

The cow can graze an area of 2826 sq.m i.e it is able to cover area of circle of 2826 sq.m

 

Therefore, πr2 =2826
r2 =(2826×7/22) = 899.18 = 900
r = 30 m.

 

Therefore, the radius of the circle implies the length of the rope = 30 mts.

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9 6285
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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3 6273
Q:

Consider the figure shown below and choose the correct option for the ratio of PA:PB.

 

A) r1 : r2 (internal) B) r1 : r2 (external)
C) r2 : r1 (internal) D) r2 : r1 (external)
 
Answer & Explanation Answer: B) r1 : r2 (external)

Explanation:
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2 6225
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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0 6056
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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2 5931
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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2 5810