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 length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

A) 40 B) 20
C) 30 D) none of these
 
Answer & Explanation Answer: D) none of these

Explanation:

Let breadth = x metres.
Then, length = (x + 20) metres.

Perimeter =5300/23.50

2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

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2 10161
Q:

The area of a square is 1024 sq.cm. What is the ratio of the length to the breadth of a rectangle whose length is twice the side of the square and breadth is 12 cm less than the side of this square ?

A) 14:9 B) 13:12
C) 16:5 D) 17:13
 
Answer & Explanation Answer: C) 16:5

Explanation:

Area of a square SxS = 1024 => S = 32 cm
Length of the rectangle = 32x2 cm = 64 cm
Breadth of the rectangle = 32 - 12 = 20 cm
Required ratio = 64 : 20 = 16 : 5

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5 10026
Q:

An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq. m ?

A) Rs. 4082.40 B) Rs. 1024.21
C) Rs. 2810.6 D) Rs. 3214
 
Answer & Explanation Answer: A) Rs. 4082.40

Explanation:

Length of the first carpet = (1.44)(6) = 8.64 cm

Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)

= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m

Cost of the second carpet = (45)(12.96 x 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40

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11 9871
Q:

The tangent at any point of the circle is _________________ to the radius through the point of contact.

 

A) parallel B) intersecting
C) perpendicular D) equal
 
Answer & Explanation Answer: C) perpendicular

Explanation:
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0 9691
Q:

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

A) 22 B) 44
C) 66 D) 88
 
Answer & Explanation Answer: D) 88

Explanation:

We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

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3 9475
Q:

Which of the following represents the sides of an acute angled triangle?

 

A) 6, 9, 16   B) 7, 8, 10  
C) 5, 12, 13   D) None of these
 
Answer & Explanation Answer: B) 7, 8, 10  

Explanation:
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1 8956
Q:

A man walked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges?

A) 10% B) 20%
C) 30% D) 40%
 
Answer & Explanation Answer: C) 30%

Explanation:

let the side of the square be x meters
length of two sides = 2x meters
diagonal = 2x= 1.414x m

 

saving on 2x meters = .59x m

 

saving % = 0.59x2x*100%

 

= 30% (approx)

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2 8939
Q:

Calculate the number of bricks, each measuring 25 cm x 15 cm x 8 cm required to construct a wall of dimensions 10 m x 4 cm x 6 m when 10% of its volume is occupied by mortar  ?

A) 720 B) 600
C) 660 D) 6000
 
Answer & Explanation Answer: A) 720

Explanation:

Let the number of bricks be 'N'

10 x 4/100 x 6 x 90/100 = 25/100 x 15/100 x 8/100 x N

10 x 4 x 6 x 90 = 15 x 2 x N => N = 720.

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6 8559