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 diagonal in cms of a rectangle of length 5 cm and width 3 cm is

 

A) √34 B) +-√34
C) 4 D) +-4
 
Answer & Explanation Answer: A) √34

Explanation:
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Q:

If each side of a square is increased by 25%, find the percentage change in its area.

A) 56.25% B) 36.25%
C) 16.25% D) 12.25%
 
Answer & Explanation Answer: A) 56.25%

Explanation:

Let each side of the square be a. Then, area = .a2
New side =125a100=5a4. New area = 5a42 = 25a216

 

Increase in area = 25a216-a2=9a216
Increase% = 9a216*1a2*100 % = 56.25%.

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39 21357
Q:

The area of a circle of radius 5 is numerically what percent its circumference?

A) 150% B) 250%
C) 350% D) 450%
 
Answer & Explanation Answer: B) 250%

Explanation:

required percentage = πr22πr×100 =250%

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26 21231
Q:

A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is

A) rs.58 B) rs.158
C) rs.258 D) rs.358
 
Answer & Explanation Answer: C) rs.258

Explanation:

area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m

cost of graveling = 344 x  (75/100) = Rs. 258

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30 20696
Q:

A rectangular grassy plot 110m by 65cm has a gravel path .5cm wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq.mt

A) 680rs B) 480rs
C) 340rs D) 320rs
 
Answer & Explanation Answer: A) 680rs

Explanation:

area of theplot = 110 * 65 = 7150 sq.m
area of the plot excluding the path = (110-5)* (65-5) = 6300 sq.m
area of the path = 7150 - 6300 =850 sq.m
cost of gravelling the path = 850 x (80/100) = 680 Rs

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11 20115
Q:

A rectangular parking space is marked out by painting three of its sides.If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?

A) 126sq.ft B) 120sq.ft
C) 130sq.ft D) 135sq.ft
 
Answer & Explanation Answer: A) 126sq.ft

Explanation:

Let l = 9 ft.

Then l + 2b = 37

=> 2b = 37 – l = 37 – 9 = 28

b = 28/2 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

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16 19910
Q:

A room 5m 55cm long and 3m 74cm broad is to be paved with square tiles.Find the least number of square tiles required to cover the floor?

A) 156 B) 166
C) 176 D) 186
 
Answer & Explanation Answer: C) 176

Explanation:

Area of the room=(544 * 374)

 

size of largest square tile= H.C.F of 544 & 374 = 34 cm

 

Area of 1 tile = (34  x  34) cm2

 

Number of tiles required== [(544 x 374) / (34 x 34)] = 176

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54 19305
Q:

The no of revolutions a wheel of diameter 40cm makes in traveling a distance of 176m is

A) 240 B) 140
C) 40 D) 340
 
Answer & Explanation Answer: B) 140

Explanation:

distance covered in 1 revolution = 2πr = 2 x (22/7) x 20 = 880/7 cm
required no of revolutions = 17600 x (7/880) = 140

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