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:

Area of the circle inscribed in a square of diagonal 6√2 cm (in sq cm) is

A) 9 Π B) 6 Π
C) 3 Π D) 9√2 Π
 
Answer & Explanation Answer: A) 9 Π

Explanation:
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1 2643
Q:

Find the area of the trapezium, whose parallel sides are 12 cm and 10 cm, and distance between the parallel sides is 14 cm?

A) 121 sq.com B) 154 sq.com
C) 186 sq.com D) 164 sq.com
 
Answer & Explanation Answer: B) 154 sq.com

Explanation:

We know that, 

Area of trapezium = 1/2 x (Sum of parallel sides) x (Distance between Parallel sides)

= 1/2 x (12 + 10) x 14

= 22 x 14/2

= 22 x 7

= 154 sq. cm

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23 2537
Q:

In the given figure, PQRS is a square of side 8 cm. PQO = 60 deg. What is the area (in sq.cm) of the triangle POQ?

A) 32√3 B) 24[(√3) – 1]
C) 48[(√3) – 1] D) 16[3 – (√3)]
 
Answer & Explanation Answer: D) 16[3 – (√3)]

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0 2515
Q:

If the radius of a circle is increased by 25%, its area increases by:

A) 50 percent B) 25 percent
C) 28.125 percent D) 56.25 percent
 
Answer & Explanation Answer: D) 56.25 percent

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4 2357
Q:

There is a square field of area ‘X’ square meters. A cylindrical ditch of radius 7 meters and depth 2 meters is dug, and the earth is taken out and spread over the remaining part of the square field, the height of square field which goes up by 0.77 meters. What is the value of ‘X’ ? 

A) 548 sq. m B) 524 sq. m
C) 518 sq. m D) 554 sq. m
 
Answer & Explanation Answer: D) 554 sq. m

Explanation:

Volume of the cylindrical ditch = πrxrh = 22/7 ×7×7×2 = 308 sq.m

Area of remaining field = (X −πrxr) sq.m

= (X − 154)

ATQ –

(X – 154) ×0.77 = 308

(X – 154) = 400

X= 554 sq.m

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1 2295
Q:

The area of a circular cricket ground is 24.64 hectares. Find the cost of making rope boundary at the rate of Rs. 5.40 per metre.

A) Rs. 9,600 B) Rs. 9,504
C) Rs. 9,802 D) Rs. 9,876
 
Answer & Explanation Answer: B) Rs. 9,504

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Q:

In the given figure, four identical semicircles are drawn in a quadrant. XA = 7 cm. What is the area (in sq.cm) of shaded region ?

 

A) 70 B) 140
C) 77 D) 84
 
Answer & Explanation Answer: D) 84

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Q:

In the circle above, chord AB is extended to meet the tangent DE at D. If AB = 12 cm and DE = 8 cm, find the length of BD.

A) 4 cm B) 5 cm
C) 6 cm D) 7 cm
 
Answer & Explanation Answer: A) 4 cm

Explanation:
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