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 altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?

A) 50 B) 60
C) 70 D) 80
 
Answer & Explanation Answer: B) 60

Explanation:

let ABC be the isosceles triangle, the AD be the altitude 

Let AB = AC = x then BC= 32-2x       [because parameter = 2 (side) + Base] 

since in an isoceles triange the altitude bisects the base so 

BD = DC = 16-x 

In a triangle ADC, AC2=AD2+DC2 

x2=82+16-x2 x=10 

BC = 32-2x = 32-20 = 12 cm 

Hence, required area = 12*BC*AD12*12*10 = 60 sq cm

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Q:

The dimensions of a room are 10m x 7m x 5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m x 3m. One window is of size 2m x 1.5m and the other 2 windows are of size 1m x 1.5m. The cost of painting the walls at Rs. 3 per sq m is

A) Rs.174 B) Rs.274
C) Rs.374 D) Rs.474
 
Answer & Explanation Answer: D) Rs.474

Explanation:

Area of 4 walls = 2(l+b)h
=2(10+7) x 5 = 170 sq m
Area of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq m
area to be planted = 170 -12 = 158 sq m

Cost of painting = Rs. 158 x 3 = Rs. 474

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Q:

ΔXYZ is right angled at Y. If sinX = 12/13, then what is the value of cosecZ ?

 

A) 5/12   B)  5/13  
C) 13/5   D)  13/12
 
Answer & Explanation Answer: C) 13/5  

Explanation:
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Q:

The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

A) 18 B) 28
C) 38 D) 48
 
Answer & Explanation Answer: A) 18

Explanation:

l2+b2=41 (or)   l2+b2=41  

Also, lb=20

l+b2=l2+b2+2lb  = 41 + 40 = 81

(l + b) = 9. 

Perimeter = 2(l + b) = 18 cm.

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Q:

The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m

A) 53800 B) 43800
C) 83800 D) 73800
 
Answer & Explanation Answer: D) 73800

Explanation:

perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126 - 6 = 120m
area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m
cost of pavement = 246*6*50 = Rs. 73800

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Q:

Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is

A) 1.967 B) 1.867
C) 1.767 D) 1.567
 
Answer & Explanation Answer: A) 1.967

Explanation:

required area = (area of an equilateral triangle of side 7 cm)- (3 * area of sector with à = 60 degrees and r = 3.5cm)
34*7*7-3*227*3.5*3.5*60360 sq cm

 

=34*49-11*0.5*3.5 sq cm

 

= 1.967 sq cm

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Q:

The length and breadth of a rectangle are 15 cm and 8 cm respectively. Calculate its perimeter (in cm).

 

A)  46 B) 92
C)  35 D)  70
 
Answer & Explanation Answer: A)  46

Explanation:
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Q:

Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq meters.

A) 56m B) 16m
C) 14m D) 76m
 
Answer & Explanation Answer: A) 56m

Explanation:

clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope.


Let the length of the rope be r mts 

then, πr2=9856  r2=9856×722=3136 r =56m

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