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 room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq.m is Rs. 270 and the cost of papering the four walls at Rs. 10 per sq.m is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room.

A) b=6; l=18; H=6 B) b=5; l=6; H=18
C) l=6; b=18; H=15 D) l=5; b=18; H=18
 
Answer & Explanation Answer: A) b=6; l=18; H=6

Explanation:

Let breadth = x metres, length = 3x metres, height = H metres. 

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m 

x×3x2=54x2=54×2x=6  

So, breadth = 6 m and length =362 = 9 m.

Now, papered area = (1720/10) =  172 sq.m 

Area of 1 door and 2 windows = 8 sq.m 

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m 

2×9+6H=180H=6

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Q:

Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13cm.

A) 30 B) 40
C) 50 D) 60
 
Answer & Explanation Answer: A) 30

Explanation:

Height of the triangle = 132-122=25 = 5 cm.

 

Its area =12*base*height12*12*530cm2.

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Q:

The ratio of the area of a square to that of the square drawn on its diagonal is

A) 1:2 B) 2;3
C) 3:1 D) 4:1
 
Answer & Explanation Answer: A) 1:2

Explanation:

ratio = a22a2

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Q:

A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?

A) 315 B) 385
C) 415 D) 485
 
Answer & Explanation Answer: A) 315

Explanation:

Let the areas of the two parts be x and (700-x) hectares

therefore, x-700-x=15x+700-x2

 2x-700=70

 x=385

So, the two parts are 385 and 315.

 Hence, Area of the smaller = 315 hectares

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Q:

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

A) 27 and 23 B) 24 and 23
C) 25 and 23 D) 22 and 23
 
Answer & Explanation Answer: A) 27 and 23

Explanation:

Let the two parallel sides of the trapezium be a cm and b cm. 

Then, a - b = 4 

And, 12×a+b×19=475=>a+b=50 

Solving (i) and (ii), we get: a = 27, b = 23. 

So, the two parallel sides are 27 cm and 23 cm.

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Q:

The base of a parallelogram is twice its height. If the area of the parallelogram is 72 sq. cm, find its height

A) 6 B) 7
C) 8 D) 9
 
Answer & Explanation Answer: A) 6

Explanation:

Let the height of the parallelogram be x. cm. Then, base = (2x) cm.

2x×x=72=>2x2=72=>x=6 

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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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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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