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 rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. Find the length of the rectangle?

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

Explanation:

Let the breadth of the given rectangle be x then length is 2x.
thus area of the given rect is 2x2

 

 

 

after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)=2x2+5x-25

since new area is 75 units greater than original area thus
2x2+75 = 2x+5x-25
5x=75+25
5x=100
therefore x=20
hence length of the rectangle is 40 cm.

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Q:

A typist uses a sheet measuring 20cm by 30cm lengthwise.If a margin of 2 cm is left on each side and a 3 cm margin on top and bottom, then percent of the page used for typing is

A) 64% B) 74%
C) 84% D) 94%
 
Answer & Explanation Answer: A) 64%

Explanation:

Area of the sheet = 20×30cm2  

Area used for typing = 20-4×30-6cm2   

required % =384600×100=64%

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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.ft, how many feet of fencing will be required?

A) 44ft B) 88ft
C) 22ft D) 11ft
 
Answer & Explanation Answer: B) 88ft

Explanation:

Given that length and area, so we can find the breadth.

 Length  x  Breadth = Area 

 => 20  x  Breadth = 680 

=> Breadth = 34 feet 

Area to be fenced = 2B + L = 2 (34) + 20 = 88 feet

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Q:

Find the area of a square, one of whose diagonals is 3.8 m long

A) 7.22 B) 6.22
C) 4.22 D) 3.22
 
Answer & Explanation Answer: A) 7.22

Explanation:

Area of the square =12*diagonal2=12*3.8*3.8=7.22m2

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Q:

An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that of the square?

A) 1:2 B) 1:3
C) 2:3 D) 3^(1/2) :2
 
Answer & Explanation Answer: D) 3^(1/2) :2

Explanation:

area of a square = a² sq cm

 

length of the diagonal = 2a cm

 

area of equilateral triangle with side  2a= 34*2a2
required ratio = 34*2a2:a2=3:2

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Q:

A circular swimming pool is surrounded by a concrete wall 4ft wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius of the pool is?

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

Explanation:

let the radius of the pool be Rft

 

Radius of the pool including the wall = (R+4)ft

 

Area of the concrete wall = πR+42-R2=πR+4+RR+4-R = 8πR+2 sq feet

 

=> 8πR+2 = 1125πR2 =>11R2 =200R+2

 

Radius of the pool R = 20ft

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Q:

A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is?

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

Explanation:

length of the diagonal = 52 x (15/60) =13m 

sum of length and breadth = 68 x (15/60) = 17m

 

l2+b2=13 or l+b =17

area =lb = 1/2[(2lb)] = 1/2[(l+b)² - (l²+b²)] = 1/2[17² -169]

=1/2 x 120 = 60 sq.m

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Q:

Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is

A) 12 B) 32
C) 42 D) 52
 
Answer & Explanation Answer: C) 42

Explanation:

required area = [14 x 14 - (4 x 1/4 x 22/7 x 7 x 7) ]sq.cm

= 196 - 154 = 42 sq.cm.

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