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:

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

A) 65.25 B) 56.25
C) 65 D) 56
 
Answer & Explanation Answer: B) 56.25

Explanation:

let each side of the square be a , then area = a2 

As given that The side is increased by 25%, then 

New side = 125a/100 = 5a/4 

 

New area = 5a42  

 

Increased area= 25a216-a2 

 

Increase %=9a2/16a2*100  % = 56.25%

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385 79777
Q:

If the radius of a circle is decreased by 50%, find the percentage decrease in its area.

A) 55% B) 65%
C) 75% D) 85%
 
Answer & Explanation Answer: C) 75%

Explanation:

Let original radius = R. 

 

New radius = 50100R50100R 

 

Original area =R2  and new area = πR2

 

3πR24*1πR2*100 

 

Decrease in area = πR22=πR24 = 75%

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85 36703
Q:

If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle.

A) 110 B) 120
C) 130 D) 140
 
Answer & Explanation Answer: B) 120

Explanation:

let length = x and breadth = y then  

2(x+y) = 46         =>   x+y = 23  

x²+y² = 17² = 289  

now (x+y)² = 23² 

=> x²+y²+2xy= 529 

=> 289+ 2xy = 529

=> xy = 120  

area = xy = 120 sq.cm

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104 32612
Q:

A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is

A) 136 B) 236
C) 336 D) 436
 
Answer & Explanation Answer: C) 336

Explanation:

let ABCD be the given parallelogram 

area of parallelogram ABCD = 2 x (area of triangle ABC) 

now a = 30m, b = 14m and c = 40m

 s=1/2 x (30+14+40) = 42 

 

Area of triangle ABC = ss-as-bs-c  

 

= 4212282= 168sq m 

area of parallelogram ABCD = 2 x 168 = 336 sq m

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56 28519
Q:

A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be

A) 7744 B) 8844
C) 5544 D) 4444
 
Answer & Explanation Answer: A) 7744

Explanation:

length of wire = 2πr= 2 x (22/7 ) x 56 = 352 cm
side of the square = 352/4 = 88cm
area of the square = 88 x 88 = 7744sq cm

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58 28087
Q:

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

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

Explanation:

Let the side of the square(ABCD) be x meters.

 

Then, AB + BC = 2x metres.

 

AC = 2x = (1.41x) m.

 

Saving on 2x metres = (0.59x) m.

 

Saving % =0.59x2x*100 = 30% (approx)

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47 27603
Q:

The area of the largest circle that can be drawn inside a rectangle with sides 18cm by 14cm is

A) 49 B) 154
C) 378 D) 1078
 
Answer & Explanation Answer: B) 154

Explanation:

The diameter is equal to the shortest side of the rectangle.

 So radius= 14/2 = 7cm. 

Therefore,  area of circle =πr2=227×49=154cm2

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41 25544
Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

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

Explanation:

Area of the park = (60 x 40) = 2400m2

Area of the lawn = 2109m2 

Area of the crossroads = (2400 - 2109) = 291m2 

Let the width of the road be x metres. Then,

60x+40x-X2=291 

x2-100x+291=0 

  (x - 97)(x - 3) = 0
   x = 3.

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28 24821