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:

Find the ratio of the areas of the incircle and circumcircle of a square.

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

Explanation:

Let the side of the square be x. Then, its diagonal = 2x2=2x 

Radius of incircle = x2 

Radius of circum circle= 2×x2=x2 

Required ratio = πx24:πx22=14:12=1:2

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Q:

A room of 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor.

A) 136 B) 146
C) 166 D) 176
 
Answer & Explanation Answer: D) 176

Explanation:

area of the room = 544 x 374 sq.cm

size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm

area of 1 tile = 34x34 sq cm

no. of tiles required = (544 x 374) / (34 x 34) = 176

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Q:

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

A) 20 B) 200
C) 300 D) 400
 
Answer & Explanation Answer: B) 200

Explanation:

Let length of plot = L meters, then breadth = L - 20 meters

and perimeter = 2[L + L - 20] = [4L - 40] meters

[4L - 40]  * 26.50 = 5300

[4L - 40] = 5300 / 26.50 = 200

4L = 240

L = 240/4= 60 meters. 

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Q:

One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field?

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

Explanation:

Other side = [(17 x 17) - (15 x 15)] = (289 - 225) = 8m
Area = 15 x 8 =120 sq. m

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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:

The perimeter of a square is 34 cm. Find its area (in cm2).

 

A) 144.5   B) 82.25  
C) 164.5   D) 72.25
 
Answer & Explanation Answer: D) 72.25

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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