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 difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

A) 2520 B) 2420
C) 2320 D) 2620
 
Answer & Explanation Answer: A) 2520

Explanation:

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) = 2520 sq.m

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Q:

The diagonal of the floor of a rectangular closet is 712 feet. The shorter side of the closet is 412 feet. What is the area of the closet in square feet?

A) 9 B) 18
C) 27 D) 36
 
Answer & Explanation Answer: C) 27

Explanation:

Other side = d2-s22254-814 = 1444 = 6 ft

 

  Area of closet = (6 x 4.5) sq. ft = 27 sq. ft.

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Q:

The inner circumference of a circular race track, 14 m wide, is 440 m. Find radius of the outer circle

A) 44 B) 22
C) 33 D) 84
 
Answer & Explanation Answer: D) 84

Explanation:

Let inner radius be r metres. Then, 2πr = 440 ; r = 440×744= 70 m.

Radius of outer circle = (70 + 14) m = 84 m.

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Q:

A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.

A) 14 B) 13
C) 12 D) 11
 
Answer & Explanation Answer: A) 14

Explanation:

Distance covered in one revolution = 88×10001000= 88m. 

 

2πR2=88  2×227×R=88  R=88×744

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Q:

A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed?

A) 65m B) 45m
C) 55m D) 56m
 
Answer & Explanation Answer: D) 56m

Explanation:

perimeter of the plot = 2(90+50) = 280m
no of poles =280/5 =56m

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Q:

The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per metre.

A) 2808 B) 3808
C) 4808 D) 5808
 
Answer & Explanation Answer: D) 5808

Explanation:

Area = (13.86 x 10000) sq.m = 138600 sq.m

 

πR2=138600R2=138600×722R=210m 

 

Circumference = 2πR2=2×227×210=1320m 

 

Cost of fencing = Rs. (1320 x 4.40) = Rs. 5808.

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Q:

What is the least number of squares tiles required to pave the floor of a room 15.17 m long and 9.02 m broad?

A) 814 B) 714
C) 614 D) 713
 
Answer & Explanation Answer: A) 814

Explanation:

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm. 

Area of each tile = 41×41cm2  

Required number of tiles  = 1517×90241×41= 814

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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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