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 and breadth of a rectangle are 15 cm and 8 cm respectively. Calculate its perimeter (in cm).

 

A)  46 B) 92
C)  35 D)  70
 
Answer & Explanation Answer: A)  46

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Q:

ΔPQR is right angled at Q. If m∠R=300, then find the value of (cosP - 1/3).

 

A) 1/6 B)  (2√2-1)/√2
C)  -1/√3 D)  (√3-4)/2√3
 
Answer & Explanation Answer: A) 1/6

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Q:

∆ABC is right angled at B. If cosA = 8/17, then what is the value of cotC ?

 

A)  15/8 B) 15/17
C)  8/17 D) 17/15
 
Answer & Explanation Answer: A)  15/8

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Q:

In the given figure, if  QR/XY = 14/9 and PY = 18 cm, then what is the value (in cm) of PQ?

 

 

A) 28 B) 18
C) 21 D) 24
 
Answer & Explanation Answer: A) 28

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Q:

Which of the following conditions is correct for a right angled triangle?

 

A) Sum of the two angles of a triangle is greater than the third angle B) Sum of the two angles of a triangle is less than the third angle
C) Sum of the two angles of a triangle is equal to the third angle D) None of these
 
Answer & Explanation Answer: C) Sum of the two angles of a triangle is equal to the third angle

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Q:

A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is

A) Rs. 258 B) Rs. 358
C) Rs. 458 D) Rs. 558
 
Answer & Explanation Answer: D) Rs. 558

Explanation:

Area to be plastered = 2l+b×h+l×b 

=225+12×6+25×12=744sq.m 

Cost of plastering = 744×75100=Rs.558                       

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Q:

If S is the midpoint of a straight line PQ and R is a point different from S, such that PR =RQ, then

 

A) ∠PRS = 90° B) ∠QRS = 90°
C) ∠PSR = 90° D) ∠QSR = 90°
 
Answer & Explanation Answer: C) ∠PSR = 90°

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Q:

Calculate the area (in cm2) of a circle whose diameter is 7 cm.

 

A) 38.5   B)  77    
C) 32.5   D) 65
 
Answer & Explanation Answer: A) 38.5  

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