# MENSURATION

Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas and Volume. In the broadest sense, it is all about the process of measurement. It is based on the use of algebraic equations and geometric calculations to provide measurement data regarding the width, depth and volume of a given object or group of objects

• Pythagorean Theorem (Pythagoras’ theorem)

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

c2 = a2 + b2 where c is the length of the hypotenuse and a and b are the lengths of the other two sides

• Pi is a mathematical constant which is the ratio of a circle’s circumference to its diameter. It is denoted by ?

??3.14?227

• Geometric Shapes and solids and Important Formulas

• Important properties of Geometric Shapes
1. Properties of Triangle
1. Sum of the angles of a triangle = 180°
2. Sum of any two sides of a triangle is greater than the third side.
• The line joining the midpoint of a side of a triangle to the positive vertex is called the median
1. The median of a triangle divides the triangle into two triangles with equal areas
2. Centroid is the point where the three medians of a triangle meet.
3. Centroid divides each median into segments with a 2:1 ratio
• Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.
• An equilateral triangle is a triangle in which all three sides are equal
1. In an equilateral triangle, all three internal angles are congruent to each other
2. In an equilateral triangle, all three internal angles are each 60°
3. An isosceles triangle is a triangle with (at least) two equal sides
• In isosceles triangle, altitude from vertex bisects the base.

2. Rectangle
1. The diagonals of a rectangle are equal and bisect each other
2. opposite sides of a rectangle are parallel
• opposite sides of a rectangle are congruent
1. opposite angles of a rectangle are congruent
2. All four angles of a rectangle are right angles
3. The diagonals of a rectangle are congruent
4. Square
• All four sides of a square are congruent
• Opposite sides of a square are parallel
1. The diagonals of a square are equal
2. The diagonals of a square bisect each other at right angles
3. All angles of a square are 90 degrees.
• A square is a special kind of rectangle where all the sides have equal length
1. Parallelogram
• The opposite sides of a parallelogram are equal in length.
• The opposite angles of a parallelogram are congruent (equal measure).
1. The diagonals of a parallelogram bisect each other.
• Each diagonal of a parallelogram divides it into two triangles of the same area
1. Rhombus
• All the sides of a rhombus are congruent
• Opposite sides of a rhombus are parallel.
• The diagonals of a rhombus bisect each other at right angles
1. Opposite internal angles of a rhombus are congruent (equal in size)
• Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)
• If each angle of a rhombus is 90°, it is a square

• The sum of the interior angles of a quadrilateral is 360 degrees
• If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)
• A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
• Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.
• Each diagonal of a parallelogram divides it into two triangles of the same area
• A square is a rhombus and a rectangle.
1. Sum of Interior Angles of a polygon
1. The sum of the interior angles of a polygon = 180(n – 2) degrees where n = number of sides Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 – 2) = 180 × 1 = 180 ° Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 – 2) = 180 × 2 = 360.

Solved Examples

Level 1

1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?
2. 4.04 %
3. 2.02 %
4. 4 %
5. 2 %

Explanation :

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100
Then the measured value = (100×(100+2))/100=102 (? error 2% in excess)

Correct Value of the area of the square = 100 × 100 = 10000
Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 – 10000 = 404
Percentage Error = (Error/Actual Value)×100=(404/10000)×100=4.04%

1. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?
2. 30 %
3. 28 %
4. 32 %
5. 26 %

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Lost 20% of length
=> New length =( Original length × (100?20))/100
=(100×80)/100=80

=(100×90)/100=90

New area = 80 × 90 = 7200

Decrease in area
= Original Area – New Area
= 10000 – 7200 = 2800

Percentage of decrease in area
=(Decrease in Area/Original Area)×100=(2800/10000)×100=28%

1. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
2. 25 % Increase
3. 25 % Decrease
4. 50 % Decrease
5. 50 % Increase

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Length of the rectangle is halved
=> New length = (Original length)/2=100/2=50

=> New breadth= Original breadth × 3 = 100 × 3 = 300

New area = 50 × 300 = 15000

Increase in area = New Area – Original Area = 15000 – 10000= 5000
Percentage of Increase in area =( Increase in Area/OriginalArea)×100=(5000/10000)×100=50%

1. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?
2. 14 metres
3. 20 metres
4. 18 metres
5. 12 metres

Explanation:

lb = 460 m2 ——(Equation 1)

Then length, l =( b×(100+15))/100=115b/100——(Equation 2)

From Equation 1 and Equation 2,
115b/100×b=460b2=46000/115=400?b=?400=20 m

1. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus?
2. equal to ½
3. equal to ¾
4. greater than 1
5. equal to 1

Explanation :

If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)

Hence greater than 1 is the more suitable choice from the given list

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Note : Proof

Consider a square and rhombus standing on the same base ‘a’. All the sides of a square are of equal length. Similarly all the sides of a rhombus are also of equal length. Since both the square and rhombus stands on the same base ‘a’,

Length of each side of the square = a
Length of each side of the rhombus = a

Area of the sqaure = a2 …(1)

From the diagram, sin ? = h/a
=> h = a sin ?

Area of the rhombus = ah = a × a sin ? = a2 sin ? …(2)

From (1) and (2)

Area of the square/Area of the rhombus= a2 /a2sin?=1/sin?

Since 0° < ? < 90°, 0 < sin ? < 1. Therefore, area of the square is greater than that of rhombus, provided both stands on same base.

(Note that, when each angle of the rhombus is 90°, rhombus is also a square (can be considered as special case) and in that case, areas will be equal.

1. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field.
2. 37500 m2
3. 30500 m2
4. 32500 m2
5. 40000 m2

Explanation :

Given that breadth of a rectangular field is 60% of its length
?b=(60/100)* l =(3/5)* l

perimeter of the field = 800 m
=> 2 (l + b) = 800
?2(l+(3/5)* l)=800?l+(3/5)* l =400?(8/5)* l =400?l/5=50?l=5×50=250 m

b = (3/5)* l =(3×250)/5=3×50=150 m

Area = lb = 250×150=37500 m2

1. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
2. 45%
3. 44%
4. 40%
5. 42%

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Increase in 20% of length.
=> New length = (Original length ×(100+20))/100=(100×120)/100=120

New area = 120 × 120 = 14400

Increase in area = New Area – Original Area = 14400 – 10000 = 4400
Percentage increase in area =( Increase in Area /OriginalArea)×100=(4400/10000)×100=44%

1. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
2. 814
3. 802
4. 836
5. 900

Explanation :

l = 15 m 17 cm = 1517 cm
b = 9 m 2 cm = 902 cm
Area = 1517 × 902 cm2

Now we need to find out HCF(Highest Common Factor) of 1517 and 902.
Let’s find out the HCF using long division method for quicker results

902)  1517  (1

-902

—————–

615)  902  (1

• 615

————–

287)  615 (2

-574

—————–

41)  287  (7

-287

————

0
————

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm
Area of each square tile = 41 × 41 cm2

Number of tiles required = (1517×902)/(41×41)=37×22=407×2=814

Level 2

1. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?
2. 126 sq. ft.
3. 64 sq. ft.
4. 100 sq. ft.
5. 102 sq. ft.

Explanation :

Let l = 9 ft.

Then l + 2b = 37
=> 2b = 37 – l = 37 – 9 = 28
=> b = 282 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

1. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?
2. 400
3. 365
4. 385
5. 315

Explanation :

Let the areas of the parts be x hectares and (700 – x) hectares.

Difference of the areas of the two parts = x – (700 – x) = 2x – 700

one-fifth of the average of the two areas = 15[x+(700?x)]2
=15×7002=3505=70

Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x – 700 = 70
=> 2x = 770
?x=7702=385

Hence, area of smaller part = (700 – x) = (700 – 385) = 315 hectares.

1. 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. What is the length of the rectangle?
2. 18 cm
3. 16 cm
4. 40 cm
5. 20 cm

Explanation :

Then length = 2x cm
Area = lb = x × 2x = 2x2

New length = (2x – 5)
New breadth = (x + 5)
New Area = lb = (2x – 5)(x + 5)

But given that new area = initial area + 75 sq.cm.
=> (2x – 5)(x + 5) = 2x2 + 75
=> 2x2 + 10x – 5x – 25 = 2x2 + 75
=> 5x – 25 = 75
=> 5x = 75 + 25 = 100
=> x = 1005 = 20 cm

Length = 2x = 2 × 20 = 40cm

1. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)?
2. 142000
3. 112800
4. 142500
5. 153600

Explanation :

l : b = 3 : 2 —-(Equation 1)

Perimeter of the rectangular park
= Distance travelled by the man at the speed of 12 km/hr in 8 minutes
= speed × time = 12×860     (? 8 minute = 860 hour)
= 85 km = 85 × 1000 m = 1600 m

Perimeter = 2(l + b)

=> 2(l + b) = 1600
=> l + b = 16002 = 800 m —-(Equation 2)

From (Equation 1) and (Equation 2)
l = 800 × 35 = 480 m
b = 800 × 25 = 320 m (Or b = 800 – 480 = 320m)

Area = lb = 480 × 320 = 153600 m2

1. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction?
2. Rs.3500
3. Rs. 4200
4. Insufficient Data
5. Rs. 4400

Explanation :
Let length and width of the rectangular plot be l and b respectively
Total area of the rectangular plot = 96 sq.m.
=> lb = 96

Width of the pathway = 2 m
Length of the remaining area in the plot = (l – 4)
breadth of the remaining area in the plot = (b – 4)
Area of the remaining area in the plot = (l – 4)(b – 4)

Area of the pathway
= Total area of the rectangular plot – remaining area in the plot
= 96 – [(l – 4)(b – 4)] = 96 – [lb – 4l – 4b + 16] = 96 – [96 – 4l – 4b + 16] = 96 – 96 + 4l + 4b – 16
= 4l + 4b – 16
= 4(l + b) – 16

We do not know the values of l and b and hence area of the pathway cannot be found out. So we cannot determine total cost of the construction.

1. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?
2. 144?3?48? cm2
3. 121?3?36? cm2
4. 144?3?36? cm2
5. 121?3?48? cm2

Explanation :
Area of an equilateral triangle = (3/?4)*a *a where a is length of one side of the equilateral triangle
Area of the equilateral ? ABC = (3/?4)*a *a = (3/?4)*24*24=144?3 cm2? (1)

Area of a triangle = 12bhwhere b is the base and h is the height of the triangle
Let r = radius of the inscribed circle. Then
Area of ? ABC
= Area of ? OBC + Area of ? OCA + area of ? OAB
= (½ × r × BC) + (½ × r × CA) + (½ × r × AB)
= ½ × r × (BC + CA + AB)
= ½ x r x (24 + 24 + 24)
= ½ x r x 72 = 36r cm2 —-(2)

From (1) and (2),
144?3=36r?r=144?3/36=4?3????(3)

Area of a circle = ?r2 where = radius of the circle
From (3), the area of the inscribed circle = ?r2=?(4?3)* (4?3)=48??(4)

Hence, area of the remaining portion of the triangle
= Area of ? ABC – Area of inscribed circle
144?3?48? cm2

1. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?
2. ?11600 cm
3. ?14400 cm
4. ?10000 cm
5. ?12040 cm

Explanation :
The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal).
Hence the length of the longest rod = AG

Initially let’s find out AC. Consider the right angled triangle ABC

AC2 = AB2 + BC2 = 402 + 802 = 1600 + 6400 = 8000
?AC = ?8000 cm

Consider the right angled triangle ACG

AG2 = AC2 + CG2
(?8000) 2+602=8000+3600=11600
=> AG = ?11600 cm
=> Length of the longest rod = ?11600cm

1. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed?
2. 30
3. 44
4. 56
5. 60