Mensuration Part 1

There are two types of problems in mensuration

  • 2 dimensional: It contains
    • Length and breadth
    • Area
    • Perimeter
  • 3 dimensional: It contains
    • Length, breadth, and height
    • Volume
    • Total surface area
    • Curved surface area

Important formulas:

Square:

Area of square = a^{2}

Perimeter of a square is= 4a

Diagno1 = \sqrt{2}a

When diagno1 is given area =\frac{d^{2}}{a}

Rectangle:

Area of rectangle = length * breadth

Perimeter = 2(Length + Breadth)

d=\sqrt{l^{2}+b^{2}}

Triangle:

\angle A+\angle B+\angle C=180

S=\frac{a+b+c}{2}

Area of a triangle = \frac{1}{2}\times base\times height

Scalene triangle

Area of the triangle when sides are given = \sqrt{s(s-a)(s-b)(s-c)}

Equilateral triangle:

Area of equilateral triangle = \frac{\sqrt{3}}{4}a^{2}

Height of a triangle = \frac{\sqrt{3}}{2}a

p=3a

Isosceles triangle

A=\frac{b}{4}\sqrt{4a^{2}-b^{2}}

h=\frac{1}{2}\sqrt{4a^{2}-b^{2}}

p=2a+b

RIGHT ANGLE TRIANGLE

Perimeter= Perpendicular + base + hypothesis

Area = \frac{1}{2}\times Perpendicular\times base

Circle :

Area of circle = \Pi r^{2}

Circumference of a circle = 2\Pi r

\Pi =\frac{22}{7}=3.14

Trapezium:

In trapezium opposite sides are equal

Area of trapezium = \frac{1}{2}(a+b)\times h

Parallelogram:

Area of parallelogram = \frac{1}{2}(h_{1}+h_{2})\times D

Perimeter = a+b+c+d

Rhombus:

In the rhombus, all sides are equal and diagonal are not equal

Area of rhombus = \frac{1}{2}\times d_{1}\times d_{2}

SECTOR:

Length of arc = 2\Pi r\frac{\Theta }{360}

Perimeter = 2r(\Pi\times \frac{\Theta }{360}+1)

Area = \Pi r^{2}\frac{\Theta }{360}

Mensuration Part 2

Problem 1: Find the area, perimeter and diagonal of the square whose side is 5m?

Solution:

Area of a square = a^{2}=5^{2}=25

Perimeter of a square = 4a =4*5= 20

Diagonal of a square= \sqrt{2}a=\sqrt{2}\times 5=5\sqrt{2}

 

Problem 2: The altitude of an equilateral triangle is \sqrt{3} cm. What is its perimeter?

Solution:

Altitude means height

h=\frac{\sqrt{3}}{2}a

\sqrt{3}=\frac{\sqrt{3}}{2}a

a=2

Perimeter = 3a=3\times 2=6

 

Problem 3: The length and breadth of a room are 6 meters and 4.5 meters respectively. Find the floor area of the room?

Solution:

Length = 6

Breadth = 4.5

Area = 6*4.5 = 27

Mensuration Part 3

Problem 1: The ratio of three angles of a quadrilateral are in the ratio 1: 6: 2. The measure of the fourth angle is 450. What is the difference between the largest and smallest angle?

Solution:

Sum of four angles in a quadrilateral is 360

x+6x+2x+45 = 360

9x = 315

x = 35

Difference between smallest and largest angle is 6x-x = 5x

5x = 5*35 = 175

Problem 2: On increasing the diameter of a circle by 75 %, what will be the percentage increase in the perimeter?

Solution:

If the radius is increased perimeter is also increases.

75%=\frac{3}{4}

Actual diameter        Increased diameter

4                                           7

16                                        49

\frac{33}{16}\times 100=87.5%

Problem 3: If the longer side of a rectangle is doubled and the other reduced to half, then find what will be the percentage increment in the area of the new rectangle?

Solution:

Let length =10     breadth = 10  ——->10*10=100

Double length =20   Half breadth=5——->20*5=100

No increment in the area.

Mensuration Part 4

Problem 1: If the perimeter of a rectangle and a square are equal and the ratio of two adjacent sides of the rectangle is 1: 2 then the ratio of the area of the rectangle and that of the square is?

Solution:

Length =x , Breadth = y

If the perimeter of a rectangle and a square are equal
2(l+b) = 4a
2(x+2x) = 4a
3x=2a
a=\frac{3}{2}x
2x^{2}:\frac{9}{4}x^{2}
8  :  9
Problem 2: The cost of fencing a circular plot at the rate of Rs.15 per meter is Rs.3300. What will be the cost of flooring the plot at the rate of Rs.100 per square meter?
Solution:
RS. 15 —–> 1meter
RS. 3300—–>?\frac{3300}{15}=220m
Perimeter = 220
2\Pi r=220
r=220\times \frac{7}{22}\times \frac{1}{2}
r=35
Area = \Pi r^{2}
\frac{22}{7}\times 35\times 35=3850 m^{2}
1m——->RS.100
3850m—->? 3850*100 = 385000

Mensuration Part 5

Problem 1: The area of a field in the shape of a trapezium measures 1440m2. The perpendicular distance between its parallel sides is 24 m. If the ratio of the parallel sides is 5 : 3, the length of the longer parallel sides is?

Solution:

Area of trapezium =\frac{1}{2}(a+b)\times h

1440=\frac{1}{2}\times (5x+3x)\times 24

1440=\frac{1}{2}\times (8x)\times 24

x=15

5x=5*x=75

Problem 2: What would be the cost of building a 7 meters wide garden around a circular field with the diameter equal to 280 meters if the cost per square meter for building the garden is Rs. 21?

Solution:

\Pi r^{2}-\Pi r^{2}

\Pi (147^{2}-140^{2})

\frac{22}{7}\times 287\times 7=6314

1m———–>21

6341m—–>? 6341*21 = 132594

Problem 3: There is a rectangular plot whose length is 36 meter and breadth is 28 meter. There are two paths middle of the plot and parallel to length and breadth of the plot. The remaining part is lawn whose area is 825 square meter. What is the area of the paths?

Solution:

Area of rectangle = 36*28 = 1008

Given area of lawn = 1008-825 = 183