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Mensuration 2D

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QUANTITATIVE APTITUDE | Mensuration 2D

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Mensuration 2D

Mensuration is mainly divided into three parts:

One is Area, second is Volume and third is Perimeter.

Triangle

Points to Remember

Perimeter = a + b + c

Area of Triangle = [½ × base × height] or

Here, s is a semi perimeter

= s s –a s – b s – c  

Q. Sides of a triangle are 12m, 13m and 11m. Find the area of triangle and height with respect to the side of 12 cm.

A. 12 13 11 36

182 2

s      m

– – –Area of Δ    s s  a s   b s   c

Area of Δ =  18×6×5×7

2Area of Δ = 6 105 m

Side = 12 m (base), height = ?

Area of Δ = ½ × base × height

6√105 = ½ × 12 × height

Height = √105 m

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Properties of equilateral triangle

Properties of equilateral triangle

All three sides are equal.

All three angles are 60°

Formulae

Perimeter = 3 × side = 3a

Area of equilateral triangle = (√3/4)a2

Height = Median = (√3/2)a

Side of a equilateral triangle =

The sum of distances of a point inside a trinagle from all the sides × 2 

3

Q. The area of an equilateral triangle is 49√3 m2, find the value of its height and side?

A. The area of an equilateral triangle = 49√3 m2

(√3/4)a2 = 49√3

a2 = 49 × 4

a = 14 m

Height = (√3/2)a

Height = (√3/2) × 14 = 7√3 m

Q. A point is situated inside equilateral triangle whose distance from three side are 2√3 m, 7√3 m and 11√3 m, find the area of this triangle.

A. The sum of distance from all the sides = 2√3 + 7√3 + 11√3 = 20√3 m

Side of a equilateral triangle =

Side of a equilateral triangle =

Area of equilateral = (√3/4)a2

Area of equilateral = (√3/4) × 402 = 400√3 m

2

The sum of distance from all the sides × 2 

3

20 3 240

3

      m

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Properties of isosceles triangle

Two sides are equal.

AB = AC

Formula:

Area of Δ = ½ × base × height

Perpendicular on unequal side = √[(equal side)2 – (unequal side/2)

2]

AD = √[AC2 – (BC/2)

2]

Q. In a isosceles triangle, if the ratio of equal side and unequal side is 5 : 8 and it’s perimeter is 36 cm. Find the area of this triangle.

A. Let the equal and unequal side be 5x and 8x

Perimeter of this triangle = 5x + 5x + 8x

36 = 18x

x = 2

Sides of this triangle are 10 cm, 10 cm and 16 cm

Perpendicular on unequal side = √[102 – (16/2)

2] = √[100 – 64]

Perpendicular on unequal side = 6 cm

Area of isosceles triangle = ½ × unequal side × perpendicular on unequal side

Area of isosceles triangle = ½ × 16 × 6 = 48 cm2

Properties of right-angle triangle

Hypotenous2 = (base)

2 + (height)

2

Area = ½ × base × height = ½ × b × p

Length of Perpendicular at hypotenuse =

 height × base 

hypotenuse

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Q. The area of a right angle triangle is 30 m and base is 7 m more than height, find the length of perpendicular on hypotenuse.

A. Let the height of the triangle be x

⇒ Base = x + 7

⇒ Area = ½ × base × height

⇒ 30= ½ × (x + 7) × x

⇒ 60 = x2 + 7x

⇒ x2 + 7x – 60 = 0

⇒ (x + 12) (x – 5) = 0

⇒ x = 5 m

⇒ Base = x + 7 = 12 m

⇒ Hypotenuse2 = (base)

2 + (height)

2

⇒ Hypotenuse2 = 144 + 25 = 169

⇒ Hypotenuse = 13 m

⇒Perpendicular on hypotenuse =

⇒Perpendicular on hypotenuse =

∴ The length of perpendicular on hypotenuse is (60/13) m.

 height × base 

hypotenuse5 12 60

13 13

   m

Square

Properties of square

Opposite sides of a square are both parallel and equal in length.

The diagonal of a square are equal.

AC = BD

The diagonals of a square bisect each other and meet at 90°.

AO = OC, BO = OD and ∠BOC = 90°

All four angles of a square are 90°.

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Formula:

Perimeter = 4 × side

Area = (side)2 = ½ × (diameter)

Diameter = √2 side

Q. Area of a square is 1352 m2. Find the perimeter and the length of diagonal.

A. Area of a square = 1352

(Side)2 = 1352

Side = 26√2 m

Perimeter = 4 × side

Perimeter = 4 × 26√2 = 104√2 m

Length of diagonal = √2 × side

Length of diagonal = √2 × 26√2 = 52 m

Rectangle

Properties of rectangle

Opposite sides of a rectangle are equal in length.

AB = DC, AD = BC

Opposite sides of a rectangle are parallel.

AB ∥ CD, AD ∥ BC

The diagonals of a rectangle bisect each other.

AO = OC, BO = OD

All four angles of a square are 90°.

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Q. In a rectangle diagonal is 9 times of its length. Find the ratio of length and breath.

A. Let the length of rectangle be x

Diagonal = 9x

Diameter = √(l2 + b

2)

9x = √(x2 + b

2)

x2 + b

2 = 81 x

2

80 x2 = b

2

Formulae:

Area = l × b

Perimeter = 2(l + b)

Diameter = √(l2 + b

2)

Here, Length = l

Breadth = b

b = (4√5)x

Length : breath = x : (4√5)x

Length : breath = 1 : 4√5

Parallelogram

A quadrilateral having opposite sides are

parallel and equal.

AB = DC and AD = BC

AB ∥ DC and AD ∥ BC

Area of parallelogram ABCD = Base × Altitude

on corresponding base = DC × h1 = BC × h2

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Q. Find the area of parallelogram PQOS given in the figure:

A. ⇒ ∠PSR = ∠QOR = 60

⇒ QR/OR = tan (60)

⇒OR = QR/tan(60)

⇒ OR = 10√3/√3 = 10 cm

⇒ OS = SR – OR = 20 – 10 = 10 cm

⇒ Area of parallelogram = Base × Altitude = 10 × 10√3 = 10 × 17.32 = 173.2 cm2

Rhombus :

A quadrilateral having all sides equal and their

diagonals are intersect at 90°.

AB = BC = CD = DA

Area of rhombus

= ½ × product of the lengths of its diagonals

= ½ × d1 × d2

d1 and d2 are diagonals

2 2

1 2Side = ½ ×  d +  d  

Q. The side of the rhombus is 20 cm and one of the diagonals of rhombus is 32 cm. Find the area of the rhombus.

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A. Using Pythagoras theorem,

⇒ (side)2 = (1/2 × one diagonal)

2 + (1/2 × other diagonal)

2

⇒ (1/2 × other diagonal)2 = 20

2 – (32/2)

2

⇒ Other diagonal = 24 cm

Area of rhombus = 1/2 × product of diagonals

= 1/2 × 24 × 32

= 384 sq.cm

Trapezium

A quadrilateral having a pair of sides is

parallel and unequal.

AB ∥ DC

Area of trapezium ABCD

= ½ × (sum of its parallel sides) × height

= ½ × (AB + DC) × h

Q. If the area of a trapezium is 250 sq.m and the lengths of the two parallel sides are 15m and 10 m respectively then distance between the parallel sides of the trapezium is

A. Given,

Area of trapezium = ½ × sum of parallel sides × distance between them

Given,

⇒ Area of trapezium = 250 sq.m

⇒ Sum of parallel sides = 15 + 10 = 25 m

Then,

⇒ 250 = 1/2 × 25 × distance between them

⇒ Distance between them = 20 m

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2

Perimeter =  4π area

θLength of arc =  ×2πr

360°

θ Area of sector= ×πr

360°

Circle

A circle is a round, two-dimensional shape. All points on the edge of the circle are at the same distance from the center.

O is a center of this circle and OA is radius.

Diameter = 2 × radius

Area = π r2

Perimeter = 2πr = πd

Q. The area of a circle is 3850 cm2. What is its circumference (in cm).

A. Area of circle with radius ‘r’ = πr2

⇒ 22/7 × r2 = 3850

⇒ r2 = 1225

⇒ r = √1225 = 35 cm

⇒ Circumference of circle = 2πr = 2 × 22/7 × 35 = 220 cm

Q. The arc of a circle is 27.5 cm and the angle made by arc at the center of the circle, is 75°, then find the circumference of the circle.

A. The angle made by arc at the center of the circle = 75° = 75 × π/180 = 5π/12

The length of the arc of the circle = 27.5

Now, Angle = Arc/Radius

Radius = Arc/Angle = 27.5/(5π/12)

Radius = (27.5 × 12 × 7)/(5 × 22)

Radius = 21 cm

So, the circumference of the circle = 2 × (22/7) × 21 = 132 cm

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Semi-circle

Area of semi-circle = ½ π r2

Perimeter of semi-circle = π r + 2r = (36/7) × r

Q. What will be the area of a semi - circle of perimeter 72 metres?

A. As we know, perimeter of semicircle = πr + 2r = (π + 2)r

Where r = radius of semicircle

(π + 2)r = 72

36/7 × r = 72

⇒ r = 14 m

⇒ Area of semi - circle = ½ × (22/7) × 14 × 14

⇒ Area of semi - circle = 308 m2

Incircle

It is the circle in side the polygon whose all the sides are tangent to the circle.

Radius of incircle = 2 × area of triangle 

 Perimeter of triangle

Radius of incircle of equilateral triangle side

2 3

  

Q. In a triangle, three sides 5m, 12m and 13m. Find the radius of incirle.

A. Radius of incircle =

Radius of incircle =

2 × area of triangle  Perimeter of triangle

12 5 12

602 25 12 13 30

    m

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Circumcircle

Circle whose circumference touches all the vertices of the polygon is called circumcircle.

Circum radius of a triangle = Product of sides/(4 × area of triangle)

Circum radius of a equilateral triangle = 3

a 

Q. In a triangle, three sides are 5cm, 6cm and 7cm. Find the circumradius of this triangle.

A. s = (5 + 6 + 7) /2 = 9cm

Area of triangle =

Area of triangle =

Circum radius of a triangle = (5 × 6 × 7)/(4 × 6√6) = 35/4√6 cm

  s s a s b s c

9 9 5 9 6 9 7  

2Area of triangle = 6 6 cm