fdaytalk · mailid : [email protected] acute angle the angle which is less than 900 right...

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Author : J. Maha Laxmaiah Mailid : [email protected]

Acute angle

The angle which is less than 900

Right angle

The angle which is equal to 900

Obtuse angle

The angle which is greater than 900

Straight angle

The angle which is 1800

Reflexive angle

The angle which is between 1800 to 3600

Complete angle

The angle which is 3600

Supplementary Angles

When sum of two angles is equal to 180◦ ( ˪𝑨 + ˪𝑩 = 𝟏𝟖𝟎0

Complementary Angles

When sum of two angles is equal to 90◦

˪A + ˪B = 900

LINES

When two lines crossed each other

∟1 = ∟3 and ∟2 = ∟4 [vertically opposite angles]

∟1 + ∟2 = 1800

∟2 + ∟3 = 1800

∟3 + ∟4 = 1800

∟4 + ∟1 = 1800

Transversal

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∟1, ∟2, ∟7 and ∟8 are exterior angles

∟3, ∟4, ∟5 and ∟6 are interior angle

∟1 and ∟7 & ∟2 and ∟8 are alternate exterior angles

∟3 and ∟5 & ∟4 and ∟6 are alternate interior angles

Corresponding angles are equal

˪1 = ˪5, ˪2 = ˪6, ˪4 = ˪8, ˪3 = ˪7

Alternate interior angles are equal

˪4 = ˪6, ˪3 = ˪5

Alternate exterior angles are equal

˪2 = ˪8, ˪1 = ˪7

And

˪4 + ˪5 = 180◦, ˪3 + ˪6 = 1800

l, m and n are three parallel lines and two transverse lines are passed across

Then, 𝑷

𝑸=

𝑿

𝒀

EXAMPLE

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ABC is a Triangle and D, E are the mid – points of AB and AC sides

Then, 𝐴𝐷

𝐷𝐵=

𝐴𝐸

𝐸𝐶

ABC is a Triangle

∟A, ∟B and ∟C are interior angles

∟A + ∟B + ∟C = 1800

And ∟D is a exterior

Exterior angle is sum of the opposite interior angles

∟D = ∟A + ∟B

TRIANGLES

Area of triangle (Hero’s formula)

= √𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)

S = 𝒂+𝒃+𝒄

𝟐

Based on sides Triangles are divided into

1 Scaline Triangle: No two sides are equal

2 Isosceless Triangle: Any two sides are equal

3 Equilateral Triangle: All sides are equal

Based on angles

1 Acute angle Triangle: The angle which is less than 900

2 Right angle Triangle: One angle should be 900

3 Obtuse angle Triangle: The angle which is greater than 900

In Triangle ABC, if ‘a’ is the largest side

a2 < b2 + c2 => Acute angle Triangle

a2 = b2 + c2 => Right angle Triangle

a2 > b2 + c2 => Obtuse angle Triangle

Congruency of Triangle

Congruency means same (same in size and shape)

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Three sides and three angles must be equal

The above two Triangles ABC & DEF have equal sides and equal angles

Therefore, ∆𝐴𝐵𝐶 ≅ ∆𝐷𝐸𝐹

1 SIDE – SIDE – SIDE (SSS)

In both two Triangles all corresponding sides are equal

The above two Triangles, ∆𝐴𝐵𝐶 ≅ 𝐷𝐸𝐹

2 SIDE – ANGLE – SIDE (SAS)

In this case, angle must be between two sides of Triangles

The above two Triangles, ∆𝐵𝐶𝐴 ≅ ∆𝐹𝐸𝐷

3 ANGLE – SIDE – ANGLE (ASA)

In this case, two angles and one side consider in both Triangles and side

must be on angles side

So, ∆𝐶𝐴𝐵 ≅ 𝐸𝐹𝐷

4 RIGHT ANGLE – HYPOTENUS – SIDE (RHS)

Here, in both the two Triangles, one angle must be 900 and Hypotenuse is

same and one side is same

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From above two Triangles, ∆𝐵𝐶𝐴 ≅ ∆𝐸𝐹𝐷

Similar Triangles

In case of Similar Triangles, shape is the same

From above two Triangles, angles are equal in both the Triangles and having

same shape

So, ∆𝐴𝐵𝐶 ~ ∆𝐷𝐸𝐹

AND

𝑎

𝑑=

𝑏

𝑒=

𝑐

𝑓=

ℎ1

ℎ2=

√∆1

√∆2

= 𝑃1

𝑃2=

𝑆1

𝑆2

Here, h1 and h2 are the heights of both the Triangles ABC & DEF

∆1 𝑎𝑛𝑑 ∆2 are the Areas of both the Triangles ABC & DEF

P1 and P2 are the Perimeters of both the Triangles ABC & DEF

S1 and S2 are the Semi – Perimeters of both the Triangles ABC & DEF

Mid – Point Theorem

In Triangle ABC, D & E are the mid – points of AB & AC and DE is

parallel to BC

Then, DE = 1

2∗ 𝐵𝐶

MEDIANS & ITS PROPERTIES:

A line segment joining from the mid – point of the side to the opposite

vertex

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In Triangle ABC, DC is the median

Here, AB, BC & CA are sides

AD, BE & CF are Medians

G is Centroid

Centroid: The point of intersection of all the three medians of a Triangle

is called Centroid

1 Median divides the Triangle into two equal parts

2 Centroid divides the Median in 2 : 1

3 Centroid always lies in inside the Triangle

4 Sum of the sides is greater than sum of the Medians

AB + BC + CA > AD + BE + CF

5 Formula for Median, AD = 1

2 × √2𝐴𝐵2 + 2𝐴𝐶2 − 𝐵𝐶2

BE = 1

2 × √2𝐴𝐵2 + 2𝐵𝐶2 − 𝐴𝐶2

CF = 1

2 × √2𝐴𝐶2 + 2𝐵𝐶2 − 𝐴𝐵2

ALTITUDE & ITS PROPERTIES

The Altitude of a Triangle is a line segment perpendicularly drawn from vertex to

the side opposite to it. The side on which the perpendicular is being drawn is called

its base

In Triangle ABC,

AB, BC & CA are sides

AD = altitude

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In Triangle ABC


AD, BE & CF are altitudes

O = Ortho – centre

ORTHO – CENTRE The point of intersection of all the three Altitudes of a

Triangle is called its Ortho – centre.

1 Sum of the sides of Triangle is greater than sum of the Altitudes

AB + BC + CA > AD + BE + CF

2 In Acute angleTriangle, Orho – centre lies in inside the Triangle

3 In Right angle Triangle, Ortho – centre lies on the right angle

4 In Obtuse angle Triangle, Ortho – centre lies on the outside the Triangle

Here, O = Ortho – centre

Then, ∟AOC = 1800 - ∟B

AND, ∟BOC = 1800 - ∟A

∟AOB = 1800 - ∟C

ANGLE BI – SECTOR & ITS PROPERTIES

The Angle Bi – sector is, the line which is bi – sects the angle equally (divided into

two equal parts)

Here

AB, BC & CA are sides of Triangle

BD = angle bi – sector line

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In Triangle ABC


AD, BE & CF are Angle bi – sector lines

I = In – centre of a Trianlge

IN – CENTRE:

∟BIC = 900 + ∟𝐴

2

∟CIA = 900 + ∟𝐵

2

∟AIB = 900 + ∟𝐶

2

EXAMPLE: ABC is a Triangle BI and CI are angle bisectors of ∟ABC and

∟ACB is ∟BAC = 500, find ∟BIC …..

SOLUTION:

SHORT METHOD:

∟BIC = 900 + 1

2∗ ∟A [Theorem]

∟BIC = 1150 (answer)

EX – CENTRE

Ex – centre is formed by two external angles bi – sector and one internal angle bi –

sector.

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1 There is only one in – centre

2 And, there are three Ex – centre formed

3 In – centre is equidistance from the sides

IN – CIRCLE

BD = Angle of bi – sector

𝐵𝐶

𝐵𝐴=

𝐶𝐷

𝐷𝐴

PERPENDICULAR BI – SECTOR & ITS PROPERTIES

Perpendicular bi – sector is, a line passes through the mid – point of opposite

vertex perpendicularly

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In Triangle ABC,

D is the mid – point of BC

CIRCUM – CENTRE

In Triangle ABC,

S = Circum – centre

The circle is formed is Circum – circle

1 Circum – centre is equidistance from all the vertices

2 In Acute angle Triangle, circum – centre lies in inside the Triangle

3 In Right angle Triangle, circum – centre lies on mid – point of Hypotenuse

4 In Obtuse angle Triangle, circum – centre lies on outside the Triangle

RIGHT ANGLE TRIANGLE

In Right angle Triangle ABC

I = In – centre

S = Circum – centre

In – radius (r) = 𝑏+𝑐−𝑎

2

Circum – radius (R) = 𝑎

2

Area of Right angle Triangle = 1

2∗ 𝑏𝑎𝑠𝑒 ∗ ℎ𝑒𝑖𝑔ℎ𝑡

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PYTHOGOROS TRIPLEX

3, 4 & 5 => 32 + 42 = 52

5, 12 & 13 => 52 + 122 = 132

7, 24 & 25 => 72 + 242 = 252

8, 15 & 17 => 82 + 152 = 172

9, 40 & 41 => 92 + 402 = 412

11, 60 & 61 => 112 + 602 = 612

ISOSCELES TRIANGLE

The Triangle which have two equal sides and other is different

Here,

AB = AC

∟B = ∟C

∆𝐴𝐵𝐷 ≅ ∆𝐴𝐶𝐷

h =

D is the mid – point of BC then, BD = DC

EQUILATERAL TRIANGLE

In Equilateral Triangle, all sides and all angles are equal

Here, AB = BC = CA = a

∟A = ∟B = ∟C = 600

h = height

From diagram, a2 = h2 + (𝑎

2)2

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By solving, we get h = √3

4 × 𝑎

Area = 1

2 × 𝑎 × ℎ =>

1

2 × 𝑎 ×

√3

4 𝑎

= √3

4 × 𝑎2

In – radius (r) = 𝑎

2√3

Circum – radius (R) = 𝑎

√3

***** In Equilateral Triangle Centroid, Ortho – centre, In – centre and Circum –

centre are coincide at the same point

QUADRILATERALS

Sum of the angles in a Quadrilateral is 3600

Quadrilateral means contain four sides

PARALLELOGRAM

In Parallelogram, opposite sides are equal and opposite angles are equal

Here, AB //DC

BC//AD

AB = DC & BC = AD

Opposite angles are equal, ∟A = ∟C & ∟B = ∟D

Adjacent angles sum = 1800

∟A + ∟B = 1800

∟B + ∟C = 1800

∟C + ∟D = 1800

∟D + ∟A = 1800

Diagonals bi – sect each other, OA = OC & OB = OD

Here, AC & BD are diagonals

Diagonals divide the Parallelogram into four equal areas

∆𝐴𝑂𝐵 ≅ ∆𝐶𝑂𝐷 ≅ ∆𝐴𝑂𝐷 ≅ ∆𝐵𝑂𝐶

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Sum of the squares of the sides = Sum of the squares of the Diagonals

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

If AB//DC, then CDE is Transversal

RECTANGLE

Rectangle has two equal opposite sides and all angles are same with 900

Here, AB = DC

AD = BC

l = length

b = breadth

d1 & d2 = diagonals

Area = l × b

Perimeter = 2(l +b)

Diagonal = √𝑙2 + 𝑏2

RHOMBUS

‘

All sides are equal

Diagonals are perpendicular to each other

Diagonals bi – sect the interior angles

AB = BC = CD = DA

d1 & d2 = diagonals

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Side = 𝟏

𝟐 × √𝒅𝟏𝟐 + 𝒅𝟐𝟐

Area = 𝟏

𝟐 × (d1 × d2)

Perimeter = 2 √𝒅𝟏𝟐 + 𝒅𝟐𝟐

The four Triangles formed are congruent to each other

∆𝐴𝑂𝐵 ≅ ∆𝐴𝑂𝐷 ≅ ∆𝐵𝑂𝐶 ≅ 𝐷𝑂𝐶

SQUARE

AB = BC = CD = DA = a = side

d = diagonal

Area = a2

Perimeter = 4a

Diagonal = √2 × 𝑎

TRAPEZIUM

Here, AB// CD

a, b are the lengths of the parallel sides

h = height between the parallel lines

Area = 1

2 × ℎ × (𝑎 + 𝑏)

Median = 𝟏

𝟐 × (a + b)

Area of ∆𝐴𝑂𝐶 = 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐵𝑂𝐷

Perimeter = AB + BC + CD + DA

EXAMPLE: The parallel sides of a Trapezium are in a ratio 2 : 3 and their

shortest distance is 12 cm. If the area of the Trapezium is 480 sq. cm., the

longer of the parallel sides is of length………..

SOLUTION:

Sides of the Trapezium (l1 & l2) = 2x and 3x cm

Height (shortest distance), h = 12 cm

Area = 𝟏

𝟐× ( l1 + l2 ) × h [Formula]

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Here, l1 & l2 are parallel sides

And h = distance between the , l1 & l2

∴ 480 = 1

2× (2𝑥 + 3𝑥) × 12

x = 16

Therefore, longer side = 3x => 3 × 16

= 48 cm (answer)

POLYGONS

Polygon is nothing but contains more than four sides

In Regular Polygon = All interior angles

Sum of interior angles of any Polygon = (n – 2) × 1800

No. of Diagonals of Polygon = 𝑛 × (𝑛−3)

2

n = no. of sides

REGULAR POLYGON

All interior angles are equal

Each exterior angle = 3600

𝑛

Each interior angle = (𝑛−2) × 1800

𝑛

Exterior angle = 1800 – interior angle

CUBE

Volume = a3 ( a = side of cube )

Lateral surface area = 4×a2

Total surface area = 6×a2

Diagonal = √𝟑 × 𝒂

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CUBOID

Volume = length (l) × breadth (b) × height (h)

Total surface area = 2× (lb + bh + lh)

Area of four walls of a room = 2h (l + b)

Diagonal = √𝒍𝟐 + 𝒃𝟐 + 𝒉𝟐

CYLINDER

Here, r = radius, h = height

Volume (v) = 𝝅r2h

Curved surface area = 2πrh

Total surface area = 2πrh + 2πr2

××××× Cylinder is nothing but no. of circles placed one by one vertically

CONE

Here, r = radius, h = height

And, l = √𝒓𝟐 + 𝒉𝟐

Volume (v) = 1/3 𝝅r2h

Curved surface area = πrl

Total surface area = πrl + πr2

××××× Cone is formed by, when Cylinder is divided into three parts

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Frustum of Cone

Here, r1 & r2 are radii of frustum

And, h= height, l = length

Then,

= 𝟏

𝟑× 𝝅 (r1

2 + r22 + r1 × r2) × h

Surface area = 𝝅 ×l (r1 + r2) + 𝝅r12 + 𝝅r2

2

SPHERE

Here, r = radius

Volume (v) = 𝟒

𝟑 πr3

Total surface area = 4 πr2

HEMISPHERE

Here, r = radius

Spherical Cell

Volume = 𝟒

𝟑 × 𝝅 (R3 – r3)

Total surface area = 4 × 𝝅 (R2 – r2)

CIRCLES

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Here, r = radius

O = Centre

Area = πr2

Perimeter = 2πr

FOR SEMICIRCLE

Area = πr2/2

Perimeter = πr + 2r or 𝒓 × 𝟑𝟔

𝟕

Area of the ring or circular path = 𝝅 (R2 – r2)

Length of the arc

= 2πr × 𝜽

𝟑𝟔𝟎

Area of sector

= πr2 × 𝜽

𝟑𝟔𝟎

Perimeter of sector

= 2πr × 𝜽

𝟑𝟔𝟎 + 2r

CHORD:

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Here, O = Centre

AB = Chord

**** largest chord is diameter

Here, AB = Chord

O = Centre

From Chord to any point on circle joins the lines, angle occurred at different places

is same

∟x = ∟x

Here, AB & CD are two equal chords (AB = CD), then

˪𝟏 = ˪2

The angle subtended by an arc of circle at the centre is double the angle subtended

by it at any point on the remaining part of the circle i. e

˪AOB = 2× ˪ACB

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Angle in the same segment of a circle is equal

˪AOB = 2× ˪ACB

˪AOB =2× ˪ADB

˪ACB = ˪ADB

The opposite pairs of angles of cyclic quadrilateral is supplementary each other

˪A + ˪C = 180◦

˪B + ˪D = 180◦

From above diagram,

AB = Chord

O = Centre

P = Mid point of A & B

r = radius

Then, r2 = y2 + (𝒙

𝟐)𝟐

TANGENTS & SECANTS

Tangent touch the circle at one point while secant touch the circle at two points

Here, AP = Tangent

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BCP = Secant and

˪A = 90◦ always

If two chords AB and CD intersect internally at a point P

PA × PB = PC × PD

If two chords AB & CD intersect externally at a point P. Then

PA × PB = PC × PD

If ABP is secant to a circle intersecting the circle at A & B, and PT is tangent, then

PA × PB = PT2 (Tangent – Secant theorem)

The side of BC of ∆ABC is produced to D

The bisector of LA meets BC at L, then

˪ABC + ˪ACD = 2× ˪ALC

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Here, ˪Y = ˪75 & ˪Z = ˪45 (Alternate angles are equal)

∴ ˪X = 60◦ [180 – (75+45)]

Here, external angle = 135◦

And, 135◦ = X + 65◦ (˪A + ˪B = 135)

X = 70◦

Here, X + 35◦ = 60◦ + 47◦

X = 72◦

Here, R = circum radius

X = side of a equilateral triangle

O = centre of circle

When an equilateral triangle is inscribed in a circle

Radius (R) = 𝒙

√𝟑

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Here, r = in-radius

X = side of a equilateral triangle


When a circle is inscribed in an equilateral triangle

Radius ( r ) = 𝒙

𝟐√𝟑

Here, r = radius


T = tangent

P = external point from center

And, d = distance between the center of circle and external point

Then, length of the Tangent = √𝒅𝟐 − 𝒓𝟐

Direct Common Tangent

If the two circles are on the same side of a line, the common tangent is said to be

direct common tangent

Here, r = radius of one circle

R = radius of another circle


And, d = distance between two centers of circles

Then, the length of the direct common tangent = √𝒅𝟐 − (𝑹 − 𝒓)𝟐

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Transverse Common Tangent

If the two circles are on the opposite side of a line, the common tangent is said to

be transverse common tangent

Here, r = radius one circle

R = radius of another circle

O = center of circles

And, d = distance between the centers

Then, the length of the Transverse common tangent = √𝒅𝟐 − (𝑹 + 𝒓)𝟐

EXAMPLE: The radii of two circles are 5 cm and 3 cm respectively and the

distance between their centres is 24 cm. Then the length of the Transverse

common tangent is ………

SOLUTION:

Transverse common tangent = √𝒅𝟐 − (𝑹 + 𝒓)𝟐

Here, d = 24 cm, r = 3 cm and R = 5 cm

= √242 − (5 + 3)2

= 16√2 cm (answer)

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Here, O = Centre

1 & 2 are Direct Common Tangents

3 & 4 are Transverse Common Tangents

Here, O = Centre

1 & 2 are Direct Common Tangent and 3 is a Transverse Common Tangent

Here, O = Centre

1 & 2 are Direct Common Tangent

Transverse Common Tangent = 0

Here, 1 is a Direct Common Tangent


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Here, O = Centre

Direct Common Tangent = 0


MENSURATION

Regular Polygon

A polygon is a regular polygon if its all sides and all angles are equal

----- Each internal angle of a regular polygon of n – sides

= (𝟐𝒏−𝟒) × 𝟗𝟎

𝒏

----- Exterior angle of a Regular Polygon of n – sides = 3600

𝑛

EXAMPLE: Each interior angle of Regular Polygon is 180 more than eight

times an exterior angle. The no. of sides of the Polygon is ………

SOLUTION:

Each internal angle of a regular polygon of n – sides

= (𝟐𝒏−𝟒) × 𝟗𝟎

𝒏

Exterior angle of a Regular Polygon of n – sides = 3600

𝑛

According to question

(2𝑛 − 4) × 90

𝑛=

360

𝑛 × 8 + 18

n = 20 (answer)

Area of regular Hexagon

= 6 × √𝟑

𝟒 × (side)2

Area of circum circle of regular hexagon = 𝝅 × (side)2

Radius of in- circle of a regular hexagon = √𝟑

𝟐 × side

Area of in- circle = 𝟑

𝟒 × 𝝅 × (side)2

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No. of diagonals in ‘n’ sides polygon

= 𝒏 (𝒏−𝟑)

𝟐

EXAMPLE: A Polygon has 54 diagonals. The number of sides in the

Polygon is …………….

SOLUTION:

No. of diagonals in ‘n’ sides polygon

= 𝒏 (𝒏−𝟑)

𝟐 [Formula]

54 = 𝑛(𝑛−3)

2

n = 12 (answer)

----- Total surface area of Prism = Curved surface area + 2 × area of base

Regular Tetrahedron

Volume = √𝟐

𝟏𝟐 × a3

Surface area = √𝟑 × a2

Here, a = side of tetrahedron

Right Pyramid

Volume = 𝟏

𝟑 × (area of the base) × vertical height

Lateral surface area = 𝟏

𝟐 × (perimeter of the base) × slant height

Total surface area = Area of the base + Lateral surface area

Volume of the Prism = Area of the base × height

1 Each edge of a regular tetrahedron is 3 cm, then its volume is

SOLUTION:

Tetrahedron is a figure composed of four triangular faces.

For a regular tetrahedron,

Surface area of each face = √𝟑

𝟒 a2 [Area of equilateral triangle]

Total surface area = 4 × √𝟑

𝟒 a2 = √𝟑 a2

Height of pyramid = √𝟐

𝟑 a

Volume = 𝟏

𝟑 ×

√𝟑

𝟒 a2 × √

𝟐

𝟑 a =

𝒂𝟑

𝟔√𝟐 , In this case volume =

𝟑𝟑

𝟔√𝟐

= 𝟗

𝟐√𝟐 cm3 (answer)

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2 If the radii of the circular ends of a truncated conical bucket which is

45cm high be 28 cm and 7 cm, then the capacity of the bucket in cubic

centimeter is (use π = 𝟐𝟐

𝟕 )

SOLUTION:

This is the case of frustum of a cone

Volume of frustum of a cone = 𝟏

𝟑 πh [r2 + Rr + R2]

Lateral surface area = πl (R + r)

In this case, volume = 𝟏

𝟑 ×

𝟐𝟐

𝟕 × 45 [282 + 28 ×7 + 72]

= 48510 cm3 (answer)

3 A solid cone of height 9 cm with diameter of its base 18 cm is cut out from

a wooden solid sphere of radius 9 cm. The percentage of wood wasted is

SOLUTION:

% of wood wasted = [

𝟒

𝟑𝝅 × 𝟗𝟑−

𝟏

𝟑 × 𝝅 × 𝟗𝟐 ×𝟗

𝟒

𝟑 𝝅 × 𝟗𝟑

] 100

= 75% (answer)

4 If the four equal circles of radius 3 cm touch each other externally, then

the area of the region bounded by the four circles is

SOLUTION:

Area of shaded region

= 62 – π × 32

= 36 - 9π = 9 (4 – π) (answer)

5 A path of uniform width runs all around the inside of rectangular field

116 m by 68 m and occupies 720 sq. m. Find the width of the path:

SOLUTION:

2 × 116x + 2 (68 – 2x) x = 720

116x + 68x – 2x2 = 360

x2 – 92x + 180 = 0

x = 2 m (answer)

r

l h

R

x 68 m 116 m

D C

B A

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6 ABCD is a square, 4 equal circles are just touching each other whose

centres are the vertices A, B, C, D of the square. What is the ratio of the

shaded to the unshaded area within square?

SOLUTION:

If radius is r, Unshaded area = πr2

Area of square = 4r2

Shaded area = 4r2 – πr2 = (4 – π) r2

Shaded : Unshaded = 4 – π : π

𝟔

𝟕 :

𝟐𝟐

𝟕

i.e. 3 : 11 (answer)

7 A goat is tied on the corner of a rectangular field of size 30 m × 20 m by a

14 m long rope. The area of the region that she graze is ……..

SOLUTION:

Required area = 𝟗𝟎

𝟑𝟔𝟎 ×

𝟐𝟐

𝟕 × 𝟏𝟒 × 𝟏𝟒

= 154 m2 (answer)

8 If each side of tetrahedron is 12 cm long then the volume of tetrahedron

will be …

SOLUTION:

Volume of tetrahedron = √𝟐

𝟏𝟐 × (𝒔𝒊𝒅𝒆)𝟑 [Formula]

= √2

12 (12)3

= 144√2 cm3 (answer)

9 In an Equilateral Triangle of side 24 cm, a circle is inscribed touching its

sides. The area of the remaining portion of the triangle is ………

SOLUTION:

Given that, Side of Equilateral Triangle (a) = 24 cm

Radius of inner circle (r) = 𝑎

2√3 [Formula]

r = 24

2√3 => r = 4√3

Remaining area of Triangle = Area of Equilateral Triangle – Area of Circle

= √3

4 𝑎2 − 𝜋𝑟2

D C

B A

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= √3

4 × 24 × 24 −

22

7 × 4√3 × 4√3

= 98. 55 sq. cm (answer)

10 A Conical cup is filled with ice – cream. The ice – cream forms a Hemi –

Spherical shape on its open top. The height of the Hemi – Spherical part is 7

cm. the radius of Hemi – Spherical part equals the height of the cone. Then

the volume of the ice – cream ……………

SOLUTION:

Given that, radius of the Hemi – Sphere (r) = 7 cm and

Height of the Cone (h) = 7 cm

Volume of the Hemi – Sphere = 2

3 𝜋𝑟3 [Formula]

= 2

3 ×

22

7 × 7 × 7 × 7

= 2156

3 cm3

Volume of the Cone = 1

3 × 𝜋𝑟2ℎ [Formula]

= 1

3 ×

22

7 × 7 × 7 × 7

= 1078

3 cm3

Therefore, Volume of the Ice – cream = Volume of the Hemi – Sphere + Volume

of the Cone

= 2156

3+

1078

3 => 1078 cm3 (answer)

11 If the radius of the Cylinder is decreased by 8% while its height is

increased by 4%. What will be the effect on volume ……..

SOLUTION:

PERCENTAGE METHOD:

Given that, Radius (r)= 8% decrease

Height (h) = 4% increase

Volume of the Cylinder = 𝜋𝑟2ℎ ≅ 𝑟2ℎ ≅ 𝑟 × 𝑟 × ℎ

∴ Effect on Volume = 100 – 100 × 92

100 ×

92

100 ×

104

100

= 11. 9744% (answer)

12 If the radius of the Cylinder is increased by 25% and its height remains

unchanged, then find the percentage increase in Volume ………..

SOLUTION:

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PERCENTAGE METHOD:

Given that, Radius (r) = 25% increase

Height (h) = Same (100)%

Volume of the Cylinder = 𝜋𝑟2ℎ ≅ 𝑟2ℎ ≅ 𝑟 × 𝑟 × h

∴ Percentage increase in Volume = 100 – 100 × 125

100 ×

125

100 ×

100

100

= 56. 25% (answer)

13 The radius of the base of a right circular Cone is increased by 15%

keeping the height fixed. The volume of the cone will be increased by

…………

SOLUTION:

PERCENTAGE METHOD:

Given that, radius (r) = 15% increase

height (h) = Same (100%)

Volume of the Cone = 1

3 × 𝜋𝑟2ℎ ≅ 𝑟2ℎ ≅ 𝑟 × 𝑟 × ℎ

∴ 100 – 100 × 115

100 ×

115

100 ×

100

100

= 32. 25% (answer)

13 On decreasing each side of and Equilateral Triangle by 2 cm, there is a

decrease of 4√𝟑 cm2 in its area. The length of the each side of the Triangle is

…….

SOLUTION:

Let the side of the Equilateral Triangle = ‘a’ cm

Area of Equilateral Triangle = √3

4 × 𝑎2 [Formula]

After decreasing each side, Area of Equilateral Triangle = √3

4 × (𝑎 − 2)2


√3

4 × 𝑎2 −

√3

4 × (𝑎 − 2)2 = 4√3

Side (a) = 5 cm (answer)

14 With vertices of Triangle ABC as centers, three circles are drawn, each

touching the other two externally. If the sides of the Triangle are 9 cm, 7 cm

and 6 cm, then the radii of the circles in centimeters are ………

SOLUTION:

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Given that, Sides of Triangle AB = 9cm, BC = 7cm & CA = 6 cm

Let the radius of the three circles are = R1 R2 and R3

From above diagram, 𝑅1 + 𝑅2 = 9 ………. (1)

𝑅2 + 𝑅3 = 7 ……….. (2)

𝑅3 + 𝑅1 = 6 ……….. (3)

From equations 1, 2 & 3, we get

R1 = 4cm, R2 = 5 cm & R3 = 2 cm (answer)

15 Three Spherical balls of radius of 1 cm, 2 cm and 3 cm are melted to

form a single Spherical ball. In the process, the loss of material is 25%. The

radius of new ball is ……..

SOLUTION:

Given that, radius of three Spherical balls = 𝑅1 = 1 𝑐𝑚, 𝑅2 = 2 𝑐𝑚 𝑎𝑛𝑑 𝑅3 =

3 𝑐𝑚

Volume of the Sphere = 4

3 × 𝜋𝑅3 [Formula]

Volume of the three Spherical balls = 4

3 × 𝜋 (13 + 23 + 33)

= 4

3 × 𝜋 × 36

Volume of the new ball = 4

3× 𝜋 × 36 ×

75

100 [25% material loss, given)

4

3 × 𝜋𝑅3 =

4

3 × 𝜋 × 36 ×

75

100

Radius (R) = 3 cm (answer)

16 The ratio of the area of Sector of a Circle to the area of the Circles is 1 :

4. If the area of the circle is 154 cm2, the perimeter of the sector is ……..

SOLUTION:

Area of Sector of Circle = 𝜋𝑟2 × 𝜃

360 [Formula]

Area of Circle = 𝜋𝑟2

Given that, 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟

𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑖𝑟𝑐𝑙𝑒 =

1

4

𝜋𝑟2 × 𝜃

360𝜋𝑟2

= 1

4

𝜃 = 900

Given that, 𝜋𝑟2 = 154 => r = 7 cm

Length of the Sector (l) = r𝜃 => 7 × 90 × 𝜋

180

= 7 × 90 × 22

7 ×

1

180

l = 11 cm

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Perimeter of the Sector = l + 2r [Formula]

= 11 + 2 × 7

= 25 cm (answer)

17 The lengths of three medians of a Triangle are 9 cm, 12 cm and 15 cm.

The area of the Triangle is ……..

SOLUTION:

Let lengths of the medians a = 9 cm, b = 15 cm & c = 12 cm

Area of the Triangle (When Medians given) =

1

3 √2 ∗ (𝑎2𝑏2 + 𝑏2𝑐2 + 𝑐2𝑎2) − (𝑎4 + 𝑏4 + 𝑐4)

= 1

3 √2 × (92 × 152 + 152 × 122 × 92) − (94 + 154 + 124)

By solving, we get

= 72 sq. cm (answer)

18 In a Triangle, distances from Centroid to vertices are respectively 4 cm,

6 cm and 8 cm. Find Medians …..

SOLUTION:

The point of intersection of Medians is called the Centroid of a Triangle

Given that, AG = 4 cm, BG = 6 cm & CG = 8 cm

So, Median AD => AG = 2

3 × 𝐴𝐷 = 4 => 𝐴𝐷 = 6 𝑐𝑚 (answer)

BG = 2

3 × 𝐵𝐸 = 6 => 𝐵𝐸 = 9 𝑐𝑚 (answer)

CG = 2

3 × 𝐶𝐹 = 8 => 𝐶𝐹 = 12 𝑐𝑚 (answer)

19 The length of each edge of a Rectangular Tetrahedron is 12 cm. The

area of the total surface area of the Tetrahedron is ……….

SOLUTION:

Given that, side of Rectangular Tetrahedron (a) = 12 cm

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Total surface area of the Tetrahedron = 4 × √3

4 × 𝑎2

= 4 × √3

4 × 12 × 12

= 144 √3 sq cm (answer)

20 The sides of a Triangle are 50 cm, 78 cm and 112 cm. The smallest

Altitude is ……….

SOLUTION:

Semi – Perimeter of Triangle (s) = 50+78+112

2= 120 𝑐𝑚

Area of the Triangle = √𝑠 (𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)

= √120(120 − 50)(120 − 78)(120 − 112)

= 1680 sq cm

The Altitude will be smallest when base is largest

∴ 1

2∗ 112 ∗ ℎ = 1680

h = 30 cm (answer)

21 The ratio of the areas of circum – circle and the in – circle of a Square

ism………

SOLUTION:

Let side of a Square is ‘x’

Radius of in – circle = r

Radius of ex – circle = R

Radius of a in – circle (r) = 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑎 𝑠𝑞𝑢𝑎𝑟𝑒

2 [from above fig.)

= 𝑥

2

Radius of ex – circle (R) = 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑜𝑓 𝑎 𝑆𝑞𝑢𝑎𝑟𝑒

2

= √2 × 𝑥

2

Required ratio =

√2 × 𝑥

2𝑥

2

= √2 ∶ 1 (answer)

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22 35 ml paint is required to paint a circular plate of 25 cm radius. How

much paint is required to paint a similar plate of radius 75 cm …….

SOLUTION:

Ratio of radius of two circular plates = 25 : 75

= 1 : 3

Ratio of Areas of two plates = 𝜋(1)2 ∶ 𝜋(3)2

= 1 : 9

i. e area of the second plate is 9 – times the area of the first plate

Therefore, paint required = 35 × 9

= 315 ml (answer)

23 An Iron Cube of edge 3 cm weighs 15 grams. What is the weight of a

similar Iron Cube whose edge is 12 cm ……

SOLUTION:

Ratio of edges of Iron Cubes = 3 : 12

= 1 : 4

Ratio of their Volumes = 13 ∶ 43

= 1 : 64

As Volume is 64 – times, weight will also be 64 – times

Therefore, weight of new Cube = 64 × 15 = 960 grams (answer)

25 The base of a right Prism is an Equilateral Triangle of area 173 cm2 and

the volume of the Prism is 10380 cm3. The area of the lateral surface of the

Prism is ….

SOLUTION:

Given that, area of Equilateral Triangle (base) = 173 cm2

Volume of the Prism = 10380 cm3

Volume of Prism = Area of Base × Height (h) [Formula]

10380 = 173 × h

h = 60 cm

Now, Area of an Equilateral Triangle = √3

4 × 𝑎2 = 173 [from question]

Side (a) = 20 cm

Perimeter of Equilateral Triangle = 3 × side=> 3 × 20 = 60 cm

Therefore, Area of the Lateral surface of the Prism = Perimeter of the Base ×

Height (h)

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= 60 × 60

= 3600 sq. cm (answer)

26 The radius of a Cylinder is 10 cm and height is 4 cm. The no. of

centimeters that may be added either to the radius or to the height to get the

same increase in the volume of the Cylinder…………

SOLUTION:

Given that, Radius of Cylinder = 10 cm

Height of Cylinder = 4 cm

Let radius be increased by ‘x’ cm

Volume of the Cylinder = 𝜋 × (10 + 𝑥)2 × 4 [V = 𝜋𝑟2ℎ]

If height of Cylinder be increased by ‘x’ cm

Volume of Cylinder = 𝜋 × 102 × (4 + 𝑥)


𝜋 × (10 + 𝑥)2 × 4 = 𝜋 × 102 × (4 + 𝑥)

x = 5 cm (answer)

27 The ratio between the no. of sides of two regular polygons is 1 : 2 and

the ratio between their interior angles is 2 : 3. The no. of sides of these

polygons is respectively ……….

SOLUTION:

Let the no. of sides of Polygon be = a & 2a

Interior angle of Polygon = (𝑛−2) × 180

𝑛 [Formula, n = side]


(𝑎 − 2) × 180𝑎

(2𝑎 − 2) × 1802𝑎

= 2

3

a = 4

So, sides = 4, 8 (answer)

28 The angles of a Triangle are in Arithmetic Progression. The ratio of the

greatest angle in degrees to the no. of radians in the greatest angle is 60 : 𝝅.

The angles in degrees are ……….

SOLUTION:

Let the angles of Triangle in A. P are = (a – d)0, a0 and (a + d)0

∴ a – d + a + a + d = 1800

a = 600


𝑎 − 𝑑

𝑎 + 𝑑=

60

𝜋

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60−𝑑

60+𝑑 =

60

180 [ 𝜋 = 1800]

d = 300

Therefore, angles = a – d, a & a + d

= 60 – 30, 60, 60 + 30

= 30, 60, 90 (answer)

29 If the length and the perimeter of a Rectangle are in the ratio 5 : 16, then

its length and breadth will be in the ratio …………..

SOLUTION:

Let the length of the Rectangle be ‘x’ cm and breadth be ‘y’ cm

Therefore, Perimeter of the Rectangle = 2(x + y)


𝑥

2𝑥 + 2𝑦=

5

16

By solving, we get

x : y = 5 : 3 (answer)

30 The circumference of a Circle is 100 cm. The measure of a side of the

square inscribed in this circle is ……..

SOLUTION:

Given that Circumference of a Circle = 𝜋𝑑 = 100 𝑐𝑚

d = 100

𝜋 𝑐𝑚 [d = diameter of Circle]

Therefore, Diagonal of inscribed Square = = 100

𝜋 𝑐𝑚

Therefore, Side of the Square = 1

√2∗ 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙

= 50√2

10 𝑐𝑚 (𝑎𝑛𝑠𝑤𝑒𝑟)

31 Find the ratio of the diameter of the Circles inscribed in an Equilateral

Triangle, the diameter circumscribing that Equilateral Triangle and the

height of the same Equilateral Triangle …….

SOLUTION:

Let the side of Equilateral Triangle = a

Then height = √3 × 𝑎

2

Diameter of the inner – circle = 2 × (1

3 ×

√3 × 𝑎

2) =

𝑎

√3

Diameter of the outer – circle = 2 × (2

3 ×

√3 × 𝑎

2) =

2𝑎

√3

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Required ratio = 𝑎

√3∶

2𝑎

√3∶

√3 × 𝑎

2

= 2 : 4 : 3 (answer)

32 Two Goats are tethered to diagonally opposites vertices of a field formed

by joining the mid – points of the adjacent sides of another square field of side

20√𝟐 𝒎. What is the total grazing area of the two Goats if the length of the

rope by which the Goats are tethered is 10√𝟐 𝒎. ?

SOLUTION:

The length of rope of Goat = 10√2 m

Then the two Goats will graze an area = Area of a Semi – circle with radius 10√2

m

So, total area grazed = 𝜋𝑟2

2= 100𝜋 𝑚2 (𝑎𝑛𝑠𝑤𝑒𝑟)

33 Find the area of the shaded region if the radius of each of the circles is 1

cm ………………

SOLUTION:

Length of the AB = 1 + 1 = 2 cm

Area of Triangle ABC (Equilateral Triangle) = √3

4 × 2 × 2

As we can see Triangle ABC is a Equilateral Tringle

∴ ∟𝐶𝐴𝐵 = ∟𝐴𝐵𝐶 = ∟𝐴𝐶𝐵 = 600

Required answer = √3 − 𝜋 × 12 × 180

360

= √3 − 𝜋

2 (answer)

34 What is the in – r5adius of the in – circle shown in the figure ?

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SOLUTION:

In – radius of the Triangle = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒

𝑆𝑒𝑚𝑖−𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒

In – radius =

1

2 × 9 × 40

(9+41+40

2)

= 4 cm (answer)

35 ∆𝑨𝑩𝑪 𝒂𝒏𝒅 ∆𝑷𝑸𝑹 both are similar and perimeter of ∆𝑨𝑩𝑪 𝒂𝒏𝒅 ∆𝑷𝑸𝑹

are 24 cm and 36 cm respectively if AB = 10 cm. Find PQ …………..

SOLUTION:

If two Triangles are similar, then

𝑅𝑎𝑡𝑖𝑜 𝑜𝑓 𝑆𝑖𝑑𝑒

𝑅𝑎𝑡𝑖𝑜 𝑜𝑓 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟=

𝑃𝑒𝑟𝑖𝑚𝑖𝑡𝑒𝑟 𝑜𝑓 ∆𝐴𝐵𝐶

𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 ∆𝑃𝑄𝑅

24

36=

𝐴𝐵

𝑃𝑄

PQ = 36

24∗ 10

PQ = 15 cm (answer)

36 A Cylindrical Iron rod whose height is eight times of its radius is melted

and cast into smaller Spheres whose radius is half of the radius of the

Cylinder. Thus the no. of Spheres so formed is ……….

SOLUTION:

Let the radius of the base of the Cylinder = ‘r’ unit

Height = 8r units

Volume of Cylinder =𝜋𝑟2ℎ [Formula]

= 𝜋 × 𝑟2 × 8𝑟

= 8𝜋𝑟3 𝐶𝑢𝑏𝑖𝑐 𝑢𝑛𝑖𝑡

Radius of Sphere = 𝑟

2 𝑢𝑛𝑖𝑡

Volume of Sphere = 4

3 × 𝜋 × 𝑟3 [Formula]

= 4

3 × 𝜋 × (

𝑟

2)3

= 𝜋𝑟3

6 𝐶𝑢𝑏𝑖𝑐 𝑢𝑛𝑖𝑡

Therefore, no. of Spheres = 8𝜋𝑟3

𝜋𝑟3

6

= 48 (answer)

37 Each of the height and the base of a right Circular Cone is increased by

100%. What will be the increase in the volume of the Cone ………….

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SOLUTION:

Let radius (base) = r

Height = h

Volume of the Right Circular Cone = = 1

3𝜋𝑟2ℎ [Formula]

PERCENTAGE METHOD:

Volume of the Right Circular Cone = 1

3𝜋𝑟2ℎ

= 1

3 × 𝜋 × [𝑟 × 𝑟 × ℎ]

= r (100% increase) × r (100% increase) × height (100% increase)

Required answer = 100 – 100 × 200

100 ×

200

100 ×

200

100

= 700% (answer)

38 Each interior angle of Regular Polygon is 1440. The number of sides of

the Polygon is ………

SOLUTION:

Let the no. of sides of Polygon = n

Then, each exterior angle = 180 – 144

= 360

Therefore, no. of sides (n) = 360

36

n = 10 (answer)

39 A wire in the form of a circle of radius 42 m is cut and again bent in the

form of a square. What is the diagonal of the square ……………

SOLUTION:

Given that, radius (r) = 42 m

Let side of the square = ‘a’ m


2𝜋𝑟 = 4 × 𝑎

2 × 22

7 × 42 = 4 × 𝑎

a = 66 m

So, the diagonal of square = √2 × 𝑎

= 66√2 (answer)

40 The circumference of the front wheel of a cart is 30 feet long and that of

the back wheel is 36 feet long. What is the distance travelled by the cart, when

the front wheel has done five more revolutions than the rear wheel ……

SOLUTION:

The circumference of the wheel is 30 feet and that of the rear wheel is 36 feet

Let the rear wheel make ‘n’ revolutions. At this time, the front wheel should have

made ‘n + 5’ revolutions

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As both the wheels would have covered the same distance

n × 36 = (n + 5) × 30

n = 25

Therefore, distance covered = 25 × 36

= 900 feet (answer)

41 The area of an Equilateral Triangle inscribed in Circle is 16√𝟑 𝒎𝟐. Find

the area of the Circle ……..

SOLUTION:

Area of Equilateral Triangle = √3

4× 𝑎2 [Formula]

√3

4× 𝑎2 = 16√3

a = 8 cm

Height of the Equilateral Triangle = √3

2 × 8 = 4√3 𝑐𝑚

Radius of the Circumcircle = 2

3 × 4√3 =

8

√3 𝑐𝑚

Therefore, area of the Circle = 𝜋 × 8

√3 ×

8

√3

= 64

3𝜋 𝑐𝑚2 (𝑎𝑛𝑠𝑤𝑒𝑟)

42 If ABCD is a Cyclic Quadrilateral in which ∟A = 4x0, ∟𝑩 = 7x0, ∟C = 5y0

and ∟D = y0, then x : y is ………

SOLUTION:

The sum of opposite angles of Cyclic Quadrilateral is 1800

∴ ∟A + ∟C = 1800

=> 4x + 5y = 1800 ………….. (1)

∟B + ∟D = 1800

7x + y = 1800 …………………. (2)

From equations (1) and (2) => (2) × 5 – (1)

x = 720

31

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y = 5580−5040

31=

540

31

Therefore, x : y = 720

31∶

540

31

= 4 : 3 (answer)

43 A is the centre of Circle whose radius is 8 and B is the centre of a Circle

whose diameter is 8. If these two Circles touch externally, then the area of the

circle with diameter AB is ……….

SOLUTION:

From above figure, Diameter AB = 8 + 4 => 12 units

Radius = 12

2= 6 𝑢𝑛𝑖𝑡𝑠

Therefore, Area of the Circle = 𝜋𝑟2

= 𝜋 × 6 × 6

= 36𝜋 sq. units (answer )

44 By decreasing 150 each angle of a Triangle the ratios of their angles are 2

: 3 : 5, the radian measure for greatest angle is ……..

SOLUTION:

Ratio of angle after decreasing 150 each angle = 2 : 3 : 5


2x + 3x + 5x = 180 – 15 × 3

x = 13. 50

Greatest angle = 5 × 13. 5 + 15

= 82. 50 or 82. 5 × 𝜋

180

= 11𝜋

24 (𝑎𝑛𝑠𝑤𝑒𝑟)

45 The in – radius of an Equilateral Triangle is of length 3 cm. Then the

length of each of its medians is ……….

SOLUTION:

In – radius of Equilateral Triangle = 𝑠𝑖𝑑𝑒

2√3=

𝑀𝑒𝑑𝑖𝑎

3

Therefore, Median = 3 × In – radius

Median = 9 cm (answer)

46 The Perimeters of two Squares are 40 cm and 32 cm. The Perimeter of

third Square whose area is the difference of the area of the two Squares is

…………

SOLUTION:

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Let sides of Squares are a, b and c

Given that, Perimeters 4a = 40 => a = 10 cm

And 4b = 32 => b = 8 cm


c2 = (10)2 – (8)2

= 36

c = 6 cm

So, Perimeter of a third Square = 4 × 6

= 24 cm (answer)

47 An angle is 300 more than the one half of its complement. Find the angle

in degrees …………

SOLUTION:

Let the angle be ‘x’, then its Complementary angle be (90 – x)


x – (90−𝑥)

2= 30

x = 500 (answer)

48 If an angle of a Parallelogram is two – third of its adjacent angle, then

the largest angle of Parallelogram …….

SOLUTION:

Since, adjacent angles of a Parallelogram are Supplementary.

Therefore, x + 2

3× 𝑥 = 1800

x = 1080

∴ 2

3 × 𝑥 =>

2

3∗ 108

= 720

Therefore, angles of Parallelogram = 1080, 720, 1080 and 720

Largest angle = 1080 (answer)

49 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 kmph

completes one round in 8 – minutes, then the area of the park is ……..

SOLUTION:

Given that, Ratio between length & breadth = 3 : 2

So, distance covered by Man in 8 – minutes = 12 × 1000 × 8

60

= 1600 meters

Then, distance = 2 × (3x + 2x) = 1600

= 160 meter

So, Area = 3x × 2x => 3 × 160 × 2 × 160

= 153600 m2 (answer)

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50 The length of the floor of a room is 20 m more than its breadth. The

area of the floor remains unchanged even if its length is decreased by 10 m

and breadth is increased by 5 m. The area of floor is …………

SOLUTION:

Let width of floor = x meter

∴ length = (x + 20) meter

∴ Area of floor = (x + 20) × x m2

In second condition

(x + 10) × (x +5) = (x + 20) × x

By solving, we get

x = 10 meter

∴ Length = x + 20 => 10 + 20

= 30 meter

∴ Area of floor = 30 × 10

= 300 m2 (answer)

51 A Cube of 384 cm2 surface area is melted to make ‘x’ number of small

cubes each of 96 cm2 surface area. The value of x is ………….

SOLUTION:

Surface area of big Cube = 6A2

A = side or edge of the big Cube

Given that, 6A2 = 384

A = 8 cm

Surface area of small cube = 6a2

6a2 = 96 [From question]

a =4 cm

∴ 83 = x × (4)3

x = 8 (answer)

52 A spherical metal of radius 10 cm is melted and made into 1000 smaller

spheres of equal sizes. In this process the surface area of the metal is increased

by:

SOLUTION:

Volume of the Sphere = 4

3∗ 𝜋𝑟3

From question, 4

3 𝜋𝑅3 = 1000 ×

4

13 𝜋𝑟3 => R = 10r

R = Radius of big sphere, r = radius of small sphere

10 = 10 × r [R = 10 cm]

r = 1 cm

Initial surface area of Sphere = 4𝜋𝑅2 => 400𝜋

Final Surface area of 1000 smaller spheres = 1000 × 4𝜋𝑟2

= 1000 × 4 × 𝜋 ∗ 12

= 4000𝜋

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∴ Increase in Surface area = 4000𝜋 – 400𝜋

∴ Increase in Surface area = 3600𝜋 i. e 9 times (answer)

53 The area of paper can be divided into 144 Squares, but if the dimensions

of each Square were reduced by 2 cm each, then the no. of Squares so formed

are 400. The area of the paper initially was ………..

SOLUTION:

Let edge of Square = x cm


114x2 = 400 (x – 2)2

By solving, we get x = 5 and 5

4

∴ edge of Square = 5 cm

∴ initial area = 144 × x2

= 144 × 52

= 3, 600 cm2 (answer)

54 A Square and an equilateral triangle have the same perimeter. If the

diagonal of the Square is 12√𝟐cm, then the area of the Triangle is ……….

SOLUTION:

Let x be the side of the Square

Diagonal (d) = √2 × 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑎 𝑆𝑞𝑢𝑎𝑟𝑒

12√2 = √2 × 𝑥 => x = 12

Perimeter of the Square as well as of the Equilateral Triangle = 4 × 12 => 48 cm

Side of the Equilateral Triangle =>3a = 48 => a = 16 cm

From above Triangle

AD2 = AC2 – CD2

= 162 – 82

=256 – 64

AD = 8√3

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∴ Area of the Triangle ABC

= 1

2 × 𝐵𝐶 × 𝐴𝐷

= 1

2× 16 × 8√3

= 64√3 cm2 (answer)

55 Three Circles of equal radius ‘a’ cm touch each other. The area of the

shaded region is

SOLUTION:

From above figure, AB = BC = CA = 2a cm

Area of Equilateral Triangle ABC = √3

4∗ 4𝑎2 = √3 a2

Area of 3 Sectors of 𝜃 = 600

= 3 × 𝜋𝑎2 × 60

360

= 𝜋𝑎2

2

Area of shaded region = area of ∆𝐴𝐵𝐶 − 𝐴𝑟𝑒𝑎 𝑜𝑓 3 𝑆𝑒𝑐𝑡𝑜𝑟𝑠

= √3 a2 – 𝜋𝑎2

2

= (2√3− 𝜋

2) 𝑎2 𝑐𝑚2 (𝑎𝑛𝑠𝑤𝑒𝑟)

56

fdaytalk · mailid : [email protected] acute angle the angle which is less than 900 right...

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