How to Prepare for Quantitative Aptitude for CAT (2014)

Block IV: Geometry

...BACK TO SCHOOL

The word GEOMETRY is derived from two words—GEO meaning earth and METRY meaning measurement. Hence, it is quite evident that geometry as a mathematical science developed mainly due to the human need of measuring land masses and distances. The major developments in the fields of Geometry and Mensuration are mainly credited to the ancient Egyptian and Greek civilisations. (In fact, this is one of the reasons why all formulae and theorems primarily carry Greek names.)

Key Geometrical Concepts

Point: The point should be visualised as a singular dot. In physical terms, a point can be defined as a single dot that can be created on a plain paper by a very sharp pencil. It could also be visualised as a singular prick on a piece of paper by a very sharp nail or pin.

Line: Mathematically, lines are defined as a group of points which are straight one after another. All lines are supposed to extend infinitely in two directions.

Segment of a Line

If a part of a line is cut out, we get a segment of a line.

Physically, the closest representations of a segment of a line would be a tight thread or the straight crease of a piece of paper.

Plane: The surface of a smooth wall or a table top is the closest representation of a portion of a plane.

Some Geometrical Properties

1. Lines Between Points: Let us take two distinct points in a plane. You can easily verify that an infinite number of lines can be drawn in the plane passing through any one of these two points. However, if you wanted to draw a line which passes through both the points, you will be able to draw only one line. This is an important result in Geometry.

Property: Given any two distinct points in a plane, there exists one and only one line containing both the points. Alternately, we can state that two distinct points in a plane determine a unique line.

2. Collinearity of Points: Collinear or non-collinear points are only defined in the context of 3 or more points.

Consider the situation of three points. There can only be two cases with respect to three points:

1.All the points lie on the same line. (Here, the points are said to be collinear).

2.All the three points do not lie on the same line (In this case the points are said to be non-collinear).

3. Points in common between distinct lines: In the case of distinct lines, there can only be two cases:

(A)There is one point in common: In such a case, the common point is called as the point of intersection, and the two lines are called as intersecting lines.

(B)There is no point in common: In such a case, the two lines are non-intersecting and are also called as parallel lines.

The property can thus be stated as:

Property: Two distinct lines in a plane cannot have more than one point in common.

4. Lines and Points:

Property: Given a line and a point in the same plane, such that the point is not on the line, there is one and only one line which passes through the point and is parallel to the given line.

5. Lines:

Property: Two intersecting lines cannot be parallel to the same line.

6. Multiple Lines: There can be four distinct possibilities in such a case.

1.No two lines intersect each other

2.Some lines intersect

3.Every line intersects all other lines, but their respective points of intersection are different from each other.

4.Every pair of lines is intersecting and all the points of intersection coincide.

The reader is required to visualise the figures for each of these situations.

The remaining formulae and results for geometry have been given within the chapter. You are required to move to the respective formulae. However, an important note while you are doing this.

Important Note

One of the least understood issues with respect to this block is:

How does a person move from:

“Cannot Solve Geometry and Mensuration!” to “Can Solve Geometry and Mensuration!”

In fact, this transition should be your sole aim while studying and solving this block of chapters.

As you are already aware, mere knowledge of formulae does not guarantee this movement. Then in that case what should a person do if he/she has to engineer this transition of abilities?

Well, the answer is really quite simple and for understanding the same you need to move out of conventional mathematical study processes. In your mind you need to create an awareness of how to think while solving a geometry/mensuration question. Here are a few tips that will help you through this transition:

1. Some formulae are more important than others: For the CAT (and indeed all other MBA and aptitude examinations) your concentration should not be on memorizing complex formulae. A close study of the past trends for questions from this block will show you that the formulae themselves can be graded into the important and the not so important formulae. In fact, if you study over 200 questions from this block that have been asked in the CAT in the past decade or more, you will realise that over 98% of the questions are simply based on a small subset of geometrical formulae and results. This important set of formulae are listed below for your convenience:

Lines: Properties related to intersecting lines and angles formed.

Polygons: Basic properties of regular polygons.

Triangles: Pythagoras theorem, 30-60-90 triangle, 45-45-90 triangle, Sine rule, Properties of Exterior angles of a triangle, equilateral triangle, isosceles triangle.

Quadrilaterals: Areas, perimeters, diagonals and angles between diagonals for all standard quadrilaterals.

Hexagon: A Hexagon can be divided into six equal triangles.

2. Length/area/volume measuring formulae versus Angle measuring formulae: In your mind, divide the formulae into length/area/volume measuring formulae on the one hand and Angle measuring formulae on the other hand. This will be very helpful, since most questions in this block can be clearly segregated as length or angle measuring questions. In such a scenario, in case you have a question requiring angle measurements, you will only require to think of angle measuring formulae.

3. The use of reverse flowcharting for solving questions: All questions in geometry and mensuration ask you to find a particular value. Very often, it makes the question much more convenient to solve if instead of trying to find what is asked, you try to find a value which is related to the required value.

Thus, for example, if in a triangle ABC angle A is given to be 40° & angle B is asked for and you are not getting a clear strategy for getting the value of angle B, then it might be wise to try to derive the value of angle C. If you can get C, then B will be got by 140-C.

This process is called as reverse flowcharting and is an extremely crucial process for solving questions in geometry.

In other words, it can also be seen to be a deflection of the original question into a more convenient question.

Pre-assessment Test

Directions for Questions 7 to 9: Answer the questions on the basis of the information given below.

Consider a cylinder of height h cm and radius r = 2/p cm as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of n turns (in other words, the string’s length is the minimum length required to wind n turns).

Chapter 11. GEOMETRY AND MENSURATION

The chapters of geometry and mensuration have their share of questions in the CAT and other MBA entrance exams. For doing well in questions based on this chapter, the student should familiarise himself/herself with the basic formulae and visualisations of the various shapes of solids and two-dimensional figures based on this chapter.

The following is a comprehensive collection of for- mulae based on two-dimensional and three-dimensional figures.

For the purpose of this chapter we have divided the theory in two parts:

• Part I consists of geometry and mensuration of two-dimensional figures
• Part II consists of mensuration of three-dimensional figures.

PART I: GEOMETRY

INTRODUCTION

Geometry and Mensuration are important areas in the CAT examination. In the Online CAT, the Quantitative Aptitude and Data Interpretation section has consisted by an average between 3–5 questions out of 30 from these chapters. Besides, questions from these chapters appear prominently in all major aptitude based exams like MBAs, Bank POs, etc.

Hence, the student is advised to ensure that he/she studies this chapter completely and thoroughly. Skills to be developed while studying and practising this chapter will be application of formula and visualisation of figures and solids.

The principal skill required for doing well in this chapter is the ability to apply the formulae and theorems.

The following is a comprehensive collection of formulae based on two-dimensional figures. The student is advised to remember the formulae in this chapter so that he is able to solve all the questions based on this chapter.

THEORY

STRAIGHT LINES

Parallel Lines: Two straight lines are parallel if they lie on the same plane and do not intersect however far produced.

Transversal: It is a straight line that intersects two parallel lines. When a transversal intersects two parallel lines then

1.Corresponding angles are equal, (that is: For the following figure)

1 = 5; 2 = 6; 4 = 8; 3 = 7

2.Alternate interior angles are equal, that is (Refer following figure)

4 = 6; 5 = 3

3.Alternate exterior angles are equal, that is

2 = 8; 1 = 7

4.Interior angles on the same side of transversal add up to 180°, that is

4 + 5 = 3 + 6 = 180°

POLYGONS

Polygons are plane figures formed by a closed series of rectilinear (straight) segments. Example: Triangle, Rectangle…

Polygons can broadly be divided into two types:

(a)Regular polygons: Polygons with all the sides and angles equal.

(b)Irregular polygons: Polygons in which all the sides or angles are not of the same measure.

Polygon can also be divided as concave or convex poly-gons.

Convex polygons are the polygons in which all the diagonals lie inside the figure otherwise it’s a concave polygon

Polygons can also be divided on the basis of the number of sides they have.

Properties

1.Sum of all the angles of a polygon with n sides = (2n – 4)p/2 or (n – 2)p Radians = (n – 2) 180° degrees

2.Sum of all exterior angles = 360°◊

i.e. In the figure below:

q₁ + q₂ + ... + q₆ = 360°

In general, q₁ + q₂ + ... + q_n = 360°

3.No. of sides = 360°/exterior angle.

4.Area = (ns²/4) × cot (180/n); where s = length of side, n = no. of sides.

5.Perimeter = n × s.

TRIANGLES (D)

A triangle is a polygon having three sides. Sum of all the angles of a triangle = 180°.

Types

1.Acute angle triangle: Triangles with all three angles acute (less than 90°).

2.Obtuse angle triangle: Triangles with one of the angles obtuse (more than 90°).

Note: we cannot have more than one obtuse angle in a triangle.

3.Right angle triangle: Triangle with one of the angles equal to 90°.

4.Equilateral triangle: Triangle with all sides equal. All the angles in such a triangle measure 60°.

5.Isosceles triangle: Triangle with two of its sides equal and consequently the angles opposite the equal sides are also equal.

6.Scalene Triangle: Triangle with none of the sides equal to any other side.

Properties (General)

• Sum of the length of any two sides of a triangle has to be always greater than the third side.
• Difference between the lengths of any two sides of a triangle has to be always lesser than the third side.
• Side opposite to the greatest angle will be the greatest and the side opposite to the smallest angle the smallest.
• The sine rule: a/sin A = b/sin B = c/sin C= 2R

(where R = circum radius.)

• The cosine rule: a²= b²+ c² – 2bc cos A

This is true for all sides and respective angles.

In case of a right –, the formula reduces to a² = b² + c²

Since cos 90 = 0

• The exterior angle is equal to the sum of two interior angles not adjacent to it.

–ACD = –BCE = –A + –B

Area

1.Area = 1/2 base × height or 1/2 bh.

Height = Perpendicular distance between the base and vertex opposite to it

2.Area = (Hero’s formula)

where S = (a,bandcbeing the length of the sides)

3.Area = rs (where r is in radius)

4.Area = 1/2 × product of two sides × sine of the included angle

= 1/2 ac sin B

= 1/2 ab sin C

= 1/2 bc sin A

4.Area = abc/4R

where R = circum radius

Congruency of Triangles Two triangles are congruent if all the sides of one are equal to the corresponding sides of another. It follows that all the angles of one are equal to the corresponding angles of another. The notation for congruency is .

Conditions for Congruency

1.SAS congruency: If two sides and an included angle of one triangle are equal to two sides and an included angle of another, the two triangles are congruent. (See figure below.)

Here, AB = PQ

BC = QR

and –B = –Q

So DABC DPQR

2.ASA congruency: If two angles and the included side of one triangle is equal to two angles and the included side of another, the triangles are congruent. (See figure below.)

Here, –A = –P

–B = –Q

and AB = PQ

So DABC DPQR

3.AAS congruency: If two angles and side opposite to one of the angles is equal to the corresponding angles and the side of another triangle, the triangles are congruent. In the figure below:

–A = –P

–B = –Q

and AC = PR

So DABC DPQR

4.SSS congruency: If three sides of one triangle are equal to three sides of another triangle, the two triangles are congruent. In the figure below:

AB = PQ

BC = QR

AC = PR

\ DABC DPQR

5.SSA congruency: If two sides and the angle opposite the greater side of one triangle are equal to the two sides and the angle opposite to the greater side of another triangle, then the triangles are congruent. The congruency doesn’t hold if the equal angles lie opposite the shorter side. In the figure below, if

AB = PQ

AC = PR

–B = –Q

Then the triangles are congruent.

i.e. DABC DPQR.

Similarity of triangles Similarity of triangles is a special case where if either of the conditions of similarity of polygons holds, the other will hold automatically.

Types of Similarity

1.AAA similarity: If in two triangles, corresponding angles are equal, that is, the two triangles are equiangular then the triangles are similar.

Corollary (AA similarity): If two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar. The reason being, the third angle becomes equal automatically.

2.SSS similarity: If the corresponding sides of two triangles are proportional then they are similar.

For DABC to be similar to DPQR, AB/PQ = BC/QR = AC/PR, must hold true.

3.SAS similarity: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

D ABC ~DPQR

If AB/BC = PQ/QR and B = –Q

Ratio of medians = Ratio of heights = Ratio of circumradii = Ratio of inradii = Ratio of angle bisectors

While there are a lot of methods through which we see similarity of triangles, the one thing that all our Maths teachers forgot to tell us about similarity is the basic real life concept of similarity. i.e. Two things are similar if they look similar!!

If you have been to a toy shop lately, you would have come across models of cars or bikes which are made so that they look like the original—but are made in a different size from the original. Thus you might have seen a toy Maruti car which is built in a ratio of 1:25 of the original car. The result of this is that the toy car would look very much like the original car (of course if it is built well!!). Thus if you have ever seen a father and son looking exactly like each other, you have experienced similarity!!

You should use this principle to identify similar triangles. In a figure two triangles would be similar simply if they look like one another.

Thus, in the figure below if you were to draw the radii OB and O¢A the two triangles MOB and MO¢A will be similar to each other. Simply because they look similar. Of course, the option of using the different rules of similarity of triangles still remains with you.

Equilateral Triangles (of side a):

1.(\ sin 60 = /2 = h/side)

h =

2.Area = 1/2 (base) × (height) = × a × = a²

3.R (circum radius) = = .

4.r (in radius) = = .

Properties

1.The incentre and circumcentre lies at a point that divides the height in the ratio 2 : 1.

2.The circum radius is always twice the in radius. [R = 2r.]

3.Among all the triangles that can be formed with a given perimeter, the equilateral triangle will have the maximum area.

4.An equilateral triangle in a circle will have the maximum area compared to other triangles inside the same circle.

Isosceles Triangle

Right-Angled Triangle

Pythagoras Theorem In the case of a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In the figure below, for triangle ABC, a² = b² + c²

Area = 1/2 (product of perpendicular sides)

R(circumradius) =

Area = rs

(where r = in radius and s = (a + b + c)/2 where a, b and c are sides of the triangle)

fi1/2 bc = r(a + b + c)/2

fi r = (bc)/(a + b + c)

In the triangle ABC,

DABC ~ DDBA ~ DDAC

(Note: A lot of questions are based on this figure.)

Further, we find the following identities:

1.DABC ~ DDBA

\ AB/BC = DB/BA

fi AB² = DB × BC

fi c² = pa

2.DABC ~ DDAC

AC/BC = DC/AC

fi AC² = DC × BC

fi b² = qa

3.DDBA ~ DDAC

DA/DB = DC/DA

DA² = DB × DC

fi AD² = pq

Basic Pythagorean Triplets

Æ 3, 4, 5 Æ 5, 12, 13 Æ 7, 24, 25 Æ 8, 15, 17 Æ 9, 40, 41 Æ 11, 60, 61 Æ 12, 35, 37 Æ 16, 63, 65 Æ 20, 21, 29 Æ 28, 45, 53. These triplets are very important since a lot of questions are based on them.

Any triplet formed by either multiplying or dividing one of the basic triplets by any positive real number will be another Pythagorean triplet.

Thus, since 3, 4, 5 form a triplet so also will 6, 8 and 10 as also 3.3, 4.4 and 5.5.

Similarity of right triangles Two right triangles are similar if the hypotenuse and side of one is proportional to hypotenuse and side of another. (RHS–similarity–Right angle hypotenuse side).

Important Terms with Respect to a Triangle

1. Median A line joining the mid-point of a side of a triangle to the opposite vertex is called a median. In the figure the three medians are PG, QFand REwhere G, E and F are mid-points of their respective sides.

• A median divides a triangle into two parts of equal area.
• The point where the three medians of a triangle meet is called the centroid of the triangle.
• The centroid of a triangle divides each median in the ratio 2 : 1.

i.e. PC : CG = 2 : 1 = QC : CF = RC : CE

Important formula with respect to a median

Æ 2 × (median)² + 2 × (1/2 the third side)²

= Sum of the squares of other two sides

fi 2(PG)² + 2 ×

= (PQ)² + (PR)²

2. Altitude/Height A perpendicular drawn from any vertex to the opposite side is called the altitude. (In the figure, AD, BFand CEare the altitudes of the triangles).

• All the altitudes of a triangle meet at a point called the orthocentre of the triangle.
• The angle made by any side at the orthocentre and the vertical angle make a supplementary pair (i.e. they both add up to 180°). In the figure below:

–A + –BOC = 180° = –C + –AOB

3. Perpendicular Bisectors A line that is a perpendicular to a side and bisects it is the perpendicular bisector of the side.

• The point at which the perpendicular bisectors of the sides meet is called the circumcentre of the triangle
• The circumcentre is the centre of the circle that circumscribes the triangle. There can be only one such circle.
• Angle formed by any side at the circumcentre is two times the vertical angle opposite to the side. This is the property of the circle whereby angles formed by an arc at the centre are twice that of the angle formed by the same arc in the opposite arc. Here we can view this as:

–QCR = 2 –QPR (when we consider arc QR and it’s opposite arc QPR)

4. Incenter

• The lines bisecting the interior angles of a triangle are the angle bisectors of that triangle.
• The angle bisectors meet at a point called the incentre of the triangle.
• The incentre is equidistant from all the sides of the triangle.

• From the incentre with a perpendicular drawn to any of the sides as the radius, a circle can be drawn touching all the three sides. This is called the incircleof the triangle. The radius of the incircle is known as inradius.
• The angle formed by any side at the incentre is always a right angle more than half the angle opposite to the side.

This can be illustrated as –QIR = 90 + 1/2 –P

• If QIand RIbe the angle bisectors of exterior angles at Q and R then,

QIR = 90 – 1/2 –P.

QUADRILATERALS

Area

(A)Area = 1/2 (product of diagonals) × (sine of the angle between, them)

If q₁ and q₂ are the two angles made between themselves by the two diagonals, we have by the property of intersecting lines Æ q₁ + q₂ = 180°

Then, the area of the quadrilateral = d₁ d₂ sin q₁ = d₁ d₂ sin q₂.

(B)Area = 1/2 × diagonal × sum of the perpendiculars to it from opposite vertices = .

(C)Area of a circumscribed quadrilateral

A =

Where S =

(where a, b, c and d are the lengths of the sides.)

Properties

1.In a convex quadrilateral inscribed in a circle, the product of the diagonals is equal to the sum of the products of the opposite sides. For example, in the figure below:

(a × c) + (b × d) = AC × BD

2.Sum of all the angles of a quadrilateral = 360°.

TYPES OF QUADRILATERALS

1. Parallelogram (|| gm)

A parallelogram is a quadrilateral with opposite sides parallel (as shown in the figure below.)

(A)Area = Base (b) × Height (h)

= bh

(B)Area = product of any two adjacent sides × sine of the included angle.

= ab sin Q

(C)Perimeter = 2 (a + b)

where a and b are any two adjacent sides.

Properties

(a)Diagonals of a parallelogram bisect each other.

(b)Bisectors of the angles of a parallelogram form a rectangle.

(c)A parallelogram inscribed in a circle is a rectangle.

(d)A parallelogram circumscribed about a circle is a rhombus.

(e)The opposite angles in a parallelogram are equal.

(f)The sum of the squares of the diagonals is equal to the sum of the squares of the four sides in the figure:

AC² + BD² = AB² + BC² + CD² + AD²

= 2(AB² + BC²)

2. Rectangles

A rectangle is a parallelogram with all angles 90°

(A)Area = Base × Height = b × h

Note: Base and height are also referred to as the length and the breadth in a rectangle.

(B)Diagonal (d) = Æ by Pythagoras theorem

Properties of a Rectangle

(a)Diagonals are equal and bisect each other.

(b)Bisectors of the angles of a rectangle (a parallelogram) form another rectangle.

(c)All rectangles are parallelograms but the reverse is not true.

3. Rhombus

A parallelogram having all the sides equal is a rhombus.

(A)Area = 1/2 × product of diagonals × sine of the angle between them.

= 1/2 × d₁ × d₂ sin 90°(Diagonals in a rhombus intersect at right angles)

= 1/2 × d₁ d₂(since sin 90° = 1)

(B)Area = product of adjacent sides × sine of the angle between them.

Properties

(a)Diagonals bisect each other at right angles.

(b)All rhombuses are parallelograms but the reverse is not true.

(c)A rhombus may or may not be a square but all squares are rhombuses.

4. Square

A square is a rectangle with adjacent sides equal or a rhombus with each angle 90°

(a)Area = base × height = a²

(b)Area = 1/2 (diagonal)² = 1/2 d² (square is a rhombus too).

(c)Perimeter = 4a (a = side of the square)

(d)Diagonal = a

(E)In radius =

Properties

(a)Diagonals are equal and bisect each other at right angles.

(b)Side is the diameter of the inscribed circle.

(c)Diagonal is the diameter of the circumscribing circle.

fiDiameter = a.

Circumradius =

5. Trapezium

A trapezium is a quadrilateral with only two sides parallel to each other.

(a)Area = 1/2 × sum of parallel sides × height = 1/2 (AB + DC) × h—For the figure below.

(b)Median = 1/2 × sum of the parallel sides (median is the line equidistant from the parallel sides)

For any line EF parallel to AB

EF =

Properties

(A)If the non-parallel sides are equal then diagonals will be equal too.

REGULAR HEXAGON

(a)Area = [(3)/2] (side)²

=

(b)A regular hexagon is actually a combination of 6 equilateral triangles all of side ‘a’.

Hence, the area is also given by: 6 × side of equilateral triangles = 6 × a²

(c)If you look at the figure closely it will not be difficult to realise that circumradius (R) = a; i.e the side of the hexagon is equal to the circumradius of the same.

CIRCLES

(a)Area = pr²

(b)Circumference = 2 pr = (r = radius)

(c)Area = 1/2 × circumference × r

Arc: It is a part of the circumference of the circle. The bigger one is called the major arc and the smaller one the minor arc.

(d)Length (Arc XY) = × 2pr

(e)Sector of a circle is a part of the area of a circle between two radii.

(f)Area of a sector = × pr²

(where q is the angle between two radii)

= (1/2)r × length (arc xy)

(\ prq/180 = length arc xy)

= × r ×

(g)Segment: A sector minus the triangle formed by the two radii is called the segment of the circle.

(h)Area = Area of the sector – Area DOAB = × pr² – × r² sin q

(i)Perimeter of segment = length of the arc + length of segment AB

= × 2pr + 2r sin

= + 2r sin

(j)Congruency: Two circles can be congruent if and only if they have equal radii.

Properties

(a)The perpendicular from the centre of a circle to a chord bisects the chord. The converse is also true.

(b)The perpendicular bisectors of two chords of a circle intersect at its centre.

(c)There can be one and only one circle passing through three or more non-collinear points.

(d)If two circles intersect in two points then the line through the centres is the perpendicular bisector of the common chord.

(e)If two chords of a circle are equal, then the centre of the circle lies on the angle bisector of the two chords.

(f)Equal chords of a circle or congruent circles are equidistant from the centre.

(g)Equidistant chords from the centre of a circle are equal to each other in terms of their length.

(h)The degree measure of an arc of a circle is twice the angle subtended by it at any point on the alternate segment of the circle. This can be clearly seen in the following figure:

With respect to the arc AB, –AOB = 2 –ACB.

(i)Any two angles in the same segment are equal. Thus, –ACB = –ADB.

(j)The angle subtended by a semi-circle is a right angle. Conversely, the arc of a circle subtending a right angle at any point of the circle in its alternate segment is a semi-circle.

(k)Any angle subtended by a minor arc in the alternate segment is acute, and any angle subtended by a major arc in the alternate segment is obtuse.

In the figure below

–ABC is acute and

–ADC = obtuse

Also q₁ = 2 –B

And q₂ = 2 –D

\q₁ + q₂ = 2(–B + –D)

= 360° = 2(–B + –D)

or –B + –D = 180°

or sum of opposite angles of a cyclic quadrilateral is 180°.

(l)If a line segment joining two points subtends equal angles at two other points lying on the same side of the line, the four points are concyclic. Thus, in the following figure:

If,q₁ = q₂

Then ABCD are concyclic, that is, they lie on the same circle.

(m)Equal chords of a circle (or of congruent circles) subtend equal angles at the centre (at the corresponding centres.) The converse is also true.

(n)If the sum of the opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

Secant: A line that intersects a circle at two points.

Tangent: A line that touches a circle at exactly one point.

(o)If a circle touches all the four sides of a quadrilateral then the sum of the two opposite sides is equal to the sum of other two

AB + DC = AD + BC

(p)In two concentric circles, the chord of the larger circle that is tangent to the smaller circle is bisected at the point of contact.

Tangents

• Length of direct common tangents is

=

where r₁ and r₂ are the radii of the circles

=

• Length of transverse common tangents is

=

=

ELLIPSE

• Perimeter = p (a+ b)
• Area = pab

STAR

Sum of angles of a star = (2n – 8) × p/2 = (n – 4)p

PART II: MENSURATION

The following formulae hold true in the area of mensuration:

1. Cuboid

A cuboid is a three dimensional box. It is defined by the virtue of it’s length l, breadth b and height h. It can be visualised as a room which has its length, breadth and height different from each other.

1.Total surface area of a cuboid = 2 (lb + bh + lh)

2.Volume of the cuboid = lbh

2. Cube of side s

A cube is a cuboid which has all its edges equal i.e. length = breadth = height = s.

1.Total surface area of a cube = 6s².

2.Volume of the cube = s³.

3. Prism

A prism is a solid which can have any polygon at both its ends. It’s dimensions are defined by the dimensions of the polygon at it’s ends and its height.

1.Lateral surface area of a right prism = Perimeter of base * height

2.Volume of a right prism = area of base* height

3.Whole surface of a right prism = Lateral surface of the prism + the area of the two plane ends.

4. Cylinder

A cylinder is a solid which has both its ends in the form of a circle. Its dimensions are defined in the form of the radius of the base (r) and the height h. A gas cylinder is a close approximation of a cylinder.

1.Curved surface of a right cylinder = 2prh where r is the radius of the base and h the height.

2.Whole surface of a right circular cylinder = 2prh + 2pr²

3.Volume of a right circular cylinder = pr²h

5. Pyramid

A pyramid is a solid which can have any polygon as its base and its edges converge to a single apex. Its dimensions are defined by the dimensions of the polygon at its base and the length of its lateral edges which lead to the apex. The Egyptian pyramids are examples of pyramids.

1.Slant surface of a pyramid = 1/2 * Perimeter of the base* slant height

2.Whole surface of a pyramid = Slant surface + area of the base

3.Volume of a pyramid = height

6. Cone

A cone is a solid which has a circle at its base and a slanting lateral surface that converges at the apex. Its dimensions are defined by the radius of the base (r), the height (h) and the slant height (l). A structure similar to a cone is used in ice cream cones.

1.Curved surface of a cone = prl where l is the slant height

2.Whole surface of a cone = prl + pr²

3.Volume of a cone =

7. Sphere

A sphere is a solid in the form of a ball with radius r.

1.Surface Area of a sphere = 4pr²

2.Volume of a sphere = pr³

8. Frustum of a pyramid

When a pyramid is cut the left over part is called the frustum of the pyramid.

1.Slant surface of the frustum of a pyramid = 1/2 * sum of perimeters of end * slant height.

2.Volume of the frustum of a pyramid = [E₁ + (E₁◊E₂)^1/2 + E₂] where k is the thickness and E₁, E₂ the areas of the ends.

9. Frustum of a cone

When a cone is cut the left over part is called the frustum of the cone.

1.Slant surface of the frustum of a cone = p(r₁ + r₂)l where l is the slant height.

2.Volume of the frustum of a cone = k(r₁² + r₁r₂ + r₂²)

WORKED-OUT PROBLEMS

Problem 11.1 A right triangle with hypotenuse 10 inches and other two sides of variable length is rotated about its longest side thus giving rise to a solid. Find the maximum possible area of such a solid.

Solution Most of the questions like this that are asked in the CAT will not have figures accompanying them. Drawing a figure takes time, so it is always better to strengthen our imagination. The beginners can start off by trying to imagine the figure first and trying to solve the problem. They can draw the figure only when they don’t arrive at the right answer and then find out where exactly they went wrong. The key is to spend as much time with the problem as possible trying to understand it fully and analysing the different aspects of the same without investing too much time on it.

Let’s now look into this problem. The key here lies in how quickly you are able to visualise the figure and are able to see that

(i)the triangle has to be an isosceles triangle,

(ii)the solid thus formed is actually a combination of two cones,

(iii)the radius of the base has to be the altitude of the triangle to the hypotenuse.

After you have visualised this comes the calculation aspect of the problem. This is one aspect where you can score over others.

In this question the figure would be somewhat like this (as shown alongside) with triangles ABC and ADC representing the cones and AC being the hypotenuse around which the triangle ABC revolves. Now that the area has to be maximum with AC as the hypotenuse, we must realise that ADB has to be an isosceles triangle, which automatically makes BCD an isosceles triangle too. The next step is to calculate the radius of the base, which is essentially the height of the triangle ABC. To find that, we have to first find AB. We know

AC² = AB² + BC²

For triangle to be one with the greatest possible area AB must be equal to BC that is, AB = BC = , since AO = 1/2AC = 5 inches.

Now take the right angle triangle ABO, BO being the altitude of triangle AOC. By Pythagoras theorem, AB² = AO ² + BO ², so BO ² = 25 inches

The next step is to find the area of the cone ABD and multiply it with two to get the area of the whole solid

Area of the cone ADB = p/3 (BO)² × AO

= (1/3) (25) × 5p = (125/3)p

Therefore, area solid ABCD = 2 × (125/3)p = (250/3)p

Problem 11.2 A right circular cylinder is to be made out of a metal sheet such that the sum of its height and radius does not exceed 9 cm can have an area of maximum.

Solution Solving this question requires the knowledge of ratio and proportion also. To solve this question, one must know that for a²b³c⁴ to have the maximum value when (a + b + c) is constant, a, b and c must be in the ratio 1 : 2 : 3.

Now lets look at this problem. Volume of a cylinder = pr²h.

If you analyse this formula closely, you will find that r and h are the only variable term. So for volume of the cylinder to be maximum, r²h has to be maximum under the condition that r + h = 9. By the information given above, this is possible only when r : h = 2 : 1, that is, r = 6, h = 3. So,

Volume of the cone = pr²h

= p × 6² × 3

= 108p

Problem 11.3 There are five concentric squares. If the area of the circle inside the smallest square is 77 square units and the distance between the corresponding corners of consecutive squares is 1.5 units, find the difference in the areas of the outermost and innermost squares.

Solution Here again the ability to visualise the diagram would be the key. Once you gain expertise in this aspect, you will be able to see that you will be able to see that the diameter of the circle is equal to the side of the innermost square that is

pr² = 77

or r = 3.5

or 2r = 7

Then the diagonal of the square is 14 sq units.

Which means the diagonal of the fifth square would be 14 + 12 units = 26.

Which means the side of the fifth square would be 13.

Therefore, the area of the fifth square = 338 sq units

Area of the first square = 98 sq units

Hence, the difference would be 240 sq. units.

Problem 11.4 A spherical pear of radius 4 cm is to be divided into eight equal parts by cutting it in halves along the same axis. Find the surface area of each of the final piece.

Fig. (a)

Solution The pear after being cut will have eight parts each of same volume and surface area. The figure will be somewhat like the above Figure (a) if seen from the top before cutting. After cutting it look something like the Figure (b).

Now the surface area of each piece = Area ACBD + 2 (Area CODB).

The darkened surface is nothing but the arc AB from side glance which means its surface area is one eighth the area of the sphere, that is, 1/8 × 4pr² = (1/2)pr².

Now CODB can be seen as a semicircle with radius 4 cm.

Fig. (b)

Therefore, 2 (Area CODB) = 2[(1/2)] p r² = p r²

fi surface area of each piece = (1/2) p r² + p r²

= (3/2) p r²

= 24p

Problem 11.5 A solid metal sphere is melted and smaller spheres of equal radii are formed. 10% of the volume of the sphere is lost in the process. The smaller spheres have a radius, that is 1/9th the larger sphere. If 10 litres of paint were needed to paint the larger sphere, how many litres are needed to paint all the smaller spheres?

Solution Questions like this require, along with your knowledge of formulae, your ability to form equations. Stepwise, it will be something like this

Step 1: Assume values.

Step 2: Find out volume left.

Step 3: Find out the number of small spheres possible.

Step 4: Find out the total surface area of each small spheres as a ratio of the original sphere.

Step 5: Multiply it by 10.

Step 1: Let radius of the larger sphere be R and that of smaller ones be r.

Then, volume = pR³ and pr³ = p(R/9)³ respectively for the larger and smaller spheres.

Step 2: Volume lost due to melting = pR³ × =

Volume left = pR³ × =

Step 3: Number of small spheres possible = Volume left/Volume of the smaller sphere

= = 9³ × 0.9

Step 4: Surface area of larger sphere = 4pR²

Surface area of smaller sphere = 4pr² = 4p (R/9)² =

Surface area of all smaller spheres = Number of small spheres × Surface area of smaller sphere

= (9³ × .9) × (4pR²)/81

= 8.1 × (4pR²)

Therefore, ratio of the surface area is = 8.1

Step 5:

8.1 × number of litres = 8.1 × 10 = 81

Problem 11.6 A solid wooden toy in the shape of a right circular cone is mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the wooden toy.

Solution Volume of the cone is given by –1/3 × pr²h

Here, r = 4.2 cm; h = 10.2 – r = 6 cm

Therefore the volume of the cone = 1/3 p × (4.2)² × 6 cm

= 110.88 cm³

Volume of the hemisphere = pr³ = 155.23

Total volume = 110.88 + 155.232 = 266.112

Problem 11.7 A vessel is in the form of an inverted cone. Its depth is 8 cm and the diameter of its top, which is open, is 10 cm. It is filled with water up to the brim. When bullets, each of which is a sphere of radius 0.5 cm, are dropped into the vessel 1/4 of the water flows out. Find the number of bullets dropped in the vessel.

Solution In these type of questions it is just your calculation skills that is being tested. You just need to take care that while trying to be fast you don’t end up making mistakes like taking the diameter to be the radius and so forth. The best way to avoid such mistakes is to proceed systematically. For example, in this problem we can proceed thus:

Volume of the cone = pr²h = p cm³

volme of all the lead shots = Volume of water that spilled out = p cm³

Volume of each lead shot = pr³ = cm³

Number of lead shots = (Volume of water that spilled out)/(Volume of each lead shot)

= = × 6 = 100

Problem 11.8 AB is a chord of a circle of radius 14 cm. The chord subtends a right angle at the centre of the circle. Find the area of the minor segment.

Solution Area of the sector ACBO =

= 154 sq cm

Area of the triangle AOB =

= 98 sq cm

Area of the segment ACB = Area sector ACBO – Area of the triangle AOB = 56 sq cm

Problem 11.9 A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. Find the diameter of the base of the cone.

Solution In questions like this, do not go for complete calculations. As far as possible, try to cancel out values in the resulting equations.

Volume of the sphere = pr³ = p(6.3)³

Volume of the cone = pr²h = r²(25.2)

Now, volume of the cone = volume of the sphere

Therefore, r (radius of the cone) = 39.69 cm

Hence the diameter = 79.38 cm

Problem 11.10 A chord AB of a circle of radius 5.25 cm makes an angle of 60° at the centre of the circle. Find the area of the major and minor segments.(Take p = 3.14)

Solution The moment you finish reading this question, it should occur to you that this has to be an equilateral triangle. Once you realise this, the question is reduced to just calculations.

Area of the minor sector = × p × 5.25²

= 14.4375 cm²

Area of the triangle = × 5.25² = 11.93 cm²

Area of the minor segment = Area of the minor sector – Area of the triangle = 2.5 cm²

Area of the major segment = Area of the circle – Area of the minor segment.

= 86.54 cm² – 2.5 cm² = 84 cm²

Problem 11.11 A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Solution Questions of this type should be solved without the use of pen and paper. A good authority over formulae will make things easier.

Volume of the cone = = Volume of a hemisphere = pr³.

Height of a hemisphere = Radius of its base

So the question is effectively asking us to find out h/r

By the formula above we can easily see that h/r = 2/1

GEOMETRY

LEVEL OF DIFFICULTY (I)

LEVEL OF DIFFICULTY (II)

MENSURATION

LEVEL OF DIFFICULTY (I)

LEVEL OF DIFFICULTY (II)

GEOMETRY

Level of Difficulty (I)

Solutions and Shortcuts

GEOMETRY

Level of Difficulty (I)

1.(a)

When the length of stick = 20 m, then length of shadow = 10 m i.e. in this case length = 2 × shadow with the same angle of inclination of the sun, the length of tower that casts a shadow of 50 m fi 2 × 50 m = 100 m

i.e. height of tower = 100 m

2.(d)

DABC ~ DEDC

Then = =

Then = = AC = 5.6 cm and

= = BC = 6.4 cm

3.(c)

For similar triangles fi (Ratio of sides)² = Ratio of areas

Then as per question = =

{Let the longest side of DDEF = x}

fi = = x = 27 cm

4.(a)

(Ratio of corresponding sides)² = Ratio of area of similar triangle

\ Ratio of corresponding sides in this question

= =

5.(a)

Ratio of corresponding sides = =

6.(c)

BC = ED = 6 m

So AB = AC – BC = 11 – 6 = 5 m

CD = BE = 12 m

Then by Pythagoras theorem:

AE ² = AB² + BE² fi AE = 13 m

7.(a)

In the DOBC; BC = 7 cm and OC = 9 cm, then using Pythagoras theorem.

OB² = OC² – BC²

OB = = 5.66 cm (approx)

8.(d)

In the DOBC, OB = 12 cm, OC = radius = 13 cm. Then using Pythagoras theorem;

BC² = OC² – OB² = 25; BC = 5 cm

Length of the chord = 2 × BC = 2 × 5 = 10 cm

9.(a)

–x = 35°; because angles subtended by an arc, anywhere on the circumference are equal.

10.(d)

–AOM = 2–ABM and

–AON = 2–ACN

because angle subtended by an arc at the centre of the circle is twice the angle subtended by it on the circumference on the same segment.

–AON = 60° and –AOM = 40°

–X = –AON + –AOM

(◊.◊ vertically opposite angles).

–X = 100°

11.(d)

DBCO and DAOD are similar (◊.◊ AC and BD one diameters). Thus

–ADO = –CBD (alternate opposite angles)

\–X = 30°

12.(a)

By the rule of tangents, we know:

6² = (5 + n)5 fi 36 = 25 + 5x fi 11 = 5x fi x = 2.2 cm

13.(b)

By the rule of tangents, we get

12² = (x + 7)x fi 144 = x² + 7x

fi x² + 7x – 144 = 0 fi x² + 16x – 9x – 144 = 0

fi x(x + 16) – 9(x + 16) fi x = 9 or –16

–16 can’t be the length, hence this value is discarded, thus, x = 9

14.(a)

By the rule of chords, cutting externally, we get

fi (9 + 6)6 = (5 + x)5 fi 90 = 25 + 5x fi 5x = 65 fi x = 13 cm

15.(b)

Inradius = area/semi perimeter

= 24/12 = 2 cm

16.(b)

–APB = 90° (angle in a semicircle = 90°)

–PBA = 180 – (90 + 25) = 65°

–TPA = –PBA(the angle that a chord makes with the tangent, is subtended by the chord on the circumference in the alternate segment).

= 65°

17.(a)

ADBC is a cyclic quadrilateral as all its four vertices are on the circumference of the circle. Also, the opposite angles of the cyclic quadrilateral are supplementary.

Therefore, –ADB = 180 – 48° = 132°

18.(a)

–POS = –QOR (vertically opposite angles)

So a = 4b

–SOT + –TOQ + –QOR = 180°

(sum of angles on a line = 180°)

4b + 2c = 180°

84 + 2c = 180°

fi2c = 96° fi c = 48°

So a = 84°, b = 21°, c = 48°

19.(b)

–ABC = 180° – 130° = 50°

(\ sum of angles on a line = 180°)

–ADC = 180° – –ABC = 130°

(◊.◊ opposite angles of a cyclic quadrilateral are supplementary).

–x = 180° – 130° = 50°

(\ sum of angles on a line = 180°)

20.(a)

–ADC = = 70° (because the angle subtended by an arc on the circumference is half of what it subtends at the centre). ABCD one cyclic quadrilateral

So –ABC = 180° – 70° = 110° (because opposite angles of a cyclic quadrilateral are supplementary).

21.(b)

OB = OA = radius of the circle

–AOB = 180 – (30 + 30)

{Sum of angles of a triangle = 180°)

fi 120°

Then –ADB = = 60°; because the angle subtended by a chord at the centre is twice of what it can subtend at the circumference. Again, ABCD is a cyclic quadrilateral;

So –ACB = 180° – 60° = 120° (because opposite angles of cyclic quadrilateral are supplementary).

22.(a)

–BAC = 30° (◊.◊ angles subtended by an arc anywhere on the circumference in the same segment are equal).

In DBAC; –x = 180° – (110° + 30°) = 40°

(◊.◊ sum of angles of a triangle = 180°)

23.(d)

As L₄ and L₃ are not parallel lines, so there can’t be any relation between 80° and x°.

Hence the answer cannot be determined.

24.(d)

Perimeter of the figure = 10 + 10 + 6 + 3p

= 26 + 3p

25.(b)CN : BR = 4 : 3

So the required answer is 3 : 5 : 2 : 7 : 4

Option (b) is correct.

Level of Difficulty (II)

1.(b)

DADE is similar to DABC (AAA property)

ED : BC = 3 : 5

Area of DADE : Area of DABC = 9 : 25

Area of trapezium = area of ABC – Area of ADE

= 25 – 9 = 16

Thus,

Area of DADE : Area of trapezium EDBC = 9 : 16

2.(a)

The area of a triangle formed by joining the mid-points of the sides of another triangle is always 1/4^th of the area of the bigger triangle.

So, the ratio is = 1 : 4

3.(a)

DDOC and DAOB are similar (by AAA property)

AB : DC = 3 : 1

So area of AOB : Area of DOC = (3 : 1)² fi 9 : 1

4.(a)

In DABC; BC = = 5

In DCDE; CE = = = 10.2

Width of street = BC + CE = 5 + 10.2 = 15.2 m

5.(c)

ABCD is a cyclic quadrilateral. Therefore

–DCB = 180––A = 180 – 60 = 120°

–ABC = 80°; therefore –BCQ = 180° – 120° = 60°

And –CBQ = 180° – 80° = 100°

(because, sum of angles on a line=180°)

Then in DBCQ; –Q = 180 – (100 + 60) = 20

(◊.◊ sum of angles of triangle = 180°)

6.(a)

–AOB = –CO¢D = –FO≤E = 120°

Distance between 2 centres = 2 m

\BC = DE = FA = 2 m

Perimeter of the figure = BC + DE + FA + circumference of sectors AOB, CO¢B and FO≤E.

But three equal sectors of 120° = 1 full circle of same radius.

Therefore, perimeter of surface

= 2pr + BC + DE + FA = (2p + 6)m

7.(a)

In DQRS; QR = RS, therefore –RQS = –RSQ (because angles opposite to equal sides are equal).

Thus –RQS + –RSQ = 180° – 100° = 80°

\ –RQS = –RSQ = 40°

–PQS = 180° – 40° = 140°

(sum of angles on a line = 180°)

Then again –QRS = –QSP

(◊.◊ angles opposite to equal sides are equal)

Thus –QPS + –QSP = 180° – 140° = 40°

And –QPS = –QSP = 20°

8.(b)

–BOC = 136°

–BAC = = 68° (because angle subtended by an arc anywhere on the circumference is half of the angle it subtends at the centre).

–BDC = 180° – 68° = 112° (\ ABCD is a cyclic quadrilateral and its opposite angles are supplementary)

9.(a)

DADO and DBOC are similar (AAA property)

Then =

fi3x – 9 = 3x² – 15x – 19x + 95

fi3x² – 37x + 104 = 0

On solving this quadratic equation, we get x = 8 or 9.

10.(b)

DAOD and DBOC are similar (AAA property)

= ; therefore = fi =

fiBC = 8 cm

11.(b)

In the given figure, DABD is similar to D ADC

Then = fi = fi DE = 2.5 cm

12.(a)

As AD bisects –CAB, so DABD is similar to D ADC

Then = fi = fi AC = 4 cm

13.(a)

Area of the quarter circle = fi 0.25p. Going by options, we have to see that the area of the inserted circle is less than the area of the quarter circle.

Option (b) fi Area = (1.5 + )p

2.9p > 0.25p. Hence discarded.

Option (c) – fi Area = p fi 0.75p > 0.25p. Hence discarded.

Option (d) 1 – fi Area = p (1 + 8 – fi 0.85p > 0.25p. Hence discarded.

Option (a) – 1 = Area p(2 + 1 – ) fi 0.20p

< 0.25p

Hence this option is correct.

14.(c)

Sum of length of all perpendiculars drawn on the sides of any equilateral triangle is constant. Perpendicular (D) = Perpendicular (E)

15.(b)

In the above question:

FE = AB = 6 cm

DADF @ DBEC; so DF = EC

Let DF = EC = x

Solving through options; e.g. option (b) 1/3; x = 6

Then by Pythagoras triplet AF = 8

Area of ABEF = 8 × 6 = 48 cm²

Area of DAFD + DBEC = 2 × × 6 × 8 fi 48 cm²

\ Area of ABCD = 48 + 48 = 96 cm². Hence the condition is proved.

16.(a)

Find the indivisible area of DABC and DACD

Add them then;

AB × BC + AC × CD fi xy + z

fi replace x, y, z by the lowest integral values.

Here as we are talking about right angles, so we have to take the smallest Pythagorean triplet, which in this case would be = 3, 4, 5

Answer fi 36 cm²

17.(d)

As the point ‘O’ is formed by the ^ bisects to the three sides of the D, so point ‘O’ is the circumcenter. This means that virtually, points A, B and C are on the circumference of the circle.

Thus m–BOC = 2m–BAC (\angle subtended by an arc at the centre of the circle is twice the angle subtended at the circumference).

18.(c)

OC = O¢D = 5 cm (radius)

CD = 24 cm

DCOE and DEO¢D are similar therefore OE = O¢E and CE = ED = 12 cm

In DCOD; OE² = CE² + OC²

= 12² + 5² = 169

OE = 13

\OO¢ = OE + EO¢ = 13 + 13 = 26 cm

19.(c)

It will always be possible to divide a circle into 360 equal parts, because the sum of angles that can be subtended at the centre of the circle = 360°

20.(d)

Area of the shaded portion = Area of circle – Area of triangle

fi Area of circle = pr² fi × 5 × 5 fi cm²

= 78.50 cm²

Area of triangle fi r ²sin q fi × 25 × sin q

fi fi 6.25 × 1.732 fi 10.8

\ Area of shaded portion = 78.50 – 10.8 = 67.7 cm²

21.(a)

OB = OA – radius of circle

fi–CAO = –OBA

(angles in alternate segments are equal).

Now, if –CAO = –OBA

\ –CAO = –OAB

\ option (a) is correct.

22.(b)

23.(a)

Angle XPA = angle ABP = x

Angle CPX = angle CDP = x + y

Anlge CDP is exterior angle of triangle PDB

So angle CDP = DBP + DPB

X + y = x + DPB

DPB = y

So angle CPA = DPB

24.(b)

DAPD ~ DCPB

\ =

i.e. PA ◊ PC = PB ◊ PD.

\ option (b)

25.(d)

AC ◊ AD = AB²

AC ◊ AD = BC²

DABC N DADB

\ =

(Corresponding sides of similar triangle

are proportional)

AC ◊ AD = AB²(i)

Also, DABC N DBDC fi =

AC ◊ CD = BC²(ii)

Both conditions of option (d) are found, therefore (d) is the answer.

Mensuration

Level of Difficulty (I)

1.(b)

Let hypotenuse = x cm

Then, by Pythagoras theorem:

x² = (48020)² + (36015)²

x fi 60025 cm

2.(d)

Let one side of the D be = a

Perimeter of equilateral triangle = 3a

\3a = 72 = a = 24 cm

Height = AC; by Pythagoras theorem

AC² = a² –

AC = 36 cm

3.(b)

Let inner radius = A; then 2pr = 440 \ p = 70

Radius of outer circle = 70 + 14 = 84 cm

\ Outer diameter = 2 × Radius = 2 × 84 = 168

4.(a)

Let inner radius = r and outer radius = R

Width = R – r = –

fi(R – r) = = 7 meters

5.(d)

Let outer radius = R; then inner radius = r = R – 7

2pR = 220 fi 35m;

r = 35 – 7 = 28 m

Area of torch = pR² – pr² fi p(R² – r²) = 1386 m²

Cost of traveling it = 1386 × = ` 693

6.(b)

Circumference of circle = 2pr = 44

= r = 7 cm

Area of a quadrant = = 38.5 cm²

7.(d)

Volume of soil removed = l × b × h

= 7.5 × 6 × 1.5 = 67.5 m³

8.(c)

The longest pole can be placed diagonally (3-dimensional)

BC = = 30

AC = = 34 m

9.(d)

Let the common ratio be = x

Then; length = 3x, breadth = 2x and height = x

Then; as per question 3x ◊ 2x ◊ x = 1296 fi 6x³ = 1296

fi x = 6 m

Breadth = 2x = 12 m

10.(d)

Data is inadequate as it’s not mentioned that what part of the cube is melted to form cylinder.

11.(b)

Let the common ratio be = x

Then, length = 4x, breadth = 3x and height = 2x

As per question;

2(4x ◊ 3x + 3x ◊ 2x + 2x ◊ 4x) = 8788

2(12x² + 6x² + 8x²) = 8788 fi 52x² = 8788

fi x = 13

Length = 4x = 52 cm

12.(b)

The total volume will remain the same, let the side of the resulting cube be = a. Then,

6³ + 8³ + 10³ = a³ fi a = = 12 cm

13.(a)

Slant length = l = = 10 cm

Then curved surface area = prl = p × 6 × 10 fi 60p

And total surface area = prl + pr² fi p((6 × 10) + 6²) = 96p

14.(b)

Volume of a cone =

Then;100p = fi r = 5 cm

Curved surface area = prl

l = fi = 13

then,prl = p × 13 × 5 = 65p cm²

15.(d)

Let the radius of the two cones be = x cm

Let slant height of 1^st cone = 5 cm and

Slant height of 2^nd cone = 7 cm

Then ratio of covered surface area = = 5 : 7

16.(c)

Radius = = = 42 cm

Diameter = 2 × Radius = 2 × 42 = 84 cm

17.(c)

Let the radius of cylinder = 1(r)

Then the radius of cone be = 2(R)

Then as per question = fi

fi fi 3 : 4

18.(c)

The perimeter would remain the same in any case.

Let one side of a square be = a cm

Then a² = 484 fi a = 22 cm \ perimeter = 4a = 88 cm

Let the radius of the circle be = r cm

Then 2pr = 88 fi r = 14 cm

Then area = pr² = 616 cm²

19.(d)

Let the radius of the circle be = p

Then 2pr – 2r = 16.8 fi r = 3.92 cm

Then 2pr = 24.6 cm

20.(d)

Let the radius of the wheel be = p

Then 5000 × 2pr = 1100000 cm fi r = 35 cm

21.(a)

Let the slant height be = l

Let radius = r

Then v = fi r = fi = 5 cm

l = = = 13 cm

22.(b)

In 4 days, the short hand covers its circumference

4 × 2 = 8 times long hand covers its circumference

4 × 24 = 96 times

Then they will cover a total distance of:-

(2 × p × 4)8 + (2 × p × 6)96 fi 3818.24 cm

23.(b)

Let the radius of the smaller sphere = r

Then, the radius of the bigger sphere = R

Let the surface area of the smaller sphere = 1

Then, the surface area of the bigger sphere = 4

Then, as per question

fi = fi = fi R = 2r

Ratio of their volumes

= × fi 1: 8

24.(b)

Inner radius(p) = = 4.5 cm

Outer radius (R) = = 5 cm

Volume of metal contained in the shell =

fi (R³ – r³)

fi 141.9 cm³

25.(c)

Let smaller radius (r) = 1

Then bigger radius (R) = 2

Then, as per question

fi = fi = 1 : 4

26.(c)

As per question fi = fi h = 4r

27.(a)

Volume of wall = 1200 × 500 × 25 = 15000000 cm³

Volume of cement = 5% of 15000000 = 750000 cm³

Remaining volume = 15000000 – 750000

= 14250000 cm³

Volume of a brick = 25 × 12.5 × 7.5 = 2343.75 cm³

Number of bricks used = = 6080

28.(a)

Let the inner radius = r

Then 2pr = 352 m. Then r = 56

Then outer radius = r + 7 = 63 = R

Now,pR² – pr² = Area of road

fip(R² – r²) = 2618 m²

29.(c)

1 hectare = 10000 m²

Height = 10 cm = m

Volume = 10000 × = 1000 m³

30.(a)

Total surface area of 7 cubes fi 7 × 6a² = 1050

But on joining end to end, 12 sides will be covered.

So there area = 12 × a² fi 12 × 25 = 300

So the surface area of the resulting figure = 1050 – 300 = 750

31.(d)

Let the rise in height be = h

Then, as per the question, the volume of water should be equal in both the cases.

Now, 90 × 40 × h = 150 × 8

h = = m = cm

= 33.33 cm

32.(d)

Slant height (l) = = 25 m

Area of cloth required = covered surface area of cone = prl = × 7 × 25 = 550 m²

Amount of cloth required = = 110 m

33.(b)

If the ratio of their diameters = 2 : 1, then the ratio of their radii will also be = 2 : 1

Let the radii of the broader cone = 2 and height be = 1

Then the radii of the smaller cone = 1 and height be = 2

Ratio of volumes = ∏

× fi 2 : 1

34.(d)

Area of base = 6 × 10 = 60 m²

Volume of tent = 30 × 10 = 300 m³

Let the radius be = r, height = h, slant height = l

pr² = 60 fi r =

300 = fi 900 = p ◊ ◊ h fi h = 15 m

35.(b)

Volume of wood used = External volume – Outer Volume

fi (10 × 8 × 6) – (10 – 1) × (8 – 1) × (6 – 1)

fi 480 – (9 × 7 × 5) = 165 cm²

36.(b)

Total volume in both the cones will be equal. Let the number of smaller cubes = x

x ◊ 3³ = 24 × 9 × 8 fi x = = 64

37.(a)

Let one side of the cube = a

Then a³ = 216 fi a = 6 m

Area of the resultant figure

= Area of all 3 cubes – Area of covered figure

fi 216 × 3 – (4 × a²) fi 648 – 144 fi 504 m²

38.(c)

Volume of metal used = –

= (12³ – 10³)

= 3047.89 cm³

Weight = volume × densityfi 4.9 × 3047.89

fi 14942.28 gm

39.(d)

Volume of cube = 7³ = 343 cm³

Radius of cone = = 3.5 cm

Height of cone = 7

Ratio of volumes = =

fi 11:42

40.(b)

The volume in both the cases will be equal. Let the height of cone be = h

4 × × (14)³ × = × ×

fi4(14)³ = h = h

=

= h = 35.84 cm

41.(b)

Diameter of circle = diagonal of square

= = = 10

◊.◊ Radius = =

Area of circle = pr² fi 50p = 50 × 3.14 = 157.14 cm³

42.(c)

Area of triangle = rS; where r = inradius

S = = 20 cm

D =

fi D

D = = 60 cm²

r = = = 3 cm

43.(c)

Circumference of the circular face of the cylinder = 2pr

fi 2 × × = 2.2 m

Number of revolutions required to lift the bucket by 11 m = = 5

44.(c) Surface area of the cube = 6a² = 6 × (20)²

= 2400

Area of 6 circles of radius 10 cm = 6pr²

= 6 × p × 100

= 1885.71

Remaining area = 2400 – 1884 = 514.28

45.(d)

x ◊ y ◊ z = lb × bh × lh = (lbh)²

(V) Volume of a cuboid = lbh

So V² = (lbh)² = xyz

46.(c)

Diameter of the circle = diagonal of rectangle

= = 10 cm

Radius = = 5 cm

Area of shaded portion = pr² – lb

= 3.14 × 5² – 8 × 6

= 30.57 cm²

47.(a)

Larger Radius (R) = 14 + 7 = 21 cm

Smaller Radius (r) = 7 cm

Area of shaded portion pR² –

fi fi 102.67 cm

48.(b)

Area of quadrilateral = Area of right angled triangle + Area of equilateral triangle x = = 16

Area of quadrilateral = + × 20 × 20

= 269 units²

49.(a)

Height = =

Volume = fi

fi

50.(b)

The perimeter would remain the same in both cases. Circumference of circle = 2pr = 2 × × 28 = 176 cm

Perimeter of square = 176

Greatest side possible = = 44 cm

Length of diagonal = = 62.216

= ◊ = 44

Level of Difficulty (II)

1.(d)

Let the angle subtended by the sector at the centre be = q

Then,

5.7 + 5.7 + (2p) × 5.7 × = 27.2

11.4 + = 27.2

fi = 0.44

Area of the sector = pr² fi (22/7) × (5.7)² × 0.44

= 44.92 approx.

2.(a)

Volume of mud dug out = 10 × 4.5 × 3 = 135 m³

Let the remaining ground rise by = h m

Then{(20 × 9) – (10 × 4.5)}h = 135

135h = 135 fi h = 1 m

3.(b)

Height of the cylinder = 13 – 7 = 6 cm

Radius of the cylinder and the hemisphere = 7 cm

Volume of the vessel = volume of cylinder + volume of hemisphere

fi pr²h + fi 3.14 × (7)² × 6 +

fi 1642.6 cm²

4.(b)

Let the original triangle be = ACD

Longest side = AC = 21cm

In the right angled DABD, by Pythagorean triplets, we get AB = 5 and BD = 12

Then, BC = 21 – 5 = 16

By Pythagoras theorem,

BD² = CD² – BC² fi BD = 12 cm

Thus, on assumption is correct.

Area of the larger DBDC = × 16 × 12 fi 96 cm²

5.(c)

Radius = = 52.5 cm

Area of the entire canvas, used for the tent

= Area of cylinder + centre of cone

= 2prh + prl