Geometry - BankerswayAn Altitude of a triangle is a straight line through a vertex and perpendicular to the opposite side or an extension of the opposite side. The three altitudes

Geometry

Geometry is one of the most important topics of Quantitative Aptitude section.

Lines and Angles

Sum of the angles in a straight line is 180°

Vertically opposite angles are always equal.

If any point is equidistant from the endpoints of a segment, then it must lie on the

perpendicular bisector.

When two parallel lines are intersected by a transversal, corresponding angles are

equal, alternate angles are equal and co-interior angles are supplementary. (All

acute angles formed are equal to each other and all obtuse angles are equal to each

other)

Fact

The ratio of intercepts formed by a transversal intersecting three parallel lines is

equal to the ratio of corresponding intercepts formed by any other transversal.

=

=

Triangles

Sum of interior angles of a triangle is 180° and sum

of exterior angles is 360°.

Exterior Angle = Sum of remote interior angles.

Sum of two sides is always greater than the third side and the difference of two

sides is always lesser than the third side.

Side opposite to the biggest angle is longest and the side opposite to the

smallest angle is the shortest.

= ½ x Base x Height

= ½ x Product of sides x Sine of included angle

= √ ( − )( − )( − ) here s is the semi perimeter

[s = (a+b+c)/2 ] = r x s [r is radius of incircle]

= 4

[R is radius of circumcircle]

Median

A Median of a triangle is a line segment joining a vertex to the midpoint of the

opposing side. The three medians intersect in a single point, called the Centroid of

the triangle. Centroid divides the median in the ratio of 2:1

Altitude

An Altitude of a triangle is a straight line through a vertex and perpendicular to the

opposite side or an extension of the opposite side. The three altitudes intersect in a

single point, alled the Orthocenter of the triangle.

Perpendicular Bisector

A Perpendicular Bisector is a line that forms a right angle with one of the triangle's

sides and intersects that side at its midpoint. The three perpendicular bisectors

intersect in a single point, called the Circumcenter of the triangle. It is the center of

the circumcircle which passes through all the vertices of the triangle.

Angle Bisector

An Angle Bisector is a line that divides the angle at one of the vertices in two equal

parts. The three angle bisectors intersect in a single point, called the Incenter of the

triangle. It is the center of the incircle which touches all sides of a triangle.

Theorems Mid Point Theorem: The line joining the midpoint of any two sides is parallel to the third

side and is half the length of the third side.

Apollonius’ Theorem: 2 + 2 = 2( 2 + 2)

Basic Proportionality Theorem: If DE || BC, then AD/DB = AE/EC

Interior Angle Bisector Theorem: AE/ED = BA/BD

Special Triangles Right Angled Triangle:

ABC ≈ ADB ≈ BDC

2 = AD x DC and AB x BC = BD X DC

Equilateral Triangle:

All angles are equal to 60°. All sides are equal also.

Isosceles Triangle: Angles equal to opposite sides are equal.

Area =

4

√4 2 − 2

30°-60°-90° Triangle

Area = √3

2

× 2

45°-45°-90° Triangle

Area = 2

2

30°-30°-120° Triangle

Area = √3

4

× 2

Similarity of Triangles

Two triangles are similar if their corresponding angles are congruent and corresponding

sides are in proportion.

Tests of similarity: (AA / SSS / SAS)

For similar triangles, if the sides are in the ratio of a:b

Corresponding heights are in the ratio of a:b

Corresponding medians are in the ratio of a:b

Circumradii are in the ratio of a:b

Inradii are in the ratio of a:b

Perimeters are in the ratio of a:b

Areas are in the ratio a2 : b2

Congruency of Triangles Two triangles are congruent if their corresponding sides and angles are congruent.

Tests of congruence: (SSS / SAS / AAS / ASA)

All ratios mentioned in similar triangle are now 1:1

Polygons

Sum of interior angles = (n - 2) x 180° = (2n - 4) x 90°

Sum of exterior angles = 360°

Number of diagonals = nC2 – n = ( − 3)

2

Number of triangles which can be formed by the vertices = nC3

Regular Polygon : If all sides and all angles are equal, it is a regular polygon.

All regular polygons can be inscribed in or circumscribed about a circle.

Area = ½ × Perimeter × Inradius {Inradius is the perpendicular from centre to any

side}

Each Interior Angle = ( − 2)180°

; Exterior = 360°

Quadrilaterals :

Sum of the interior angles = Sum of the exterior angles = 360°

Area for a quadrilateral is given by 1

2

d1 d2 Sin

Cyclic Quadrilateral

If all vertices of a quadrilateral lie on the circumference of a circle, it is known as a

cyclic quadrilateral.

Opposite angles are supplementary

Area = √( − )( − )( − )( − ) where s is the semi perimeter s = + + +

2

Parallelogram

Opposite sides are parallel and congruent.

Opposite angles are congruent and consecutive angles are supplementary.

Diagonals of a parallelogram bisect each other.

Perimeter = 2(Sum of adjacent sides);

Area = Base x Height = AD x BE

Facts

Each diagonal divides a parallelogram in two triangles of equal area.

Sum of squares of diagonals = Sum of squares of four sides

o 2 + 2 = 2 + 2 + 2 + 2

A Rectangle is formed by intersection of the four angle bisectors of a parallelogram.

Rhombus

A parallelogram with all sides equal is a Rhombus. Its diagonals bisect at 90°.

Perimeter = 4a; Area = 1

2

d1 d2

Area = d x √ 2 − (

)2 2

Rectangle

A parallelogram with all angles equal (90°) is a Rectangle. Its diagonals are congruent.

Perimeter = 2(l+b)

Area = lb

Square

A parallelogram with sides equal and all angles equal is a square. Its diagonals are

congruent and bisect at 90°.

Perimeter = 4a

Area = 2

Diagonals = a √2

Fact: From all quadrilaterals with a given area, the square has the least perimeter. For all

quadrilaterals with a given perimeter, the square has the greatest area.

Kite

Two pairs of adjacent sides are congruent.

The longer diagonal bisects the shorter diagonal at 90°.

Area =

2

Trapezium / Trapezoid

A quadrilateral with exactly one pair of sides parallel is known as a Trapezoid. The

parallel sides are known as bases and the non-parallel sides are known as lateral

sides.

Area = 1

2 × (Sum of parallel sides) × Height

Median, the line joining the midpoints of lateral sides, is half the sum of parallel

sides.

Fact

Sum of the squares of the length of the diagonals = Sum of squares of lateral sides +

2 Product of bases.

2 + 2 = 2 + 2 + 2 × AB × CD

Isosceles Trapezium

The non-parallel sides (lateral sides) are equal in length. Angles made by each parallel side

with the lateral sides are equal.

Facts: If a trapezium is inscribed in a circle, it has to be an isosceles trapezium. If a circle

can be inscribed in a trapezium, Sum of parallel sides = Sum of lateral sides.

Hexagon (Regular)

Perimeter = 6a; Area = 3 √3

× 2 2

Sum of Interior angles = 720°.

Each Interior Angle = 120°. Exterior = 60°

Number of diagonals = 9 {3 big and 6 small}

Length of big diagonals (3) = 2a

Length of small diagonals (6) = √3 a

Area of a Pentagon = 1.72 2

Area of an Octagon = 2(√2 + 1) 2

Facts: A regular hexagon can be considered as a combination of six equilateral triangles.

All regular polygons can be considered as a combination of ‘n’ isosceles triangles.

Circles

Diameter = 2r; Circumference = 2πr; Area = π 2

Chords equidistant from the centre of a circle are equal. A line from the centre,

perpendicular to a chord, bisects the chord. Equal chords subtend equal angles at the centre.

The diameter is the longest chord of a circle. A chord /arc subtends equal angle at any point

on the circumference and double of that at the centre.

Chords / Arcs of equal lengths subtend equal angles.

[

Chord AB divides the circle into two parts: Minor Arc AXB and Major Arc AYB

Measure of arc AXB = ∠AOB =

Length (arc AXB) =

360°

× 2πr

Area (sector OAXB) =

360°

× π 2

Area of Minor Segment = Shaded Area in above figure

Area of Sector OAXB - Area of ∆OAB

2

360° -

2

Properties of Tangents, Secants and Chords

The radius and tangent are perpendicular to each other. There can only be two tangents from

an external point, which are equal in length PA = PB

]

2

2

2

PA × PB = PC × PD

= 1 [m(Arc AC) – m(Arc BD)]

PA × PB = PC × PD

= 1 [m(Arc AC) + m(Arc BD)]

Properties

PA × PB = 2

= 1 [m(Arc AC) - m(Arc BC)]

Alternate Segment Theorem

The angle made by the chord AB with the tangent at A (PQ) is equal to the angle that it

subtends on the opposite side of the circumference.

∠BAQ = ∠ACB

Common Tangents

Two Circles No. of Common Tangents Distance Between Centers

(d)

One is completely inside

other

0 < r1 - r2

Touch internally 1 = r1 - r2 Intersect 2 r1 - r2 < d < r1 + r2

Touch externally 3 = r1 + r2

One is completely outside

other

4 > r1 + r2

Length of the Direct Common Tangent (DCT)

AD = BC = √ 2 − ( 1 − 2)2

Length of the Transverse Common Tangent (TCT)

RT = SU = √ 2 − ( 1 + 2)2

Example with Solution

Example 1: In following figure, CE is perpendicular to AB, ∠ACE = 20° and ∠ ABD =

50°. Find ∠ BDA:

Solution: To find: angle BDA

For this what we need, ∠ BAD

Because, Sum of all angles = 180°

Consider, ∆ECA,

∠ CEA + ∠ EAC + ∠ ACE = 180°

i.e. 90° + 20° + ∠ EAC = 180°

Therefore, ∠ EAC = 70° Now, come to ∆ABD,

∠ ABD + ∠ BDA + ∠ BAD = 180°

70° + 50° + ∠ BAD = 180°

Therefore, ∠ BAD = 60°

Example 2: In given figure. BC is produced to D and ∠ BAC = 40° and ∠ ABC =70°.Find

∠ ACD:

Solution: In above figure, ∠ ACD is an exterior angle, and according to property,

Exterior angle = Sum of interior angles

Therefore, ∠ ACD = 70° + 40°

=110°

This is not the end of this chapter. These are just the basics. In next session, I will

discuss some important results, properties (congruency, similarity) and much more.

Always remember, Geometry needs practice and time.

Exercise

1) If (5, 1), (x, 7) and (3, -1) are 3 consecutive verticles of a square then x is

equal to :

a) - 3 b) - 4 c) 5

c) 6 e) None of these

2) What is the area of an obtuse angled triangle whose two sides are 8 and 12

and the angle included between two sides is 150°?

a) 24 sq units b) 48 sq units c) 24 √3

d) 48 √3 e) Such a triangle does not exist

3) What is the measure of the radius of the circle that circumscribes a triangles

whose sides measure 9, 40 and 41?

a) 6 b) 4 c) 24.5

d) 20.5 e) 12.5

4) Verticles of a quadrilateral ABCD are A (0, 0), B (4, 5), C (9, 9) and D (5, 4). What

is the shape of the quadrilateral?

a) Square b) Rectangle but not a square

c) Rhombus d) Parallelogram but not a rhombus e) None of these

5) If the sum of the interior angles of a regular polygon measures upto 1440 degrees,

how many sides does the polygon have?

a) 10 sides b) 8 sides c) 12 sides

d) 9 sides e) None of these

6) What is the radius of the in circle of the triangle whose sides measure 5, 12 and 13

units?

7)

8) If 10, 12 and ‘x’ are sides of an acute angled triangle, how many integer values of

‘x’ are possible?

a) 7 b) 12 c) 9 d) 13 e) 11

9) Find the length of the hypotenuse of a right triangle if the lengths of the other two

sides are 6 inches and 8 inches.

a) 10 inches b) 11 inches c) 18 inches

d) 20 inches e) None of these

a) 2units

d) 6 units

b) 12 units

e) 7.5 units

c) 6.5 units

How many diagonals does a 63 sided convex polygon have? a) 3780 b) 1890 c) 3843

d) 3906 e) 1953

10) Find the length of one side of a right triangle if the length of the hypotenuse is 15

inches and the length of the other side is 12 inches.

a) 8 inches b) 7 inches c) 9 inches

d) 13 inches e) None of these

11) Find the length of the hypotenuse of a right triangle if the lengths of the other two

sides are both 3 inches.

a) 5 inches b) 3√4 inches c) 6 inches

d) 3√2 inches e) None of these

12) Find the lengths of the other two sides of a right triangle if the length of the

hypotenuse is 4√2 inches and one of the angles is 45°. a) 4 inches b) 9 inches c) 8 inches d) 7 inches e) None of these

13) Find the length of the hypotenuse of a right triangle if the lengths of the other two

sides are 4 inches and 4√3 inches. a) 8 inches b) 9 inches c) 10 inches d) 11 inches e) None of these

14) Find the lengths of the other two sides of a right triangle if the length of the

hypotenuse is 8 inches and one of the angles is 30°.

a) 4, 4√3 inches b) 5, 6 inches c) 2, 4√2

d) 3, 4√2 inches e) None of these

15) What is the area of the following square, if the length of BD is 2√2 ? a) 1 b) 2 c) 3 d) 4 e) 5

16) In the figure below, what is the value of y?

a) 40 b) 50 c) 60

d) 100 e) 120

17) Two circles both of radii 6 have exactly one point in common. If A is a point on one

circle and B is a point on the other circle, what is the maximum possible length for

the line segment AB?

a) 12 b) 15 c) 18

d) 20 e) 24

18) A right circular cylinder has a radius of 3 and a height of 5. Which of the following

dimensions of a rectangular solid will have a volume closest to the cylinder?

a) 4, 5, 5 b) 4, 5, 6 c) 5, 5, 5 d) 5, 5, 6 e) 5, 6, 6

19) Note: Figures not drawn to scale

In the figures above, x = 60, How much more is the perimeter of triangle ABC

compared with the triangle DEF.

a) 0 b) 2 c) 4 d) 6 e) 8

20) A right triangle has one other angle that is 35°. What is the size of the third angle?

a) 55° b) 65° c) 90°

e) 180° e) None of these

21) An equilateral triangle has one side that measures 5 in. What is the size of the angle

opposite that side?

a) 55° b) 70° c) 110°

d) 60° e) None of these

22) An isosceles triangle has one angle of 96°. What are the sizes of the other two

angles?

23)

24)

25)

26) The parallelogram shown in the figure below has a perimeter of 44 cm and an area

of 64 2. Find angle T in degrees?

a) 43.4° b) 44.2° c) 34.8°

d) 48.1° e) None of these

27) Find the area of the quadrilateral shown in the figure. (note: Figure not drawn to

scale)

a) 169 b) 185 c) 199

d) 144 e) None of these

a) 24° b) 34° c) 42°

d) 96° e) None of these

Find the circumference of the circle with a diameter of 8 inches? a) 25 inches b) 25.163 inches c) 29.45 inches

d) 35.62 inches e) None of these

Find the area of the circle with a diameter of 10 inches? a) 55.78 sq. inches b) 99.75 sq. inches c) 92 inches

d) 78.55 sq. inches e) None of these

Find the area of the circle with a radius of 10 inches? a) 314.2 sq. inches b) 115 inches c) 320.29 sq. inches

d) 56.12 sq. inches e) None of these

28) In the figure below triangle OAB has an area of 72 and triangle ODC has an area of

288. Find x and y.

a) x = 20, y = 14 b) x = 14, y = 20 c) x = 41, y = 2

d) x = 24, y = 1 e) None of these

29) Find the circumference of a circular disk whose area is 100π square centimeters?

a) 40 π b) 10 π c) 20π

d) 30π e) None of these

30) The semicircle of area 1250 π centimeters is inscribed inside a rectangle. The

diameter of the semicircle coincides with the length of the rectangle. Find the area

of the rectangle?

a) 4000 b) 5000 c) 3000 d) 2000 e) None of these

31) If in a triangle ABC cos

=

cos =

cos

, then what can be said about the triangle?

a) Right angled triangle b) Isosceles triangle c) Equilateral triangle d) Obtuse triangle e) None of these

Solutions:

1. Option A

For the verticles to form a square, we know that the length of each side of the

square should be equal. Therefore,

( − 5)2 + (7 − 1)2 = ( − 3)2 + (7 + 1)2

[ 2 + 52 - 2 (x) (5) ] + [36] = [ 2 + 32 - 2 (x) (3) ] + [64] [52 + 36] - [9 + 64] = (10 - 6) x

x = - 12

4

= - 3

This gives the side of the square x = - 3.

2. Option A

If two sides of a triangle and the included angle ‘y’ is known, then the area of the

triangle can be found using the formula 1

* (product of sides) * sin y 2

Substituting the values in the formula, we get 1

2

* 8 * 12 * sin 150 = 24 sq units

3. Option D

From the measure of the length of the sides of the triangle 9, 40 and 41 we can infer

that the triangle is a right angled triangle. 9, 40, 41 is a Pythagorean triplet.

In a right angled triangle, the radius of the circle that circumscribes the triangle is

half the hypotenuse. In the given triangle, the hypotenuse = 41

Therefore, the radius of the circle that circumscribes the triangle = 41

2

= 20.5 units

4. Option C

The lengths of the four sides AB, BC, CD and DA are all equal to √41 Hence, the given quadrilateral is either a Rhombus or a Square.

Now let us compute the lengths of the two diagonals AC and BD.

The length of AC is √162 and the length of BD is √2 As the diagonals are not equal and the sides are equal, the given quadrilateral is a

Rhombus.

5. Option A

We know that the sum of an exterior angle and an interior angle of a polygon =

180°

We also know that sum of all the exterior angles of a polygon = 360° The question states that the sum of all interior angles of the given polygon = 1440°

Therefore, sum of all the interior and exterior angles of the polygon = 1440 + 360 =

1800

If there are ‘n’ sides to this polygon, then the sum of all the exterior and interior

angles = 180 × n = 10

6. Option A

The triangle given is a right angled triangle as its sides are 5, 12 and 13 which is one

of the Pythagorean triplets.

Note: In a right angled triangle, the radius of the incircle is given by the following

relation − h

2

As the given triangle is a right angled triangle, radius of its incircle = 5 + 12 − 13

= 2 1

units

7. Option B

The number of diagonals of an n-sided convex is ( − 3)

2

This polygon has 63 sides. Hence, n = 63

Therefore, number of diagonals = 63 × 60

= 1890 2

8. Option C

For any triangle sum of any two sides must be greater than the third side.

The sides are 10, 12 and ‘x’.

From Rule 2, x can take the following values : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,

15, 16, 17, 18, 19, 20, 21 – A total of 19 values.

When x = 3 or x = 4 or x = 5 or x = 6, the triangle is an OBTUSE angled triangle.

The smallest value of x that satisfies both conditions is 7. (102 + 72 > 122)

The highest value of x that satisfies both conditions is 15. (102 + 122 + 152) When x = 16 or x = 17 or x = 18 or x = 19 or x = 20 or x = 21, the triangle is an

OBTUSE angled triangle.

Hence, the values of x that satisfy both the rules are x = 7, 8, 9, 10, 11, 12, 13, 14,

15. A total of 9 values.

9. Option A

Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.

6 : 8 : ?= 3 (2) : 4 (2) : ?

Yes, it is a 3-4-5 triangle for n =

Calculate the third side 5n = 5 × 2 = 10

The length of the hypotenuse is 10 inches.

10. Option C

Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.

? : 12 : 15 = ? : 4 (3) : 5 (3) Yes, it is a 3-4-5 triangle for n = 3

Calculate the third side 3n = 3 × 3 = 9

The length of the side is 9 inches.

11. Option D

This is a right triangle with two equal sides so it must be a 45° - 45° - 90° triangle.

You are given that the both the sides are 3. If the first and second value of the ratio

n : n : n√2 is 3 then the length of the third side is 3√2

The length of the hypotenuse is 3√2 inches.

12. Option A

This is a right triangle with a 45° so it must be a 45° - 45° - 90° triangle.

You are given that the hypotenuse is 4√2 . If the third value of the ratio n : n : n√2

is 4√2 then the lengths of the other two sides must 4. The lengths of the two sides are both 4 inches.

13. Option A

Test the ratio of the lengths to see if it fits the n : n√3 : 2n ratio.

4 : 4√3 : ? n : n√3 : 2n Yes, it is a 30° - 60° - 90° triangle for n = 4 Calculate the third side

2n = 2 × 4 = 8

The length of the hypotenuse is 8 inches.

14. Option A

This is a right triangle with a 30° angle so it must be a 30° - 60° - 90° triangle.

You are given that the hypotenuse is 8. Substituting 8 into the third value of the

ratio n: n√3 : 2n, we get that 2n = 8 n = 4

Substituting n = 4 into the first and second value of the ratio we get that the other

two sides are 4 and 4√3

The lengths of the two sides are 4 inches and 4√3 inches.

15. Option D

We need to find the length of the side of the square in order to get the area.

The diagonal BD makes two 45° - 45° - 90° triangles with the sides of the square.

Using the 45° - 45° - 90° special triangle ratio n: n : n√2. If the hypotenuse is 2√2 then the legs must be 2. So, the length of the side of the square is 2.

Area of square = 52 = 22 = 4

16. Option C

Vertical angles being equal allows us to fill in two angles in the triangle that y° belongs to.

Sum of angles in a triangle = 180°

So, y° + 40° + 80° = 180°

y° + 120° = 180°

y° = 60°

17. Option E

Sketch the two circles touching at one point. The furthest that A and B can be would be at the two ends as shown in the above

diagram.

If the radius is 6 then the diameter is 2 × 6 = 12 and the distance from A to B would

be 2 × 12 = 24

18. Option E

Write down formula for volume of cylinder V = πr2h

Plug in the values

V = 45 × 3.142 = 141.39

We now have to test the volume of each of the rectangular solids to find out which is the closest

to 141.39

(A) 4 × 5 × 5 = 100

(B) 4 × 5 × 6 = 120

(C) 5 × 5 × 5 = 125

(D) 5 × 5 × 6 = 150

(E) 5 × 6 × 6 = 180

19. Option A

Note: Figures not drawn to scale

Since x = 60°, triangle ABC is an equilateral triangle with sides all equal.

The sides are all equal to 8.

Perimeter of triangle ABC = 8 + 8 + 8 = 24

Triangle DEF has two angles equal, so it must be an isosceles triangle.

The two equal sides will be opposite the equal angles.

V = π × 32 × 5 = 45 π

So, the length of DF = length of DE = 10

Perimeter of triangle DEF = 10 + 10 + 4 = 24

Subtract the two perimeters.

24 - 24 = 0

20. Option A

A right triangle has one angle = 90°. Sum of known angles is 90° + 35° = 125° The sum of all the angles in any triangle is 180°. Subtract sum of known angles

from 180°. 180° 125° = 55°

The size of the third angle is 55°

21. Option D

Since it is an equilateral triangle all its angles would be 60°. The size of the angle

does not depend on the length of the side.

The size of the angle is 60°.

22. Option C

Since it is an isosceles triangle it will have two equal angles. The given 96° angle

cannot be one of the equal pair because a triangle cannot have two obtuse angles.

Let x be one of the two equal angles. The sum of all the angles in any triangle is

180°.

x + x + 96° = 180°

2x = 84°

x = 42° The sizes of the other two angles are 42° each.

23. Option B

Formula C = πd

C = 8π

The circumference of the circle is 8π = 25.163 inches

24. Option D

Formula A = 2

Change diameter to radius r = 1

2 d =

1

2 × 10 = 5

Plug in the value: A = 52 = 25 The area of the circle is 25 78.55 sq. inches

25. Option A

Formula A = 2

7

2

2

Plug in the value A = 102 = 100 The area of the circle is 100 314.2 sq. inches

26. Option C

44 = 2 (3x + 2) + (2 (5x + 4), solve for x x = 2

height =

= 64 14

= 32

cm 7

Sin(T = (32/7) / 8 = 32

= 4

56 7

T = arcsin [ 4 ] = 34.8°

27. Option D

ABD is a right triangle: hence 2 = 152 + 152 = 450

Also 2 + 2 = 212 + 32 = 450 The above means that triangle BCD is also a right triangle and the total area of the

quadrilateral is the sum of the areas of the two right triangles.

Area of quadrilateral = [ 1 ] × 15 × 15 + [ 1 ] × 21 × 3 = 144 2 2

28. Option A

Area of OAB = 72 = [ 1 ] sin (AOB) * OA * OB

Solve the above for sin (AOB) to find sin (AOB) = [ 1 ]

2

2

2

sin

Area of ODC = 288 = [ 1 ] sin (DOC) * OD * OD

Note that sin(DOC) = sin(AOB) = [ 1 ] , OD = 18 + y and OC = 16 + x and

substitute in the above to obtain the first equation in x and y

1152 = (18 + y) (16 + x) We now use the theorem of the intersecting lines outside a circle to write a second

equation in x and y

16 × (16 + x) = 14 × (14 + y) Solve the two equations simultaneously to obtain

x = 20 and y = 14

29. Option C

Let r be the radius of the disk. Area is known and equal to 100π ; hence

100π = π 2

r.r = 10

Circumference = 2π = 20π

30. Option B

Let the radius of the semicircle. Area of the semicircle is known; hence

1250π = 1

π 2

r.r = 50

Length of rectangle = 2r = 100

Width of rectangle = r = 50

Area = 100 × 50 = 5000

31. Option C

The sine rule of triangle says that

=

sin

=

= k sin

Therefore, a = k(sin A), b = k(sin B) and c = k(sin C)

Hence, we can rewrite cos

=

cos =

cos as

cos

sin = cos sin =

cos sin

Or Cot A = Cot B = Cot C Or A =

B = C

Or the triangle is an equilateral triangle