May 9, 2013

Free Geometry Shortcuts workshop at Cetking… Register here…

Register here for FREE Geometry shortcut workshop at Cetking

Cetking launches advance shortcut workshops

Geometry Mensuration Shortcuts… 

Topic this week:
Geometry Mensuration Shortcuts workshops

Download handout
These questions are going to be covered please go through them.
Download Cetking Geometry Handout >>>
Handout is shared because i want you to see the questions before you come so that we can focus on shortcuts only.

Homework before workshop:
Do basics of Geometry Mensuration or watch basic lectures of Geometry on

All India – Online
Mumbai in class-  Cetking centres in Thane Dadar Vashi

Sample shortcut video  
Shortcuts in Geometry 1

Shortcuts in Geometry 2

How to prepare for Geometry


Getting started with Geometry

Agenda and topics covered: 
Topic covered in workshop.
Triangles – Theorems
Triangles – Similar/Cong
Mensuration – Areas
Mensuration – Volume
Trigonometry Geometry
Coordinate Geometry


Cetking’s 10 shortcut concepts for Geometry

1.Uniform Figures
2.Triple Delight
3.Draw Lines
4.Scale drawing
6.Drawing to scale

Download Cetking Geometry Handout >>>


Importance of Geometry Mensuration in CAT – 6-8 questions
Importance of Geometry Mensuration in CET – 4 ques
Importance of Geometry Mensuration in NMAT – 6 ques
Importance of Geometry Mensuration in SNAP – 6 ques
Importance of Geometry Mensuration in Bank PO – 2 ques

CAT 2014 paper downloadCAT 2014 paper download

[section_title]Other workshops coming soon….  [/section_title]



Cetking Courses….  

Scope of Geometry Shortcuts
Download Cetking Geometry Handout >>>

Do this handout the way you can and in workshops we will teach you how to solve all these questions using shortcuts..

Join workshops online call – 09594441448
for classroom call:
Thane – 09930028086
Vashi & VileParle – 09820377380
Dadar & Borivali – 09167917984

Concept Test Topic: Geometry Test

1.  Find the equation of a line whose intercepts are twice of the line 3x – 2y – 12 = 0

(1) 3x – 2y = 24 (2) 2x – 3y = 12
(3) 2x – 3y = 24 (4) None of these

Answer: Option 1

Put x = 0, then y = – 6
Therefore, y intercept is – 6
Put y = 0, then x = 4
Therefore, x intercept is 4
Hence the intercepts of the required line are 8 and – 12
Hence the equation of the line is (x/8) + (y/-12) = 1
-12x + 8y = -96
3x – 2y = 24.


2.  Find the number of triangles in an octagon.

(1) 326 (2) 120
(3) 56 (4) Cannot be determined

 Answer : option 3


No. of triangles =nC3 where n is the no. of points.

Here n = 8 => number of triangles =8C3 = 56.

3. Each interior angle of a regular polygon is 120 degrees greater than each exterior angle. How many sides are there in the polygon?

(1) 6 (2) 8 (3) 12 (4) 3

Answer : option 3




Let an exterior angle Ao. Then each interior angle will be 120 + Ao.
We know that in any regular polygon, the sum of an exterior and interior angle is always = 180o.
Therefore, A + 120 + A = 180 => A = 30o.
No. of sides of a polygon =  360  /  each exterior angle = 360/30  = 12 sides.

4.  What is the circum radius of a triangle whose sides are 7, 24 and 25 respectively?

(1) 18 (2) 12.5
(3) 12 (4) 14

 Answer :  option 2


7, 24, 25 is a Pythagorean triplet. Therefore, the given triangle is a right angled triangle.
In a right-angled triangle, the median to the hypotenuse is half the hypotenuse and is also the circum radius of the triangle.
As the hypotenuse is 25, the circum radius is 12.5


5.  If 5 and 7 are the lengths of two sides of a triangle. which of the following could be the length of the third side?

I. 2

II. 5

III. 7

(1)   I only

(2)  III only

(3)  I and II only

(4)  II and III only

(5)  I, II and III

Answer:  option  4


The triangle inequality says that no one side of a triangle can be longer than or equal to the sum of the other two.


So is this true for each of the options?


I. Gives us sides with lengths 2, 5 and 7. 7 is equal to 2+5 therefore one side is equal to the sum of the other two and so these sides cannot form a triangle.

II. Gives us two sides with length 5 and one with 7. 5 is less than 7+5 and 7 is less than 5+5 therefore no one side is longer than or equal to the length of the other two and so these sides can form a triangle.

III. Gives us two sides with length 7 and one with 5. 5 is less than 7+7 and 7 is less than 5+7 therefore no one side is longer than or equal to the length of the other two and so these sides can form a triangle.

This means that the correct answer is 4.



A square ABCD is inscribed within a circle. If the perimeter of the square is 16, what is the approximate diameter of the circle?

(1) 4        (2) 6        (3) 8              (4) 12

 Answer : option  2


If the perimeter of the square is 16 then each side must have a length of 4.

6 andThis means that the diameter of the circle (marked in red below) is the hypotenuse of an isosceles right-angled triangle which has short sides of length 4.

 From this diagram we should be able to eliminate many of the answers.

We can see that the diameter is longer than than one of the sides of the square and so we know that it must be greater than 4 which eliminates answers 1  which is equal to 4.

We can also see the diameter is shorter than two of the sides of the square added together and so we know it must be less than 8 which eliminates 3 and 4 which are greater than or equal to 8.

This only leaves answer 2 which must be the correct answer.


If you want to double check this answer then since you know that the ratios of the sides of an isosceles right-angled triangle are 1:1:, , you know that the diameter has a length of  4. Use the Pythagoras theorem to check this if you wish.

You should also know that


Which means that


Again this is answer 2.


7. What is x + y equal to?

7 q

(1) 80         (2) 100      (3) 280       (4)300

 Answer : option 3


We know that the angle of a straight line is 180° so we can fill in the interior angles in the triangle.


And we know that the interior angles of a triangle add up to 180°, so


Simplifying this gives us




Which means the answer is 3.


8. In the figure above, what is the value of x?

8 q

(1) 30       (2) 52       (3) 60         (4) 120

Answer:  option 1


The angles next to each other when lines cross add up to a straight line i.e. 180°.

Therefore 1 is the correct answer.


9 q

In the figure above angles ADC and BCD are right-angles and AB has length 13, AD length 16 and CD length 5.

What is the area of quadrilateral ABCD?

(1) 30       (2) 60       (3) 40       (4) 50

 Answer: option  4


Firstly we should draw the figure and add in all the information we are given in the question, the two right-angles and the 3 lengths.


We don’t know how to calculate the area of an irregular quadrilateral like this but we can divide it into a rectangle and a triangle which we do know how to work with.


Triangle ABE is a right-angled triangle with a short side length 5 and a hypotenuse of length 13. You should recognize this as a 5-12-13 right-angled triangle so you know the final side is 12. This enables you to work out that the width of the rectangle BCDE is 4.

Now we can work out the area of the triangle.


And the area of the rectangle.


And sum them to get the area of the quadrilateral


Area of quadrilateral ABCD = Area of triangle ABE + Area of Rectangle BCDE

= 30 + 20 = 50


10. What is the distance from O (0,0) to P (4,-2)?

(1)        (2) 6         (3) 10          (4)

Answer : option  4


You can see that the line OP is the hypotenuse of a right angled triangle with base and height 4 and 2 respectively.


This means that we can find the distance OP using the Pythagoras theorem.


Take the square root of both sides  = This means that the correct answer is 4.


11. What is the area of the above triangle?

11 q

(1) 120       (2) 130      (3) 240       (4) 260

 Answer : option 1


You should recognise this triangle as equivalent to the 5-12-13 triangle.

This triangle has been scaled up by a factor of 2, so the base is 10 (i.e. ) and hypotenuse 26 (i.e. ).

Therefore the height is .

The base of the triangle is 10 and the height is 24 so we can calculate the area of the triangle.

Area of triangle = ½( base ) ( height) = 120

This means that the correct answer is 1.


12.   What is the surface area, in square inches, of a cuboid which is 8 inches long, 4 inches wide and 2 inches high?

(1)64            (2) 80          (3) 96        (4) 112

Answer : option 4

Explanation: The surface area of a cuboid with length l, width w and height h can be calculated using the following formula.

Since we know that the length of this cuboid is 8 inches, the width is 4 inches and the height is 2 inches, we can put these values into the formula to calculate the surface area.

Surface area = 2( 8*4 +  8*2 + 4*2 )

= 2( 32 + 16 + 8 )

= 2(56)

= 112

Therefore the correct answer is 4.


13. What is the surface area, in centimetres squared, of a cube that has a volume of 125 centimeters cubed?

(1) 30       (2)100      (3)150       (4)200

Answer : option 3


If the volume of the cube is  then the length of one of it’s edges is the cube root of this.

So we can work out the area of one of it’s sides

A cube has 6 sides so we multiply this by 6 to be the total surface area

Therefore the surface area of the cube is  and the correct answer is 3.


14.  In a circular pond a fish starts from a point on the edge of the pond, swims 50 mts in the south direction and reaches another point on the edge of the pond. From here it turns west, swims 120 mts and reaches yet another point on the edge of the pond. Find the radius of the pond.

(1) 60          (2) 65         (3) 70          (4) 85

Answer : 2


14 a

With a question like this it is really helpful to draw a diagram to keep track of the information you are given in the question.

If label the point where the fist starts as P, and the point where it reaches the edge of the pond after swimming 50m south as Q and the point after swimming 120m as R then we get a diagram like the one below.


We know that angle PQR = 90° because the directions south and west are at right angles.

Since the angle made by the chord PR on the circumference of the circle is , PR will be the diameter of the circle. (Angle in a semi-circle is .)

Now you should recognize the triangle as similar to the 5-12-13 triangle and so the length of PR (and diameter of the fish pond) is 130 m.

Hence the radius of the pond = .

The correct answer is 2.


15   What is the approximate perimeter of a square that has a diagonal of length 6?

(1) 17             (2) 13         (3) 24          (4) 20

 Answer  : option 1


We can see that the diagonal, length 6, forms a right angled triangle with two of the sides of the square, length x, and so we can use the Pythagoras theorem to determine the length of those sides.

We know that the perimeter is equal to  therefore we can estimate that the perimeter is somewhere around 16 or 17.

None of the answers is 16 and C is 17, so 1 is the correct answer.

Note: A quick check with a calculator reveals the answer to be 16.97… and so our estimate is very good.


16   What is the area of a triangle with two sides each of length 13 and a total perimeter of 36 ?

(1) 32.5             (2) 60           (3) 65            (4) 130

 Answer : option 2


Firstly we should draw a picture of the triangle described. We are told that it is

a triangle with a perimeter of 36 and two sides of length 13

So we can draw

16 a

We know that the total perimeter is 36 so we can work out the length of the base.


So we know that the base of the triangle has length 10. Adding this information to our diagram gives us.


The area of a triangle  equal to   .  We already know the base is 10 so to find the area of this triangle we need to work out the height.

If we draw in the height of the triangle into the diagram you can see that it will splits the figure into two right-angled triangles with base 5 and hypotenuse 13.


We could use the Pythagoras theorem to find the height but those students who have prepared themselves will save precious minutes because they will recognize that a right-angled triangle with base length 5 and a hypotenuse length 13 is a 5-12-13 triangle, one of the triangles that commonly appears in the questions.

Therefore the height of the triangle is 12.


Now we can work out the area of this triangle which has a base of length 10 and a height of 12.

Area of a triangle = ½ ( base) ( height)

= ½ (10)(12)

= 60

Therefore the correct answer is 2.


17.    The sum of the areas of two circles, which touch each other externally , If the sum of their radii is 15, find the ratio of the larger to the smaller radius.

(1) 4:1         (2) 2:1             (3)  3:1            (4) none of these

Answer : option 1

Explanation: Let the radii of the 2 circles be and , then = 15 (given)

and = (given)

= 153

solving we get, = 12 and = 3

ratio of the larger radius to the smaller one is 12 : 3 = 4 : 1 hence option (1) is the answer.

18. In the adjoining figure, points A, B, C and D lie on the circle. AD= 24 and BC = 12. what is the ratio of the area of CBE to that of the triangle ?

(1)1:4                        (2) 1:2                              (3) 1:3                           (4)  Data Insufficient

Answer : option 2

Explanation: In and , ∟ CBA = ∟ CDA.

( a chord of a circle subtends equal angel at all points on the circumference, lying in the same segment)

similarly ∟ BCD = ∟ BAD and ∟ BEC = ∟ AED

Therefore ( AAA similarity rule)


Hence hence option 2

19.   and are two concentric circles while BC and CD are tangents to at point P and respectively. AB to the circle , which of the following is tru




(4) data insufficient

Answer : option 3

Explanation : l ( cp) = l (cr) ( since tangents drawn from the same point are equal)

hence l(BC) = l( CD) ( points P and R are midpoints of BC and CD as OP is ┴BC and OR ┴ CD)

Therefore m( arc BXC) = m(arc CYD)
hence quadrilateral OPCR is a cyclic quadrilateral.

Hence answer is option c)

But m(arc BXC ) = ( ∟ABC) = ( tangent secant theorem)

therefore m(arc BZD) =

Therefore m(BCD) =

Hence m∟PCR =

m∟POR = 2m ∟PQR = (angle at center)



m∟OPC + m∟ORC =

Hence m∟POR + m∟PCR =


20. Two circles of radius 3 units and 4 units are at some distance such that length of the transverse common tangents and the length of their direct common tangents are in the ratio 1: 2 . What is the distance between the centres of those circles?

(1) units

(2) units

(3)   8 units

(4)   Cannot be determined .

Answer : option 3

Explanation : Let x = distance between the centres of the circles.

T= length of the transverse common tangent.

D = length of direct common tangent.


And = 4 units and = 3 units

then, t = ————— 1)

also d =

——————- 2)








                                                             hence x =

from 1) and 2 ) we get

d = 2t

d = 8 units

Hence answer is option 3.