Geometry CAT Previous Year Questions with Answer PDF

This is the equation of a circle with radius 4 units and centered at (-2, 3)
From a point L we drop two tangents on the circle such that the angle between the tangents is 60^o.


Therefore the locus of the point L, is a circle centered at (-2, 3) and has a radius of (4 + x = 8) units.

The equation of this locus is thus, (x + 2)² + (y - 3)² = 82

When X = 6, we have, (8)² + (y - 3)² = 8², that is y = 3

The circle, (x + 2)² + (y - 3)² = 8², touches the line x = 6 at (6, 3).

∠DAC=∠DBC
(Angles subtended by the chord DC on the same side.)
∠ADB=∠ACB
(Angles subtended by the chord AB on the same side.)
∠AED=∠BEC
(Vertically Opposite angles.)
Therefore, △AED∼△BEC

(Angles subtended by the chord AD on the same side.)
∠BAC=∠BDC
(Angles subtended by the chord BC on the same side.)
∠AEB=∠DEC
(Vertically Opposite angles.)
Therefore, △AEB∼△DEC

Therefore, AE : EC = 8 : 5

Let BP = x, PQ = y and QC = z.

Since the angle subtended by the diameter on the circle is a right-angle, such a triangle inscribed in a circle with the diameter as one of its sides will be right angled.

“…and the other two sides have their lengths in the ratio a: b.”
Let the two sides be ax and bx.

Let the dimensions of the rectangle inscribed in the circle be a and b.
We are being asked to express the Area of the rectangle in terms of Perimeter and Radius.
Area of the rectangle = a * b
Perimeter, P = 2(a + b)

Since the vertices of the rectangle subtend right angle on the circle,

Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangle forming the hexagon. Hence, its length should be twice the side of the hexagon, in this case, 4 cm.

Now, AD divided the hexagon into two symmetrical halves. Hence, AD bisects angle D, and hence, angle ADC is 60^∘.
We can find out the value of AT using cosine formula:

Area of a regular hexagon = 
Area of an equilateral triangle =  ; where a = side of the triangle
Since the area of the two figures are equal, we can equate them as folllows:

On simplifying: x² = 24 
∴ x = 2√6

We know that angle APB = AQB = 90^o
We know PB = 6, AB = 10 and AP = 2AQ
From using Pythagorean triplet, we find that AP = 8 cms
So, AQ = 4cms
AQ² + QB² = AB²
16 + QB² = 100
QB² = 84
QB = 9.3

We know that intersection of two medians is called a Centroid
We know, a Centroid divides the median in the ratio 2:1
AD = 12cms,
AG = 8cms, GD = 4cms
And similarly, BG = 6cms, GE = 3cms
Now, area (ΔABE) can be easily found out
area (ΔABE) = 1 / 2 x BE X AG
area (ΔABE) = 1 / 2 x 9 x 8 = 36 cm²
area (ΔABE) = area (ΔBEC)
area (ΔABC) = 2 x 36 = 72 cm²

For three circles to have a common tangent, the third circle must lie in between the two circles. Let the radius of the third circle be r cms
Now, Connect the centres of the circles with one another and with the intersection points. Let them be A,B,C,D,E respectively.
Now, from observing the constructed diagram, we find
AB = 8cms
AC = (4 + r) cms
BC = (4 + r) cms
OC = (4 - r) cms
Now, consider the right triangle OBC,
Applying Pythagoras theorem,
(4 + r)² = 4² + (4 - r)²
16 + r² + 8r = 16 + 16 + r² - 8r
16r = 16
r = 1 cms

Now, the length of AP would be maximum, when AB = AC (ABC is an Isosceles Right Triangle)
So, AB = AC = 10√2cms
Applying Pythagoras theorem to ΔAPB, we get
AB² = AP² + PB²
(10√2)² = 10² + x²
x² = 100
x = 10 cms

Given ∠AOB = 60°
Area of Sector AOB = 
Given OC = OD => ∠OCD = ∠ODC = 60°
△OCD is an Equilateral Triangle with side = a
Area(△OCD) = 
Its given that Area(OCD) = 1/2 × Area(OAB)

As the triangle progresses infinitely and the side length decreases, it follows an infinite GP series
As the sides decrease by half, their areas decrease by 1 / 4
We know, Area of an Equilateral Triangle = 
Area of T1 = 

Sum of an Infinite GP = a / 1 − r where a = 144 √3 , r = 1 / 4
Sum of areas ( T1, T2, T3,..) = 
Therefore, Sum of areas= 192 √3 sq cms.

Since, ABCD is a rectangle inscribed inside a circle
ABC must be a Right triangle
Given that radius of Circle = 13 cms
5, 12, 13 forms a Pythagorean Triplet
10, 24, 26 is also a Pythagorean triplet
So, 10 and 24 are possible length and breadth of ABCD

Given, Area(EFGH) = 62.5% Area(ABCD)
Area(EFGH) = 5 / 8 Area(ABCD)
Let EB = 1 and CG = r
Similarly, the rest other dimensions are also of lengths 1 or r units
Applying pythagoras theorem on △DHG we get GH = √(1 + r²)
Area(ABCD) = (1 + r)² and Area(EFGH) = (√(1 + r²))²
=> (1 + r²) = 5 / 8(1 + r)²
1 + r² = 5 / 8 (1 + r² + 2r)
8 + 8r² = 5 + 5r² + 10r
3r² – 10r + 3 = 0
(3r - 1) × (r - 3) = 0
r = 1 / 3 or r = 3
As CG > EB
r = 3 and ratio (EB : CG)= (1 : r) = (1 : 3)

Given that Chords lie on the same side of diameter with lengths 4 cms and 6 cms
Draw a perpendicular from the origin to both the chords and mark the points of intersection as P and Q respectively
Consider radius of circle as ‘r’ and distance OP as ‘x’
Draw lines from origin to the end of the chord and mark the points as D and B respectively
Thus, △OQB and △OPD form a right Triangle
Applying Pythagoras Theorem on both triangles,
(x+1)² +2² = r² ---(1)
x² +3² = r² ---(2)
We find that there is an increase and decrease by 1 in both equations
So, x² = 2² , x = 2
r² = 2² + 3² = 4 + 9 = 13
r = √13 cms

Given that a chord of length 5 cm subtends an angle of 60° at the centre of a circle.
We have to find the length of a chord that subtends an angle of 120° at the centre of the same circle
sin 60° = BC / OB
⟹ √3 / 2 = BC / 5
⟹ BC = 5√3 / 2 and AC = 5√3 / 2
So AB = 5√3 / 2 + 5√3 / 2 = 5√3
The length of a chord that subtends an angle of 120° at the centre of the same circle is 5√3

Given that from a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. Here GBC is the one third of the area of the triangle.
We can join AG and GD which is the median. Each of the shaded triangle has the same area and therefore the remaining area is two-thirds of ABC
The ratio of the sides are 8 : 5 : 7
Area of triangle = √(s(s − a)(s − b)(s − c))
where, semi perimeter (s) = 8 + 5 + 7 / 2 = 10
Area = √(10(10 − 8)(10 − 5)(10 − 7))
Area = √(10(2)(5)(3))
Area = 10√3
Area of ABC = 25 × 10 √3 (As 8 : 5 : 7 multiplied by 5 gives the sides of triangle ABC)
Therefore area of the remaining traingle = 2 / 3 × 250√3 = 500 / √3

Given that ABCDEF be a regular hexagon with each side of length of 1 cm.
We have to find the area of a square with AC as one side.
ABO is the equilateral triangle with each side having length of 1 cm and ABCO is the rhombus.
Altitude of an equilateral triangle = √3/2 a
So AP = √3/2 × 1
Where AP = PC
⟹ AC = AP + PC
⟹ AC = √3 / 2 × 2
⟹ AC = √3
Area of a square = a² sq.units
Area of the square with AC as one side = √3 × √3 = 3 sq.units

We have to find the total area of all six surfaces of the pillar which is 4 rectangles + 2 trapeziums one at the top and one at the bottom
Area of the rectangle = b × h
⟹ Area of rectangle with l = 20 b = 10
⟹ Area = 200 sq.cm
⟹ Area of rectangle with l = 20 b = 20
Area = 400 sq.cm
⟹ Area of 2 rectangles with l = 20 b = 13 is 2 × 13 × 20 = 520 sq. cm
Sum of areas of 4 rectangles = 520 + 200 + 400 = 1120 sq.cm
Area of the trapezium = 1/2 × 12 (10 + 20)
Area of the trapezium = 6 × 30 = 180 sq.cm
Area of both the trapezium = 2 × 180 = 360 sq.cm
Sum of areas of 4 rectangles = 1120 sq.cm.
Area of both the trapezium = 360 sq.cm
Total surface area of the pillar = 4 rectangles + 2 trapeziums
Total surface area of the pillar = 1120 + 360
Total surface area of the pillar = 1480 sq.cm