Geometry - Circles - Previous Year CAT/MBA Questions

Answer: Option C

Explanation :

Length of the diagonal = √(30²+40² ) = 50 m.

Each circle on the end of the diagonal will touch sides of the rectangular field

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Using Pythagoras' theorem, the distance between the vertex of the rectangle and centre of the first circle drawn on the diagonal (OC) = 1.25√2

Distance between the vertex of the rectangle and circumference of the first circle drawn on the diagonal (OD) = 1.25√2 - 1.25 = 0.51 meters

Space that cannot be used to draw circle otherwise they will go outside rectangle on every diagonal = 0.51 × 2 = 1.02 meters

Space that can be used to draw circles = length of diagonal - unused space = 50 - 1.02 = 48.98 meters

On every diagonal, maximum number of such circles = usable length/diameter of each circle = 48.98/2.5 = 19.6

∴ Maximum 19 circles can be drawn on a diagonal.

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Now, on every diagonal, one circle will be at the centre (intersection of diagonals) and 9 circles will be on each half of the diagonal

⇒ The circle in centre will be common for both diagonals. 

∴ Total circles = 19 + 19 – 1 = 37.

Hence, option (c).

Answer: Option D

Explanation :

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In a circle a chord which subtends an angle of 60° at the center is equal to the radius of the circle.

∴ OA = OB = AB = 12

∆OAB is an equilateral triangle
OP (height of an equilateral triangle) = √3/2 × 12 = 6√3

∴ PM = OM = OP = 12 - 6√3

Area of AQBMA = Area of Triangle ABQ – Area of segment AMBP
Area of AQBMA = Area of Triangle ABQ – (Area of minor arc AMBO - Area of ∆OAB)

Now, PQ = MQ + PM = 12 + (12 - 6√3) = 24 - 6√3 

Area of triangle ABQ = 1/2 × 12 × (24 - 6√3) = 6(24 - 6√3) = 81.64

Also, Area of minor arc AMB – Area of OAB 

= 60/360 × π × 144 - 1/2 × 12× 6√3 = 24π - 36√3 = 13.07

∴ Area of AQBMA = 68.57 ≈ 69.

Hence, option (d).

Answer: Option D

Explanation :

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Length of the diagonal AC = 2√2
⇒ AX = ½ of diagonal = √2

Length of AC₁ = 2/3 × √2 = 2√2/3

⇒ C₁X = AX – AC₁ = √2 - 2√2/3 = √2/3

Using Pythagoras theorem:
XM = $\sqrt{{\left(\frac{2}{3}\right)}^2-{\left(\frac{\sqrt{2}}{3}\right)}^2}$ = $\frac{\sqrt{2}}{3}$

In ∆C₁XM,
C₁X = XM = √2/3 and ∠C₁XM = 90°
⇒ ∠XC₁M = 45°

And ∠LC₁M = 90°

∴ Area of ∆LC₁M = ½ × 2/3 × 2/3 = 2/9

Area of minor arc LM = Area of sector LC₁M - Area of ∆LC₁M

= $\frac{90}{360}\times \mathrm{\pi }\times {\left(\frac{2}{3}\right)}^2-\frac{2}{9}$ = $\frac{\mathrm{\pi }-2}{9}$

∴ Area of overlapping region =  $2\left(\frac{\mathrm{\pi }-2}{9}\right)$

Hence, option (d).

Answer: Option B

Explanation :

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Ptolemy’s theorem:
Product of diagonal of a cyclic quadrilateral is equal to sum of product of opposite pair of sides.

∴ XP × YZ = PY × XZ + PZ × XY

⇒ XP × a = PY  × a + PZ × a

⇒ XP = PY + PZ

Hence, option (b).

Answer: Option C

Explanation :

l₁: 2x – y = 1
l₂: x + 2y = 18

Slope for l₁ i.e., m₁ = -2/-1 = 2 and slope for l₂ i.e., m₂ = -1/2

Here, m₁ × m₂ = -1, hence the two given lines are perpendicular

Let us first represent the figure and the 2 lines l₁ and l₂.

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Let O (a, b) be the centre of the circle and A be the point of intersection of the 2 lines l₁ and l₂. 

The coordinates of A can be found out by solving the simultaneous equations of the lines l₁(2x – y = 1) and l₂ (x + 2y = 18). 

Solving both these equations we get the value of x and y as 4 and 7 respectively. 

∴ Coordinates of A are (4, 7).

Also, distance OA is the radius of the circle i.e., √20 units.

∴ (a - 4)² + (b - 7)² = 20   …(1)

Also (a, b) lies on l₁ hence, 2a − b = 1 
⇒ b = 2a – 1   ...(2)

From (1) and (2) we get

(a - 4)² + (2a - 1 - 7)² = 20

⇒ 5a² – 40a + 60 = 0

⇒ a² – 8a + 12 = 0

⇒ a = 2 or 6.

If a = 2, b = 3, hence a + b = 5

If a = 6, b = 11, hence a + b = 17

∴ (a + b) is either 5 or 17.

From 5 and 17 only 17 is listed in option (c).

Hence, option (c).

Answer: Option E

Explanation :

If we take one point at the centre of the circle, the remaining points can only be at the circumference of the circle as the minimum distance between any 2 points is at least 1 m (which is the radius of the circle). 

Also, the remaining points on the circumference of the circle have to be such that 2 points are at a distance of less than one.

Now circumference of the circle = 2 × 22/7 × 1 = 44/7 ≈ 6.28 m

As the circumference or the length of the boundary is 6.28 m, we can have a maximum of 6 points on the circumference such that the distance between any 2 points is at least 1 m.

So, in all we can have a maximum of 6 points on the circumference and 1 point at the centre of the circle, making a total of 7 points.

Hence, option (e).

Answer: Option A

Explanation :

Let ∠CAB and ∠ACB be 2x and 5x respectively.

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From the alternate segment thorem (In any circle, the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment), we can deduce that ∠AEC = ∠CAB = 2x.
 
In ∆AEC, ∠AEC = 2x, ∠EAC = 30 × 2 = 60° and ∠ACE = 180 − 5x.
 
∴ 2x + 60 + (180 − 5x) = 180.  
 
∴ x = 20. 
 
∠ABC = 180 − (∠BAC + ∠ACB) = 180 − (2x + 5x) = 180 − 7x = 180 − (7 × 20) = 40°.
 
Hence option (a).

Answer: Option C

Explanation :

Let O is the center of the bigger circle. So, OA = 2R. Let P be the center of the smaller circle. So, PA = R√2.

In ∆OAM, ∠AOM = 30° and OA = 2R, so AM =OA × Sin 30 = 2R/2 = R. Also, OM = R√3

In ∆PAM, let ∠APM = θ and PA = R√2 and AM = R, so Sin θ = AM/PA = R/ R√2 = 1/√2, so θ = 45°

Therefore ∠APB = 2∠APM = 2 × 45 = 90°. Also, PM = PA × Cos 45 = R√2 × Cos 45 = R

Common area between the two circles is the area of the smaller circle minus the area of the shaded region.

Area of the shaded region = Area of quadrant APB + Area ∆OPA + Area ∆OPB – Area of sector AOB

Area of quadrant APB = π × (R√2)²/4

Area ∆OPA + Area ∆OPB = Area ∆AOB – Area ∆PAB = [(1/2) × OM × AB) – [(1/2) × PM × AB]

OM = R√3 and AB = AM × 2 = R × 2 = 2R

So, Area ∆OPA + Area ∆OPB = [(1/2) × R√3 × 2R) – [(1/2) × R × 2R] = (√3 – 1)R²

Area of sector AOB = (π/6)(2R)²

Hence, Area of the shaded region = [π × (R√2)²/4] + [(√3 – 1)R²] +[(π/6)(2R)²]

= [(√3 – 1)R²] – [πR²/6]

Common area between the two circles = Area of the smaller circle – Area of the shaded region

= π(R√2)² – {[(√3 – 1)R²] – [πR²/6] = (13π/6) + 1 – √3) R²

Hence, option (c).

Answer: Option B

Explanation :

Since Joseph takes less time while crossing the path diametrically rather than in a semi-circular manner, the distance travelled along the diameter is less.

Let the radius of the path be r m.

∴ Difference in distance = πr – 2r

= (π – 2)r m

Also, difference in distance = difference in time × constant speed

∴ (π – 2)r = 48 × (50/60)

∴ 1.14r = 40 i.e. r = 40/(8/7) = 35 m

∴ Diameter = 2r = 70 m

Hence, option (b).

Answer: Option B

Explanation :

Let us draw the diagram using the given conditions.

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AB = 24 cm and P is the mid-point of AB. 

∴ AP = PB = 12 cm.

MN is perpendicular to AB and passes through P. 

PM < PN. 

∴ M should be closer to A and B than N.

MN and AB are 2 perpendicular chords intersecting at P. 

Therefore, according to the intersecting chords theorem, AP × PB = PM × PN

⇒ 12 × 12 = 8 × PN

⇒ PN = 18 cm.

Hence, option (b).

Answer: Option C

Explanation :

Area of sector O-APB = (x/360) × π × (13)² = 169πx/360 cm²

Let the perpendicular from O to AB meet AB at M. Hence, OM is the perpendicular bisector of AB. 

Hence, AM = 24/2 = 12 cm and OM = 5 cm (5, 12, 13 form a pythagorean triplet in right triangle OAM). 

∴ Area of triangle OAB = (1/2) × OM × AB = (1/2) × 5 × 24 = 60 cm²  

∴ Required area = [(169πx/360) − 60] cm²

Hence, option (c).

Answer: Option B

Explanation :

In quadrilateral PQOT, ∠QOT = 360 − ∠QPT − ∠PQO − ∠PTO 

Since PQ and PT are tangents to the circle at Q and T, ∠PQO = ∠PTO = 90°

∴ ∠QOT = 360 − 55 − 90 − 90 = 125° 

∴ ∠QOS + ∠SOT  = ∠QOT = 125° 

Now, when two tangents are drawn from a point to the circle, the line joining the external point and the centre of the circle is the angle bisector of the central angle subtended by the two tangential points at the centre.   
 
For instance, when PQ and PT are tangents from P to the circle with centre O, then line PO is the angle bisector of ∠QOT. 

Similarly, AQ and AS are tangents from A to the circle with centre O. Hence, AO is the angle bisector of ∠QOS. 

∴ ∠QOA = ∠AOS = (∠QOS)/2

Similarly, for quadrilateral BSOT, ∠TOB = ∠BOS = (∠SOT)/2 

Now, ∠AOB = ∠AOS + ∠BOS 

= (∠QOS)/2 + (∠SOT)/2 = (∠QOS + ∠SOT)/2 

= 125/2 = 62.5°

Hence, option (b).

Answer: Option A

Explanation :

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From the diagram, we can say that Y makes an angle of 60° and X an angle of 90°.

For Y ; ⅙ × 2πr = 10π.( r is radius of circle of which Y is a part).

∴ r = 30.

From the diagram ;

the radius of other circle = 15√2.

∴ Length of X = ¼× 2× π ×15√2 = 15π/√2.

Hence, option (a).

Answer: Option C

Explanation :

Y is the centre of the circle.

Let r be the radius of the circle.

∴ Length of arc XZ =  $\frac{1}{4}\left(2\pi r\right)=10\pi$

∴ r = 20

Arc of sector XYZ =  $\frac{1}{4}\left(\pi r^2\right)=100\pi$

Hence, option (c).

Answer: Option A

Explanation :

Radius of P₁ = r ⇒A(P₁) = πr²

Diameter = d = 2r

Side of Q₁ = $\frac{2r}{\sqrt{2}}$ = r$\sqrt{2}$ ⇒ A(Q₁) = 2r²

Radius of P₂ = $\frac{r\sqrt{2}}{2}$ = $\frac{r}{\sqrt{2}}$ ⇒ A(P₁) = $\frac{\pi r^2}{2}$

Side of Q₂ = $\left(\frac{r}{\sqrt{2}}\right)$ × $\sqrt{2}$ = r ⇒ A(Q₂) = r² ... and so on.

i.e., Areas of circles are in G.P. with common ratio = $\frac{1}{2}$

Also, areas of squares are in G.P. with common ratio = $\frac{1}{2}$

S_N = Q₁ – P₂ + Q₂ – P₃ + Q₃ – P₄ + …

= (Q₁ + Q₂ + Q₃ + … ) – (P₂ + P₃ + P₄ + … )

= 2r² $\left(1+\frac{1}{2}+\frac{1}{4}+...\right)$ - $\frac{\pi r^2}{2\left(1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}+...\right)}$

= $\left(1+\frac{1}{2}+\frac{1}{4}+...\right)$$\left(2r^2-\frac{\pi r^2}{2}\right)$

= r² × $\left[\frac{1}{1-{\displaystyle \frac{1}{2}}}\right]$ × $\frac{4-\pi }{2}$

= (4 - π)r²

S_N: A(Q₁) = (4 – π)r² : 2r² = (4 – π)/2

Hence, option (a).

Answer: Option A

Explanation :

Let the length of each side of the hexagon be y.

Area of the hexagon = $\frac{3\sqrt{3}}{2}$y²

Area of the triangle = $\frac{\sqrt{3}}{16}$y²

Area of the shaded region = $\frac{23\sqrt{3}}{16}$ y² = X

y² = $\frac{16}{23\sqrt{3}}$X

Area of the circle = πy²

Thus, area of the circle in terms of X

= $\frac{16\pi }{23\sqrt{3}}$X

Hence, option (a).

Answer: Option C

Explanation :

Volume of the cylinder = π × 6² × 24 = 864π cu. units

Diameter of the cylinder = 2 × 6 = 12 units

Side of the square base = $\frac{12}{\sqrt{2}}$ = 6$\sqrt{2}$ units

Volume of the rectangular solid

= (6$\sqrt{2}$)² × 20 = 1440 cu. units

∴ Volume of the cylinder = 844π – 1440 = 288(3π – 5) cu. units

Hence, option (c).

Answer: Option C

Explanation :

Area of a circular field = πr²

Area of tank = 130 × 110 = 14300 m²

∴ πr² – 14300 = 20350

∴ πr² = 34650

r² = 1575 × 7 = 225 × 49

r = 15 × 7 = 105 m

Hence, option 3.

Answer: Option C

Explanation :

Let the height of the cylindrical vessel be 2h.

∴ Radius of the cylindrical vessel is 3h.

Also the radius of the hemispherical bowl = 3h

∴ Volume of the hemispherical bowl = $\left(\frac{2}{3}\right)$ × π × (3h)³ = 18πh³

Volume of the cylindrical vessel = π × (3h)² × (2h) = 18πh³

Hence the cylindrical vessel will be completely filled when the contents are transferred.

Hence, option 3.

Answer: Option B

Explanation :

The number of triangles formed using the 8 points

= ⁸C₃ = 56

The number of quadrilaterals formed using the 8 points = ⁸C₄ = 70

∴ The difference = 14

Hence, option (b).

Answer: Option D

Explanation :

Cylinder is formed by rolling the paper along its length.

Hence, circumference of cylinder = 22

∴ 2πr = 22

∴ r = 7/2

∴ Volume of cylinder = πr²h = $\frac{22}{7}$ × ${\left(\frac{7}{2}\right)}^2$ × 10

= 385 cm³

Hence, option (d).

Answer: Option A

Explanation :

i. Volume of parallelepiped = 5 × 4 × 3 = 60 cm³

ii. Volume of cube = 4 × 4 × 4 = 64 cm³

iii. Volume of cylinder = πr²h = π × 3 × 3 × 3 > 81 cm³

iv. Volume of sphere = 4/3 × πr³ = 36π cm³ > 81 cm³

∴ Required order is iv, iii, ii, i.

Hence, option (a).

Answer: Option C

Explanation :

The given scenario can be depicted in the diagram below.

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As, OC ⊥ AB and OC = AO/2,

ΔAOC is a 30-60-90 triangle.

∴ Angle subtended by the arc AB = 2 × 60 = 120°

∴ Angle subtended by half the arc is 60°

∴ Triangle formed by the chord of ‘half the arc’ and two radii is equilateral.

Hence, length of chord AM = 42 cm

Hence, option (c).

Answer: Option A

Explanation :

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Let the height of the cylinder be 5h than that of the conical part be 3h.
∴ Height of hole = $\frac{1}{3}$(3h + 5h)
= $\frac{8h}{3}$

Radius of cone = radius of cylinder = 8

Let radius of hole = r

Now, 1/2(Total volume – Volume of hole) = Volume of hole

∴ Total volume = 3Volume of hole

∴ π × 8² × 5h + $\frac{1}{3}$ × π × 8² × 3h = 3πr² $\left(\frac{8h}{3}\right)$

∴ 8² × 6 = 8r²

∴ r² = 8 × 6

∴ r = 4$\sqrt{3}$ cm

Hence, option (a).

Answer: Option A

Explanation :

By the cosine rule, in a ΔABC with ∠ACB = θ and AB = c, BC = a and AC = b,

cos θ = $\frac{a^2+b^2-c^2}{2ab}$

∴ 2 ab cos θ  = a² + b² – c²

∴ c² = a² + b² - 2ab cos θ

Hence, option (a).