Geometry - Circles - Previous Year CAT/MBA Questions

Answer: Option D

Text Explanation :

Answer: Option E

Text Explanation :

Answer: Option C

Video Explanation

Text 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

Video Explanation

Text 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

Video Explanation

Text 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

Video Explanation

Text 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

Video Explanation

Text 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

Video Explanation

Text 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 C

Text 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

Text 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 A

Text 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).