# Geometry - Circles - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Geometry - Circles. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2020 QADI | Geometry - Circles**

A rectangular field is 40 meters long and 30 meters wide. Draw diagonals on this field and then draw circles of radius 1.25 meters, with centers only on the diagonals. Each circle must fall completely within the field. Any two circles can touch each other but should not overlap.

What is the maximum number of such circles that can be drawn in the field?

- A.
39

- B.
40

- C.
37

- D.
36

- E.
38

Answer: Option C

**Explanation** :

Length of the diagonal = √(30^{2}+40^{2} ) = 50 m.

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

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.

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).

Workspace:

**XAT 2020 QADI | Geometry - Circles**

In the figure given below, the circle has a chord AB of length 12 cm, which makes an angle of 60° at the center of the circle, O. ABCD, as shown in the diagram, is a rectangle. OQ is the perpendicular bisector of AB, intersecting the chord AB at P, the arc AB at M and CD at Q. OM = MQ. The area of the region enclosed by the line segments AQ and QB, and the arc BMA, is closest to (in cm^{2}):

- A.
137

- B.
63

- C.
35

- D.
69

- E.
215

Answer: Option D

**Explanation** :

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).

Workspace:

**XAT 2020 QADI | Geometry - Circles**

Mohanlal, a prosperous farmer, has a square land of side 2 km. For the current season, he decides to have some fun. He marks two distinct points on one of the diagonals of the land. Using these points as centers, he constructs two circles. Each of these circles falls completely within the land, and touches at least two sides of the land. To his surprise, the radii of both the circles are exactly equal to 2/3 km. Mohanlal plants potatoes on the overlapping portion of these circles.

- A.
5( 𝜋 + 4)/27

- B.
2(2 𝜋 − 3 √3 )/27

- C.
4(𝜋 − 3 √3 )/27

- D.
𝟐(𝝅 − 2)/9

- E.
(𝝅 − 2)/9

Answer: Option D

**Explanation** :

Length of the diagonal AC = 2√2

⇒ AX = ½ of diagonal = √2

Length of AC_{1} = 2/3 × √2 = 2√2/3

⇒ C_{1}X = AX – AC_{1} = √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 ∆C_{1}XM,

C_{1}X = XM = √2/3 and ∠C_{1}XM = 90°

⇒ ∠XC_{1}M = 45°

And ∠LC_{1}M = 90°

∴ Area of ∆LC_{1}M = ½ × 2/3 × 2/3 = 2/9

Area of minor arc LM = Area of sector LC_{1}M - Area of ∆LC_{1}M

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

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

Hence, option (d).

Workspace:

**XAT 2020 QADI | Geometry - Circles**

XYZ is an equilateral triangle, inscribed in a circle. P is a point on the arc YZ such that X and P are on opposite sides of the chord YZ. Which of the following MUST always be true?

- A.
XZ + YP = XY + PZ

- B.
XP = YP + PZ

- C.
XP + PZ = XY + YP

- D.
XP = XY

- E.
XP = XY + YZ

Answer: Option B

**Explanation** :

**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).

Workspace:

**XAT 2019 QADI | Geometry - Circles**

Let C be a circle of radius √20 cm. Let l_{1}, l_{2} be the lines given by 2x − y −1 = 0 and x + 2y −18 = 0, respectively. Suppose that l_{1} passes through the center of C and that l_{2} is tangent to C at the point of intersection of l_{1} and l_{2}.

If (a, b) is the center of C, which of the following is a possible value of a + b?

- A.
11

- B.
14

- C.
17

- D.
8

- E.
20

Answer: Option C

**Explanation** :

l_{1}: 2x – y = 1

l_{2}: x + 2y = 18

Slope for l_{1} i.e., m_{1} = -2/-1 = 2 and slope for l_{2} i.e., m_{2} = -1/2

Here, m_{1} × m_{2} = -1, hence the two given lines are perpendicular

Let us first represent the figure and the 2 lines l_{1} and l_{2}.

Let O (a, b) be the centre of the circle and A be the point of intersection of the 2 lines l_{1} and l_{2}.

The coordinates of A can be found out by solving the simultaneous equations of the lines l_{1}(2x – y = 1) and l_{2} (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 l_{1} hence, 2a − b = 1

⇒ b = 2a – 1 ...(2)

From (1) and (2) we get

(a - 4)^{2} + (2a - 1 - 7)^{2} = 20

⇒ 5a^{2} – 40a + 60 = 0

⇒ a^{2} – 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).

Workspace:

**XAT 2019 QADI | Geometry - Circles**

What is the maximum number of points that can be placed on a circular disk of radius 1 metre (some of the points could be placed on the bounding circle of the disk) such that no two points are at a distance of less than 1 metre from each other?

- A.
5

- B.
8

- C.
6

- D.
9

- E.
7

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).

Workspace:

**IIFT 2019 QA | Geometry - Circles**

AB is the tangent on the circle at point A. The line BC meets the circle at points C and E. Line AD bisects the angle EAC. If angle EAC = 60° and angle BAC : angle ACB = 2: 5. Find angle ABC.

- A.
40°

- B.
60°

- C.
30°

- D.
None of these

Answer: Option A

**Explanation** :

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

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).

Workspace:

**XAT 2018 QADI | Geometry - Circles**

Two circles with radius 2R and √2R intersect each other at points A and B. The centers of both the circles are on the same side of AB. O is the center of the bigger circle and ∠AOB is 60°. Find the area of the common region between two circles.

- A.
(√3 – π – 1)R

^{2} - B.
(√3 – π)R

^{2} - C.
(13π/6 + 1 - √3)R

^{2} - D.
(13π/6 + √3)R

^{2} - E.
None of the above

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)R^{2}

Area of sector AOB = (π/6)(2R)^{2}

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

= [(√3 – 1)R^{2}] – [πR^{2}/6]

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

= π(R√2)^{2} – {[(√3 – 1)R^{2}] – [πR^{2}/6] = (13π/6) + 1 – √3) R^{2}

Hence, option (c).

Workspace:

**IIFT 2018 QA | Geometry - Circles**

Joseph diametrically crosses a semi-circular playground and takes 48 seconds less than if he crosses the playground along the semi-circular path. If he walks 50 metres in one minute, the diameter of playground is

- A.
54 metres

- B.
70 metres

- C.
85 metres

- D.
35 metres

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).

Workspace:

**XAT 2017 QADI | Geometry - Circles**

AB is a chord of a circle. The length of AB is 24 cm. P is the midpoint of AB. Perpendiculars from P on either side of the chord meets the circle at M and N respectively. If PM < PN and PM = 8 cm. then what will be the length of PN?

- A.
17 cm

- B.
18 cm

- C.
19 cm

- D.
20 cm

- E.
21 cm

Answer: Option B

**Explanation** :

Let us draw the diagram using the given conditions.

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).

Workspace:

**IIFT 2017 QA | Geometry - Circles**

A chord AB of length 24 cm is drawn in a circle of radius 13 cm. Find the area of the shaded portion APB.

- A.
13π x cm

^{2} - B.
$\frac{13\pi x}{180}$cm

^{2} - C.
$\frac{169\pi x}{360}-60c{m}^{2}$

- D.
$\frac{169\pi x}{180}-60c{m}^{2}$

Answer: Option C

**Explanation** :

Area of sector O-APB = (x/360) × π × (13)^{2} = 169π*x*/360 cm^{2}

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 cm^{2}

∴ Required area = [(169π*x*/360) − 60] cm^{2}

Hence, option (c).

Workspace:

**IIFT 2017 QA | Geometry - Circles**

Two tangents are drawn from a point P on the circle with centre at O, touching the circle at point Q and T respectively. Another tangent AB touches the circle at point S. If angle QPT=55°, find the angle AOB=?

- A.
125°

- B.
62.5°

- C.
97.5°

- D.
95°

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).

Workspace:

**XAT 2016 QADI | Geometry - Circles**

In the figure below, two circular curves create 60° and 90° angles with their respective centres. If the length of the bottom curve Y is 10, find the length of the other curve.

- A.
15π/√2

- B.
20π√2/3

- C.
60π/√2

- D.
20π/3

- E.
15π

Answer: Option A

**Explanation** :

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).

Workspace:

**IIFT 2015 QA | Geometry - Circles**

If in the figure below, angle XYZ=90° and the length of the arc XZ=10π, then the area of the sector XYZ is:

- A.
10π

- B.
25π

- C.
100π

- D.
None of the above

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).

Workspace:

**IIFT 2014 QA | Geometry - Circles**

Let P_{1 }be the circle of radius r. A square Q_{1} is inscribed in P_{1} such that all the vertices of the square Q_{1} lie on the circumference of P_{1}. Another circle P_{2} is inscribed in Q_{1}. Another Square Q_{2} is inscribed in the circle P_{2}. Circle P_{3} is inscribed in the square Q_{2} and so on. If S_{N} is the area between Q_{N} and P_{N+1}, where N represents the set of natural numbers, then the ratio of sum of all such S_{N} to that of the area of the square Q_{1} is:

- A.
$\frac{4-\pi}{2}$

- B.
$\frac{2\pi -4}{\pi}$

- C.
$\frac{\pi -2}{2}$

- D.
None of the above

Answer: Option A

**Explanation** :

Radius of P_{1} = r ⇒A(P_{1}) = πr^{2}

Diameter = d = 2r

Side of Q_{1} = $\frac{2r}{\sqrt{2}}$ = r$\sqrt{2}$ ⇒ A(Q_{1}) = 2r^{2}

Radius of P_{2} = $\frac{r\sqrt{2}}{2}$ = $\frac{r}{\sqrt{2}}$ ⇒ A(P_{1}) = $\frac{\pi {r}^{2}}{2}$

Side of Q_{2} = $\left(\frac{r}{\sqrt{2}}\right)$ × $\sqrt{2}$ = r ⇒ A(Q_{2}) = r^{2 }... 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} = Q_{1} – P_{2} + Q_{2} – P_{3} + Q_{3} – P_{4} + …

= (Q_{1} + Q_{2} + Q_{3} + … ) – (P_{2} + P_{3} + P_{4} + … )

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

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

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

= (4 - π)r^{2}

S_{N}: A(Q_{1}) = (4 – π)r^{2 }: 2r^{2} = (4 – π)/2

Hence, option (a).

Workspace:

**IIFT 2014 QA | Geometry - Circles**

ABCDEF is a regular hexagon and PQR is an equilateral triangle of side a. The area of the shaded portion is X and CD : PQ : : 2 : 1. Find the area of the circle circumscribing the hexagon in terms of X.

- A.
$\frac{16\pi}{23\sqrt{3}}$x

- B.
$\frac{42\pi}{5\sqrt{3}}$x

- C.
$\frac{2\pi}{3\sqrt{3}}$x

- D.
2$\sqrt{3}$πx

Answer: Option A

**Explanation** :

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

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

Area of the triangle = $\frac{\sqrt{3}}{16}$y^{2}

Area of the shaded region = $\frac{23\sqrt{3}}{16}$ y^{2} = X

y^{2} = $\frac{16}{23\sqrt{3}}$X

Area of the circle = πy^{2}

Thus, area of the circle in terms of X

= $\frac{16\pi}{23\sqrt{3}}$X

Hence, option (a).

Workspace:

**IIFT 2014 QA | Geometry - Circles**

A right circular cylinder has a radius of 6 and a height of 24. A rectangular solid with a square base and a height of 20, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. If water is then poured into the cylinder such that it reaches the rim, the volume of water is:

- A.
288(π – 5)

- B.
288(2π – 3)

- C.
288(3π – 5)

- D.
None of the above

Answer: Option C

**Explanation** :

Volume of the cylinder = π × 6^{2} × 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).

Workspace:

**IIFT 2012 QA | Geometry - Circles**

In a circular field, there is a rectangular tank of length 130 m and breadth 110 m. If the area of the land portion of the field is 20350 m2 then the radius of the field is

- A.
85 m

- B.
95 m

- C.
105 m

- D.
115 m

Answer: Option C

**Explanation** :

Area of a circular field = π*r*^{2}

Area of tank = 130 × 110 = 14300 m^{2}

∴ π*r*^{2 }– 14300 = 20350

∴ π*r*^{2 }= 34650

*r*^{2} = 1575 × 7 = 225 × 49

*r* = 15 × 7 = 105 m

Hence,** **option 3.

Workspace:

**IIFT 2012 QA | Geometry - Circles**

A hemispherical bowl is filled with hot water to the brim. The contents of the bowl are transferred into a cylindrical vessel whose radius is 50% more than its height. If diameter of the bowl is the same as that of the vessel, the volume of the hot water in the cylindrical vessel is

- A.
60% of the cylindrical vessel

- B.
80% of the cylindrical vessel

- C.
100% of the cylindrical vessel

- D.
None of the above

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 × (3*h*)^{3} = 18π*h*^{3}

Volume of the cylindrical vessel = π × (3*h*)^{2} × (2*h*) = 18π*h*^{3}

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

Hence,** **option 3.

Workspace:

**IIFT 2012 QA | Geometry - Circles**

Eight points lie on the circumference of a circle. The difference between the number of triangles and the number of quadrilaterals that can be formed by connecting these points is

- A.
7

- B.
14

- C.
32

- D.
84

Answer: Option B

**Explanation** :

The number of triangles formed using the 8 points

= ^{8}C_{3} = 56

The number of quadrilaterals formed using the 8 points = ^{8}C_{4 }= 70

∴ The difference = 14

Hence, option (b).

Workspace:

**IIFT 2011 QA | Geometry - Circles**

A rectangular piece of paper is 22 cm. long and 10 cm. wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder.

- A.
175 cm

^{3} - B.
180 cm

^{3} - C.
185 cm

^{3} - D.
None of the above

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 = πr^{2}h = $\frac{22}{7}$ × ${\left(\frac{7}{2}\right)}^{2}$ × 10

= 385 cm^{3}

Hence, option (d).

Workspace:

**IIFT 2011 QA | Geometry - Circles**

Consider the volumes of the following objects and arrange them in decreasing order:

i. A parallelepiped of length 5 cm, breadth 3 cm and height 4 cm

ii. A cube of each side 4 cm.

iii. A cylinder of radius 3 cm and length 3 cm

iv. A sphere of radius 3 cm

- A.
iv, iii, ii, i

- B.
iv, ii, iii, i

- C.
iv, iii, i, ii

- D.
None of the above

Answer: Option A

**Explanation** :

i. Volume of parallelepiped = 5 × 4 × 3 = 60 cm^{3}

ii. Volume of cube = 4 × 4 × 4 = 64 cm^{3}

iii. Volume of cylinder = πr^{2}h = π × 3 × 3 × 3 > 81 cm^{3}

iv. Volume of sphere = 4/3 × πr^{3} = 36π cm^{3} > 81 cm^{3}

∴ Required order is iv, iii, ii, i.

Hence, option (a).

Workspace:

**IIFT 2011 QA | Geometry - Circles**

In a circle, the height of an arc is 21 cm and the diameter is 84 cm. Find the chord of ‘half of the arc’

- A.
45 cm

- B.
40 cm

- C.
42 cm

- D.
None of the above

Answer: Option C

**Explanation** :

The given scenario can be depicted in the diagram below.

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).

Workspace:

**IIFT 2010 QA | Geometry - Circles**

In a rocket shape firecracker, explosive powder is to be filled up inside the metallic enclosure. The metallic enclosure is made up of a cylindrical base and conical top with the base of radius 8 centimeter. The ratio of height of cylinder and cone is 5:3. A cylindrical hole is drilled through the metal solid with height one third the height of the metal solid. What should be the radius of the hole, so that volume of the hole (in which gun powder is to be filled up) is half of the volume of metal solid after drilling?

- A.
4$\sqrt{3}$ cm

- B.
4.0 cm

- C.
3.0 cm

- D.
None of these

Answer: Option A

**Explanation** :

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 × 8^{2} × 5h + $\frac{1}{3}$ × π × 8^{2} × 3h = 3πr^{2} $\left(\frac{8h}{3}\right)$

∴ 8^{2} × 6 = 8r^{2}

∴ r^{2} = 8 × 6

∴ r = 4$\sqrt{3}$ cm

Hence, option (a).

Workspace:

**IIFT 2010 QA | Geometry - Circles**

What is the value of *c*^{2} in the given figure, where the radius of the circle is ‘*a*’ unit.

- A.
c

^{2}= a^{2}+ b^{2}– 2ab cos θ - B.
c

^{2}= a^{2}+ b^{2}– 2ab sin θ - C.
c

^{2}= a^{2}– b^{2}+ 2ab cos θ - D.
None of these

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 = a^{2} + b^{2} – c^{2}

∴ c^{2} = a^{2} + b^{2} - 2ab cos θ

Hence, option (a).

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**