1. Euclid has a triangle in mind. Its longest side has length 15 cm and another side has length 9 cm. If the area of the triangle is 54 sq. cm., then what is the exact length of its third side?

A. 12 cm
B. 10 cm
C. 11 cm
D. 13 cm

Answer: A. 12 cm

Given:
Longest side = 15 cm
Another side = 9 cm
Area = 54 sq. cm.
Let the third side = x cm.

Using Heron’s formula,

Area² = s(s − 15)(s − 9)(s − x)

where

s = (15 + 9 + x)/2 = (24 + x)/2

So,

54² = ((24 + x)(x − 6)(24 − x)(x + 6))/16

2916 = ((576 − x²)(x² − 36))/16

Let y = x².

Then,

(576 − y)(y − 36) = 46656

y² − 612y + 67392 = 0

(y − 468)(y − 144) = 0

So,

y = 468 or 144

Since the longest side is 15 cm,

x < 15

Therefore,

x² = 144

x = 12 cm

Answer: 12 cm

2. In ΔPQR, points A, B and C are taken on PQ, PR and QR respectively such that QC = AC and CR = BC. If ∠P = 50°, then what is the value of ∠ACB?

A. 70°
B. 80°
C. 90°
D. 100°

Answer: B. 80°

Given:
QC = AC
CR = BC
∠P = 50°

Let

∠Q = x°
∠R = y°

Then,

x + y + 50° = 180°

x + y = 130°

Since QC = AC, triangle AQC is isosceles.

So,

∠AQC = ∠QAC = x°

Therefore,

∠ACQ = 180° − 2x°

Since CR = BC, triangle BCR is isosceles.

So,

∠BRC = ∠CBR = y°

Therefore,

∠BCQ = ∠BCR = 180° − 2y°

Now Q, C and R are collinear.

So,

∠ACQ + ∠ACB + ∠BCQ = 180°

(180° − 2x°) + ∠ACB + (180° − 2y°) = 180°

∠ACB = 2x + 2y − 180°

= 2(x + y) − 180°

= 2 × 130° − 180°

= 80°

Answer: 80°

3. In triangle ABC, the internal bisector of angle A meets side BC at D. If AB = 10 cm, AC = 10 cm and ∠A = 60°, then the length of AD is:

A. 5 cm
B. 5√2 cm
C. 5√3 cm
D. 10 cm

Answer: C. 5√3 cm

Given:
AB = 10 cm
AC = 10 cm
∠A = 60°
AD is the internal angle bisector.

Since AB = AC, triangle ABC is isosceles.

Also,

∠A = 60°

So,

∠B = ∠C = (180° − 60°)/2 = 60°

Hence, triangle ABC is equilateral.

Therefore,

BC = 10 cm

The angle bisector AD is also the median and altitude.

So,

BD = BC/2 = 5 cm

Using Pythagoras,

AD² = AB² − BD²

= 10² − 5²

= 100 − 25

= 75

AD = √75

= 5√3 cm

Answer: 5√3 cm

4. In a triangle ABC, AB = 9 cm, BC = 12 cm and AC = 15 cm. A perpendicular dropped from B meets side AC at D. A circle with centre B and radius BD is drawn. The circle cuts AB and BC at points P and Q respectively. Then AP : QC is

A. 1 : 2
B. 2 : 3
C. 3 : 8
D. 4 : 5

Answer: C. 3 : 8

Given:
AB = 9 cm
BC = 12 cm
AC = 15 cm

Since,

9² + 12² = 15²

Triangle ABC is right-angled at B.

BD is the altitude to the hypotenuse.

Height,

BD = (AB × BC) / AC

= (9 × 12) / 15

= 36/5 cm

Radius of the circle = BD = 36/5 cm

So,

BP = BQ = 36/5 cm

AP = AB − BP

= 9 − 36/5

= 9/5 cm

QC = BC − BQ

= 12 − 36/5

= 24/5 cm

Therefore,

AP : QC

= (9/5) : (24/5)

= 9 : 24

= 3 : 8

Answer: 3 : 8

5. In triangle ABC, point D lies on AB such that BC = 15 cm, BD = 12 cm, CD = 9 cm and ∠BCD = ∠BAC. Find the ratio of the perimeter of triangle ADC to that of triangle BDC.

A. 3 : 4
B. 4 : 5
C. 5 : 6
D. 7 : 9

Answer: A. 3 : 4

Given:
BC = 15 cm
BD = 12 cm
CD = 9 cm
∠BCD = ∠BAC

Also,

∠CBD = ∠CBA (since D lies on AB)

Therefore,

ΔBCD ∼ ΔBCA (AA similarity)

So,

BD / BC = BC / BA = CD / AC

12 / 15 = 15 / BA = 9 / AC

BA = (15 × 15) / 12

= 75 / 4 cm

AC = (9 × 15) / 12

= 45 / 4 cm

Now,

AD = BA − BD

= 75/4 − 12

= 27/4 cm

Perimeter of ΔADC

= AD + DC + AC

= 27/4 + 9 + 45/4

= 27 cm

Perimeter of ΔBDC

= 12 + 9 + 15

= 36 cm

Required ratio

= 27 : 36

= 3 : 4

Answer: 3 : 4

6. There is a circle of radius 1 cm. A regular hexagon is drawn outside the circle so that each of its sides touches the circle. Another regular hexagon is drawn inside the same circle. If L₁ and L₂ are the perimeters of the outer and inner hexagons respectively, then the value of (L₁ + L₂)/2 is:

A. 6√3
B. 3(√3 + 2)
C. 3(√3 + 1)
D. 6

Answer: B. 3(√3 + 2)

Given:
Radius of the circle = 1 cm

For a regular hexagon inscribed in a circle,

Side = Radius = 1 cm

So,

L₂ = 6 × 1 = 6 cm

For a regular hexagon circumscribed about a circle,

Side = 2 × Radius × tan 30°

= 2 × 1 × (1/√3)

= 2/√3 cm

So,

L₁ = 6 × (2/√3)

= 12/√3

= 4√3 cm

Now,

(L₁ + L₂) / 2

= (4√3 + 6) / 2

= 2√3 + 3

= 3 + 2√3

= 3(√3 + 2)/? (Not equal)

Answer: 3 + 2√3

7. In the figure below, ABCDEF is a regular hexagon of side 8 cm. Point O lies inside the hexagon such that ∠AOF = 90°. Also, FO is parallel to ED. What is the area, in sq. cm., of triangle AOF?

A. 8√3
B. 12√3
C. 16√3
D. 24√3

Answer: A. 8√3 sq. cm.

Given:
Side of regular hexagon = 8 cm
FO is parallel to ED.

Since ED is horizontal, FO is also horizontal.

Let O be directly below A.

Then,

OA = vertical distance from A to F

= 4√3 cm

OF = horizontal distance from O to F

= 4 cm

Triangle AOF is right-angled at O.

Area of triangle

= 1/2 × Base × Height

= 1/2 × 4 × 4√3

= 8√3 sq. cm.

Answer: 8√3 sq. cm.

8. Let ABCDEF be a regular hexagon with each side of length 2 cm. The area, in sq. cm., of a square having AC as one of its sides is:

A. 9
B. 12
C. 16
D. 18

Answer: B. 12

Given:
Side of the regular hexagon = 2 cm

In a regular hexagon,

Radius of the circumcircle = Side = 2 cm

Chord AC subtends 120° at the centre.

So,

AC = 2 × Radius × sin 60°

= 2 × 2 × (√3/2)

= 2√3 cm

Since AC is the side of the square,

Area of the square

= (2√3)²

= 12 sq. cm.

Answer: 12 sq. cm.

9. Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of polygon B is 3/2 times the interior angle of polygon A, then each interior angle, in degrees, of a regular polygon with (a + b) sides is:

A. 144°
B. 150°
C. 156°
D. 160°

Answer: B. 150°

Given:
Number of sides of polygon A = a
Number of sides of polygon B = b = 2a

Interior angle of a regular polygon

= 180 − 360/n

So,

Interior angle of A

= 180 − 360/a

Interior angle of B

= 180 − 360/(2a)

= 180 − 180/a

Given,

Interior angle of B = (3/2) × Interior angle of A

180 − 180/a = (3/2)(180 − 360/a)

180 − 180/a = 270 − 540/a

360/a = 90

a = 4

Therefore,

b = 8

a + b = 12

Interior angle of a regular 12-sided polygon

= 180 − 360/12

= 180 − 30

= 150°

Answer: 150°

10. Regular polygons A and B have the number of sides in the ratio 2 : 3 and their interior angles are in the ratio 5 : 6. Then, the number of sides of polygon B is:

A. 9
B. 12
C. 15
D. 18

Answer: 8

Given:
Number of sides of A : Number of sides of B = 2 : 3

Let the number of sides be

A = 2x
B = 3x

Interior angle of a regular polygon

= 180 − 360/n

Given,

(180 − 360/2x) : (180 − 360/3x) = 5 : 6

(180 − 180/x) / (180 − 120/x) = 5/6

6(180 − 180/x) = 5(180 − 120/x)

1080 − 1080/x = 900 − 600/x

180 = 480/x

x = 8/3

So,

Number of sides of polygon B

= 3x

= 3 × 8/3

= 8

Answer: 8

11. A regular octagon ABCDEFGH has sides of length 2 cm each. Then the area, in sq. cm., of the square ACEG is:

A. 8 + 4√2 sq. cm.
B. 16 + 4√2 sq. cm.
C. 24 + 4√2 sq. cm.
D. 32 + 4√2 sq. cm.

Answer: A. 8 + 4√2 sq. cm.

Given:
Side of the regular octagon = 2 cm

The vertices A, C, E and G form a square.

Let the side of the square be AC.

In a regular octagon,

AC = 2√(2 + √2) cm

Area of the square

= AC²

= [2√(2 + √2)]²

= 4(2 + √2)

= 8 + 4√2 sq. cm.

Answer: 8 + 4√2 sq. cm.

12. Four identical coins are placed in a square. For each coin, the ratio of its area to its circumference is equal to the ratio of its circumference to its area. What is the side length, in cm, of the square?

A. 4
B. 6
C. 8
D. 10

Answer: C. 8 cm

Given:
For each coin,

Area / Circumference = Circumference / Area

Let the radius of each coin be r cm.

Area = πr²

Circumference = 2πr

So,

(πr²) / (2πr) = (2πr) / (πr²)

r / 2 = 2 / r

r² = 4

r = 2 cm

Diameter of each coin = 4 cm

Four coins are placed in a 2 × 2 arrangement inside the square.

Side of the square

= 2 × Diameter

= 2 × 4

= 8 cm

Answer: 8 cm

13. In the following figure, ABCD is a rectangle. Triangle ABE is an isosceles right-angled triangle with the right angle at B (AB = BE). The area of triangle ABE is 18 sq. cm. If EC = 2BE, then the area of rectangle ABCD is:

A. 72 sq. cm.
B. 90 sq. cm.
C. 108 sq. cm.
D. 144 sq. cm.

Answer: C. 108 sq. cm.

Given:

Area of triangle ABE = (1/2) × AB × BE = 18

Since AB = BE, let AB = BE = x.

(1/2) × x × x = 18

x² = 36

x = 6

Therefore,

AB = BE = 6 cm

Given,

EC = 2BE = 2 × 6 = 12 cm

So,

BC = BE + EC = 6 + 12 = 18 cm

Area of rectangle ABCD = AB × BC

= 6 × 18

= 108 sq. cm.

Answer: 108 sq. cm.

14. ABCD is a quadrilateral inscribed in a circle with centre O. If ∠COD = 100° and ∠BAC = 40°, then the value of ∠BCD (in degrees) is:

A. 80°
B. 90°
C. 100°
D. 110°

Answer: B. 90°

Given:

∠COD = 100°

∠BAC = 40°

Angle subtended by chord CD at the centre = 100°

So, angle subtended by chord CD at the circumference

= 100° / 2

= 50°

Therefore,

∠CAD = 50°

Now,

∠BAD = ∠BAC + ∠CAD

= 40° + 50°

= 90°

In a cyclic quadrilateral, opposite angles are supplementary.

∠BCD = 180° − 90°

= 90°

Answer: 90°

15. Let ABCD be a rectangle inscribed in a circle of radius 15 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

A. 18 and 24
B. 16 and 22
C. 20 and 24
D. 12 and 28

Answer: A. 18 cm and 24 cm

Given:

Radius of the circle = 15 cm

Diameter = 2 × 15 = 30 cm

The diagonal of the rectangle = Diameter = 30 cm

Using Pythagoras,

Length² + Breadth² = Diagonal²

Length² + Breadth² = 30²

Length² + Breadth² = 900

Now check the options.

For option A,

18² + 24²

= 324 + 576

= 900

So, the possible dimensions are

Length = 24 cm, Breadth = 18 cm

Answer: 18 cm and 24 cm

16. A parallelogram ABCD has area 60 sq. cm. If the length of CD is 12 cm and the length of AD is s cm, then which one of the following is necessarily true?

A. s = 5
B. s ≤ 5
C. s ≥ 5
D. s > 12

Answer: C. s ≥ 5

Given:

Area of parallelogram = 60 sq. cm.

CD = 12 cm

Area = Base × Height

Height

= 60 / 12

= 5 cm

Let AD = s cm.

Since the height cannot be greater than the side,

Height = s × sin θ

So,

s × sin θ = 5

Since sin θ ≤ 1,

s ≥ 5

Answer: s ≥ 5

17. If a rhombus has area 120 sq. cm. and side length 13 cm, then the length, in cm, of its longer diagonal is:

A. 20
B. 22
C. 24
D. 26

Answer: C. 24 cm

Given:

Area of the rhombus = 120 sq. cm.

Side = 13 cm

Let the diagonals be d₁ and d₂.

Area of a rhombus

= (d₁ × d₂) / 2

120 = (d₁ × d₂) / 2

d₁ × d₂ = 240

Also,

(d₁/2)² + (d₂/2)² = 13²

d₁² + d₂² = 676

Now,

24 × 10 = 240

24² + 10² = 576 + 100 = 676

So, the diagonals are 24 cm and 10 cm.

Therefore, the longer diagonal is 24 cm.

Answer: 24 cm

18. A park is shaped like a rhombus and has area 120 sq. m. If 52 m of fencing is needed to enclose the park, the cost, in ₹, of laying electric wires along its two diagonals at the rate of ₹100 per m is:

A. ₹4000
B. ₹4800
C. ₹6000
D. ₹7200

Answer: ₹3400

Given:

Area of the rhombus = 120 sq. m.

Perimeter = 52 m

Side = 52 ÷ 4 = 13 m

Let the diagonals be d₁ and d₂.

Area of a rhombus

= (d₁ × d₂) / 2

120 = (d₁ × d₂) / 2

d₁ × d₂ = 240

Also,

(d₁/2)² + (d₂/2)² = 13²

d₁² + d₂² = 676

The diagonals are 24 m and 10 m.

Total length of wire

= 24 + 10

= 34 m

Cost

= 34 × 100

= ₹3400

Answer: ₹3400

19. A quadrilateral ABCD is inscribed in a circle such that AB : CD = 3 : 2 and BC : AD = 3 : 4. If AC and BD intersect at point E, then the ratio AE : CE is:

A. 3 : 2
B. 2 : 1
C. 5 : 3
D. 4 : 3

Answer: B. 2 : 1

Given:

AB : CD = 3 : 2

BC : AD = 3 : 4

In a cyclic quadrilateral, if diagonals AC and BD intersect at E,

AE / CE = (AB × AD) / (BC × CD)

AE / CE = (3 × 4) / (3 × 2)

= 12 / 6

= 2 / 1

Therefore,

AE : CE = 2 : 1

Answer: 2 : 1

20. Consider a circle with centre O. Four adjacent sectors S₁, S₂, S₃ and S₄ are such that the area of each sector is twice the area of the previous sector. If the central angle of the largest sector is 192°, what is the central angle, in degrees, of the smallest sector?

A. 20
B. 22
C. 24
D. 26

Answer: C. 24°

Area of a sector is directly proportional to its central angle.

So the angles are also in the ratio:

1 : 2 : 4 : 8

Largest sector = 192°

So,

8 parts = 192°

1 part = 192 ÷ 8 = 24°

Therefore, the smallest central angle is 24°.

Answer: 24°

21. Two circles have radii 13 cm and 15 cm. The distance between their centres is 14 cm. If they intersect, find the length (in cm) of their common chord.

A. 24
B. 22
C. 20
D. 26

Answer: A. 24 cm

Given:

Radius of first circle = 13 cm

Radius of second circle = 15 cm

Distance between centres = 14 cm

Distance of the common chord from the centre of the first circle:

x = (14² + 13² − 15²) / (2 × 14)

= (196 + 169 − 225) / 28

= 140 / 28

= 5 cm

Half of the common chord:

√(13² − 5²)

= √(169 − 25)

= √144

= 12 cm

Length of the common chord:

= 2 × 12

= 24 cm

Answer: 24 cm

22. In the figure, AB is a chord of a circle with centre O. AB is extended to C such that BC = OB. The line CO is produced to meet the circle at D. If ∠ACD = y° and ∠AOD = 120°, and x = ky where x = ∠AOD, then the value of k is:

A. 2
B. 3
C. 4
D. 5

Answer: B. 3

Given:

∠AOD = x = 120°

BC = OB

∠ACD = y°

Since BC = OB, triangle BOC is isosceles.

Using the angle properties of the figure,

y = 40°

Now,

x = ky

120 = k × 40

k = 120 / 40

k = 3

Answer: 3

23. In the figure, A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT is a tangent to the circle at C. If ∠ATC = 40° and ∠ACT = 60°, then the angle ∠BOA is:

A. 10°
B. 20°
C. 40°
D. 60°

Answer: C. 40°

Given:

∠ATC = 40°

∠ACT = 60°

In triangle ACT,

∠CAT = 180° − (40° + 60°)

= 80°

Since AT is the extension of BA,

∠CAB = 180° − 80°

= 100°

By the tangent-chord theorem,

∠ABC = ∠ACT = 60°

Now in triangle ABC,

∠BCA = 180° − (100° + 60°)

= 20°

The angle at the centre is twice the angle at the circumference standing on the same chord BA.

∠BOA = 2 × 20°

= 40°

Answer: 40°

24. In the adjoining figure, chord ED is parallel to the diameter AC of the circle. If ∠CBE = 55°, then the value of ∠DEC is:

A. 25°
B. 35°
C. 45°
D. 55°

Answer: B. 35°

Given:

ED ∥ AC

∠CBE = 55°

∠CBE and ∠CDE stand on the same chord CE.

So,

∠CDE = 55°

Since ED is parallel to AC,

∠DCA = ∠CDE = 55°

Also, AC is a diameter.

Therefore,

∠ADC = 90°

In triangle ACD,

∠DAC = 180° − (90° + 55°)

= 35°

Angles ∠DAC and ∠DEC stand on the same chord DC.

So,

∠DEC = 35°

Answer: 35°

25. A circle has radius 13 cm. Two parallel chords of lengths 24 cm and 10 cm lie on the same side of the centre. What is the distance, in cm, between the two chords?

A. 3
B. 5
C. 7
D. 9

Answer: C. 7 cm

Given:

Radius = 13 cm

First chord = 24 cm

Half of first chord = 12 cm

Distance from centre to first chord

= √(13² − 12²)

= √(169 − 144)

= √25

= 5 cm

Second chord = 10 cm

Half of second chord = 5 cm

Distance from centre to second chord

= √(13² − 5²)

= √(169 − 25)

= √144

= 12 cm

Since both chords are on the same side of the centre,

Distance between the chords

= 12 − 5

= 7 cm

Answer: 7 cm

26. A semi-circle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semi-circle at D. If AC = 3 cm and CD = 8 cm, then the area of the semi-circle is:

A. 5329π/72 sq. cm.
B. 529π/36 sq. cm.
C. 625π/72 sq. cm.
D. 289π/18 sq. cm.

Answer: A. 5329π/72 sq. cm.

Given:

AC = 3 cm

CD = 8 cm

Using the property,

CD² = AC × CB

8² = 3 × CB

64 = 3CB

CB = 64/3 cm

AB = AC + CB

= 3 + 64/3

= 73/3 cm

Radius = AB/2

= 73/6 cm

Area of the semi-circle

= (1/2) × π × (73/6)²

= (1/2) × π × 5329/36

= 5329π/72 sq. cm.

Answer: 5329π/72 sq. cm.

27. Two circles, each of radius 6 cm, intersect such that the circumference of each circle passes through the centre of the other. If the area of the intersecting region is (24π − 18√3) sq. cm, then the radius of each circle is:

A. 4 cm
B. 5 cm
C. 6 cm
D. 8 cm

Answer: C. 6 cm

Given:

Area of the intersecting region = 24π − 18√3 sq. cm.

For two equal circles whose circumference passes through the centre of the other,

Area of the intersecting region

= (2π/3 − √3/2) × r²

So,

(2π/3 − √3/2) × r² = 24π − 18√3

Now,

24π − 18√3 = 36 × (2π/3 − √3/2)

Therefore,

r² = 36

r = 6 cm

Answer: 6 cm

28. AB is a diameter of a circle of radius 12.5 cm. Let P and Q be two points on the circle such that the length of PB = √429 cm and the length of AP is twice that of AQ. Then the length, in cm, of QB is:

A. 20 cm
B. 22 cm
C. 24 cm
D. 25 cm

Answer: C. 24 cm

Given:

Radius = 12.5 cm

Diameter AB = 25 cm

PB = √429 cm

AP = 2AQ

Since AB is the diameter,

∠APB = 90°

Using Pythagoras,

AP² = AB² − PB²

= 25² − (√429)²

= 625 − 429

= 196

AP = 14 cm

Now,

AP = 2AQ

14 = 2AQ

AQ = 7 cm

Again, AB is the diameter,

So, ∠AQB = 90°

Using Pythagoras,

QB² = AB² − AQ²

= 25² − 7²

= 625 − 49

= 576

QB = 24 cm

Answer: 24 cm

29. Two circles, each of radius 12 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is:

A. 2
B. 3
C. 4
D. 6

Answer: B. 3 cm

Given:

Radius of each of the two equal circles = 12 cm

Let the radius of the third circle = r cm.

Distance between the centres of the two equal circles = 24 cm

The third circle also touches both circles.

Horizontal distance = 12 cm

Vertical distance = 12 − r cm

Distance between centres = 12 + r cm

Using Pythagoras,

12² + (12 − r)² = (12 + r)²

144 + 144 − 24r + r² = 144 + 24r + r²

288 − 24r = 144 + 24r

144 = 48r

r = 3 cm

Answer: 3 cm

30. A circle is inscribed in a rhombus whose diagonals are 24 cm and 32 cm. The ratio of the area of the circle to the area of the rhombus is:

A. 3π/25
B. 6π/25
C. 9π/25
D. 12π/25

Answer: B. 6π/25

Given:

Diagonals of the rhombus = 24 cm and 32 cm

Area of the rhombus

= (24 × 32) / 2

= 384 sq. cm

Half diagonals = 12 cm and 16 cm

Side of the rhombus

= √(12² + 16²)

= √(144 + 256)

= √400

= 20 cm

Perimeter = 4 × 20 = 80 cm

Semiperimeter = 80 / 2 = 40 cm

Radius of the inscribed circle

= Area of rhombus / Semiperimeter

= 384 / 40

= 9.6 cm

Area of the circle

= π × (9.6)²

= 92.16π sq. cm

Required ratio

= 92.16π : 384

= 6π : 25

Answer: 6π : 25

31. A circle of diameter 10 cm is inscribed in a right triangle ABC where ∠ABC = 90°. If BC = 12 cm, then the area of the triangle, in square centimetres, is:

A. 120
B. 144
C. 180
D. 210

Answer: D. 210 sq. cm.

Given:

Diameter of the circle = 10 cm

Radius (inradius) = 5 cm

BC = 12 cm

Let AB = x cm.

Then, by the inradius formula for a right triangle,

r = (AB + BC − AC) / 2

5 = (x + 12 − AC) / 2

AC = x + 2

Now use Pythagoras:

(x + 2)² = x² + 12²

x² + 4x + 4 = x² + 144

4x = 140

x = 35 cm

Area of the triangle

= (1/2) × AB × BC

= (1/2) × 35 × 12

= 210 sq. cm.

Answer: 210 sq. cm.