Search Previous Year Questions

Shortcut Trick

The unequal side (base) b = 36 − (2 × 15) = 6 cm.

Divide the triangle into two right-angled triangles; the semi-base is 3 cm.

Height (h) = √(15² − 3²) = √(225 − 9) = √216 = 6√6 cm.

Area = (1/2) × base × height = (1/2) × 6 × 6√6 = 18√6 cm².

∴ The correct answer is 18√6 cm².

Alternate Method

Given:

Perimeter of the isosceles triangle = 36 cm

Length of equal sides (a) = 15 cm

Formula Used:

Heron's Formula: Area = √(s × (s − a) × (s − b) × (s − c))

Where s is the semi-perimeter = (a + b + c) ÷ 2

Calculations:

⇒ Let the sides be a = 15 cm, b = 15 cm, and c (unequal side).

⇒ c = 36 − (15 + 15) = 6 cm

⇒ Semi-perimeter (s) = 36 ÷ 2 = 18 cm

⇒ Area = √(18 × (18 − 15) × (18 − 15) × (18 − 6))

⇒ Area = √(18 × 3 × 3 × 12)

⇒ Area = √(162 × 12) = √1944

⇒ Area = √(324 × 6) = 18√6 cm²

∴ The correct answer is 18√6 cm².

Additional Information

Area of Isosceles Triangle

If equal sides are 'a' and base is 'b', Area = (b/4) × √(4a² − b²).

Altitude Property

In an isosceles triangle, the altitude drawn from the vertex to the unequal base bisects the base at 90°.

Heron's Formula

Used for any triangle with sides a, b, c: Area = √(s(s−a)(s−b)(s−c)) where s = (a+b+c)/2.

Shortcut Trick

The circumference (C) of a circle is directly proportional to its radius (r) because C = 2πr.

In any linear relationship (where power of variable is 1), the percentage change in the result is exactly the same as the percentage change in the variable.

Since the radius increases by 20%, the circumference must also increase by 20%.

∴ The correct answer is 20%.

Alternate Method

Given:

Percentage increase in radius = 20%

Formula Used:

Circumference of a circle (C) = 2 × π × r

Calculations:

⇒ Let the initial radius of the circle be r.

⇒ Initial circumference (C₁) = 2 × π × r

⇒ New radius (r_new) = r + 20% of r = r + 0.2r = 1.2r

⇒ New circumference (C₂) = 2 × π × (1.2r) = 2.4πr

⇒ Increase in circumference = C₂ − C₁ = 2.4πr − 2πr = 0.4πr

⇒ Percentage Change = (Increase / Original Circumference) × 100

⇒ Percentage Change = (0.4πr / 2πr) × 100 = 0.2 × 100 = 20%

∴ The correct answer is 20%.

Additional Information

Change in Area of Circle

Area (A) = πr². If the radius increases by x%, the Area increases by (2x + x²/100)%. For 20%, Area increase = 44%.

Linear Dimension Property

In any 2D figure, if all linear dimensions (radius, side, diagonal, perimeter) are scaled by k%, the perimeter or circumference also changes by k%.

Ratio Method

If radius ratio is r₁:r₂, then circumference ratio is also r₁:r₂, but area ratio is r₁²:r₂².

Shortcut Trick

Area of a square = (side)² = 28 × 28 = 784 m².

Area of the inscribed circle = (π/4) × Area of Square.

Remaining Area = (1 − π/4) × Area of Square.

Remaining Area = (1 − 3.14/4) × 784 = (1 − 0.785) × 784 = 0.215 × 784 = 168.56 m².

∴ The correct answer is 168.56 m².

Alternate Method

Given:

Side of square courtyard (s) = 28 m

Value of π = 3.14

Formula Used:

Area of Square = s²

Area of Circle = πr²

Where r is the radius of the circle.

Calculations:

⇒ Area of the square courtyard = 28 × 28 = 784 m²

⇒ Since the circle touches all four sides, its diameter = side of the square = 28 m

⇒ Radius (r) of the circular fountain = 28 ÷ 2 = 14 m

⇒ Area of the circular fountain = 3.14 × 14 × 14

⇒ Area of the circular fountain = 3.14 × 196 = 615.44 m²

⇒ Remaining area = Area of square − Area of circle

⇒ Remaining area = 784 − 615.44 = 168.56 m²

∴ The correct answer is 168.56 m².

Additional Information

Inscribed Circle (Incircle) of a Square

When a circle is inscribed in a square of side 'a', its radius r = a/2 and its area is πa²/4.

Circumscribed Circle of a Square

When a circle passes through all vertices of a square of side 'a', its radius R = a/√2 and its area is πa²/2.

Area of Remaining Corners

The area of the four corners outside the incircle of a square is approximately 21.5% of the total square area (when π ≈ 3.14).

Shortcut Trick

Angle in radians is directly given by the ratio of Arc length to Radius.

θ = Arc / Radius

θ = 10 / 75 = 2/15 rad

∴ The correct answer is 2/15 rad.

Alternate Method

Given:

Length of the pendulum (Radius, r) = 75 cm

Length of the arc (s) = 10 cm

Formula Used:

Arc length (s) = Radius (r) × Angle (θ) [where θ is in radians]

Calculations:

⇒ s = r × θ

⇒ 10 = 75 × θ

⇒ θ = 10 / 75

⇒ θ = 2 / 15 rad

∴ The correct answer is 2/15 rad.

Additional Information

Radian to Degree Conversion

To convert radians to degrees, multiply the value by 180/π. Conversely, to convert degrees to radians, multiply by π/180.

Area of a Sector

The area of the sector formed by the swing is given by the formula A = 1/2 × r² × θ, where θ is in radians.

Pendulum Motion

For small angles, the motion of a simple pendulum is approximately Simple Harmonic Motion (SHM), and the length of the string acts as the radius of the circular path.

Shortcut Trick

When a solid is melted into n equal smaller solids, the relation between the radii is: R = r × n^1/3.

Here, R = 8 cm and n = 8.

8 = r × 8^1/3

8 = r × 2

r = 8 ÷ 2 = 4 cm.

∴ The correct answer is 4 cm.

Alternate Method

Given:

Radius of the original large hemisphere (R) = 8 cm

Number of smaller hemispheres (n) = 8

Formula Used:

Volume of a hemisphere = (2/3) × π × radius³

Calculations:

The total volume remains constant during recasting.

⇒ Volume of Original Hemisphere = 8 × Volume of each Smaller Hemisphere

⇒ (2/3) × π × (8)³ = 8 × (2/3) × π × r³

⇒ 512 = 8 × r³

⇒ r³ = 512 ÷ 8

⇒ r³ = 64

⇒ r = ³√64

⇒ r = 4 cm

∴ The correct answer is 4 cm.

Additional Information

Total Surface Area of Hemisphere

The total surface area (TSA) includes the curved base and the flat circular top: TSA = 3 × π × r².

Curved Surface Area of Hemisphere

The area of the curved part only is exactly half of a full sphere: CSA = 2 × π × r².

Conservation of Volume

When any 3D shape is melted and recast into another shape, the total volume remains unchanged provided there is no wastage.

Volume of a Sphere

A full sphere has exactly double the volume of a hemisphere of the same radius: V = (4/3) × π × r³.

Shortcut Trick

For a hollow hemisphere with outer radius R and inner radius r:

Total Surface Area (TSA) = π(3R² + r²)

Given: R = 10 cm, r = 6 cm, π = 22/7

TSA = (22/7) × (3 × 10² + 6²) = (22/7) × (300 + 36) = (22/7) × 336

TSA = 22 × 48 = 1056 cm²

∴ The correct answer is 1056 cm².

Alternate Method

Given:

Outer radius (R) = 10 cm

Inner radius (r) = 6 cm

Value of π = 22/7

Formula Used:

1. Outer Curved Surface Area = 2πR²
2. Inner Curved Surface Area = 2πr²
3. Area of ring-shaped base = π(R² − r²)

Calculations:

⇒ Total Surface Area = Outer CSA + Inner CSA + Base Area

⇒ TSA = 2πR² + 2πr² + π(R² − r²)

⇒ TSA = π [2R² + 2r² + R² − r²]

⇒ TSA = π [3R² + r²]

⇒ TSA = (22/7) × [3(10)² + (6)²]

⇒ TSA = (22/7) × [300 + 36]

⇒ TSA = (22/7) × 336

⇒ TSA = 22 × 48

⇒ TSA = 1056 cm²

∴ The correct answer is 1056 cm².

Additional Information

Volume of Hollow Hemisphere

The volume of the material in a hollow hemisphere is (2/3)π(R³ − r³), where R is the outer radius and r is the inner radius.

Surface Area of Solid Hemisphere

For a solid hemisphere of radius r, the Curved Surface Area is 2πr² and the Total Surface Area (including flat base) is 3πr².

Area of a Ring (Annulus)

The area between two concentric circles is π(R² − r²). In a hollow hemisphere, this forms the area of the base ring.

Shortcut Trick

The Total Surface Area (TSA) of a pyramid is simply the sum of its Lateral Surface Area (LSA) and the Base Area.

LSA = TSA − Base Area

LSA = 784 − 400 = 384 cm²

∴ The correct answer is 384 cm².

Alternate Method

Given:

Total Surface Area (TSA) of square pyramid = 784 cm²

Area of the square base = 400 cm²

Formula Used:

Total Surface Area (TSA) = Lateral Surface Area (LSA) + Area of Base

Calculations:

⇒ TSA = LSA + Base Area

⇒ 784 = LSA + 400

⇒ LSA = 784 − 400

⇒ LSA = 384 cm²

∴ The correct answer is 384 cm².

Additional Information

Volume of a Square Pyramid

The volume is given by (1/3) × Base Area × Vertical Height (h). V = (1/3) × a² × h.

Lateral Surface Area (LSA) Formula

LSA = 2 × side (a) × slant height (l), or (1/2) × Perimeter of base × Slant height.

Slant Height Relation

For a square pyramid, the relationship is l² = h² + (a/2)², where l is slant height and h is vertical height.

Shortcut Trick

The ratio of Height to Shadow remains constant for objects at the same time.

Ratio = 6 / 4 = 1.5

Tower Height = 1.5 × Tower Shadow

Tower Height = 1.5 × 28 = 42 m

∴ The correct answer is 42 m.

Alternate Method

Given:

Stick Height (h₁) = 6 m

Stick Shadow (s₁) = 4 m

Tower Shadow (s₂) = 28 m

Formula Used:

Based on the similarity of triangles (AAA criteria), the ratio of corresponding sides is equal:

Height₁ / Shadow₁ = Height₂ / Shadow₂

Calculations:

⇒ 6 / 4 = H / 28

⇒ H = (6 ÷ 4) × 28

⇒ H = 1.5 × 28

⇒ H = 42 m

∴ The correct answer is 42 m.

Additional Information

Similarity of Triangles

Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.

Sun Altitude

At any given time, the Sun's rays strike the ground at the same angle for all objects in the same location, creating similar right-angled triangles.

Properties of Ratios

If a : b = c : d, then the product of extremes equals the product of means (a × d = b × c).

Shortcut Trick

Assertion (A) represents a fundamental Circle Theorem: A line from the center to the midpoint of a chord is always perpendicular (90°) to it.

Reason (R) claims every intersection between a radius and a chord is 90°, which is False because this only happens at the midpoint.

Since the theorem correctly validates A and the generalization in R is incorrect, A is true and R is false.

∴ The correct answer is A is true, but R is false.

Alternate Method

Given: A circle with center O and a chord AB. Let M be the midpoint of AB.

Property Used: SSS (Side-Side-Side) Congruence Criterion.

Step-by-step validation of Assertion (A):

⇒ Consider ΔOAM and ΔOBM.

⇒ OA = OB (Radii of the same circle).

⇒ AM = MB (Given M is the midpoint of chord AB).

⇒ OM = OM (Common side).

⇒ By SSS congruence, ΔOAM ≅ ΔOBM.

⇒ ∴ ∠OMA = ∠OMB (By CPCT).

⇒ Since ∠OMA + ∠OMB = 180° (Linear pair), each angle must be 90°.

⇒ This proves Assertion (A) is true.

Validation of Reason (R):

⇒ A radius can intersect a chord at any point (not just the midpoint).

⇒ If the intersection is not at the midpoint, the angle is not 90°.

⇒ ∴ Reason (R) is false.

∴ The correct answer is A is true, but R is false.

Additional Information

Perpendicular Bisector Property

The perpendicular bisector of any chord of a circle always passes through the center of the circle.

Equal Chords and Distance

Chords of a circle which are equidistant from the center are equal in length.

Angle in a Semicircle

A chord that is also a diameter subtends a right angle (90°) at any point on the circumference.

Radius-Tangent Relationship

A radius is always perpendicular to the tangent at the point of contact, which is different from a chord intersection.

Shortcut Trick

The radius, tangent, and distance from center form a right-angled triangle.

Identify the Pythagorean Triplet: (7, 24, 25).

Given: Radius = 7 cm and Hypotenuse (distance) = 25 cm.

The missing side (tangent length) is 24 cm.

∴ The correct answer is 24 cm.

Alternate Method

Given:

Radius of the circle (OA) = 7 cm

Distance of point P from center (OP) = 25 cm

Formula Used:

In a right-angled triangle ΔOAP, by Pythagoras Theorem:

OP² = OA² + AP²

Calculations:

⇒ AP² = OP² − OA²

⇒ AP² = 25² − 7²

⇒ AP² = 625 − 49

⇒ AP² = 576

⇒ AP = √576

⇒ AP = 24 cm

∴ The correct answer is 24 cm.

Additional Information

Tangent-Radius Property

A tangent at any point of a circle is perpendicular to the radius through the point of contact (Angle OAP = 90°).

Length of Tangents

The lengths of tangents drawn from an external point to a circle are always equal.

Common Pythagorean Triplets

Frequently used triplets in exams include (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), and (9, 40, 41).