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Shortcut Trick

Since the cylinder and hemisphere have equal volumes on the same base radius, their height ratio h : r = 2 : 3.

Total height in terms of ratio units = 2 + 3 = 5 units.

Given total combined height = 30 cm.

Radius of hemisphere (3 units) = (30 ÷ 5) × 3 = 18 cm.

∴ The correct answer is 18 cm.

Alternate Method

Given:

Total combined height = h + r = 30 cm

Volume of the Cylinder = Volume of the Hemisphere

Formula Used:

Volume of Cylinder = πr²h

Volume of Hemisphere = (2/3)πr³

30 cm

h

r

r

Cylinder and Hemisphere

Calculations:

⇒ πr²h = (2 ÷ 3) × πr³

⇒ h = (2 ÷ 3)r

⇒ Substituting the value of h in the combined height equation:

⇒ (2 ÷ 3)r + r = 30

⇒ (5 ÷ 3)r = 30

⇒ r = 30 × 3 ÷ 5

⇒ r = 18 cm

∴ The correct answer is 18 cm.

Additional Information

Volume and Surface Area of a Cylinder

The volume is given by V = πr²h. The total surface area is 2πr(r + h), where r is the base radius and h is the height.

Volume and Surface Area of a Hemisphere

The volume is given by V = (2/3)πr³. The curved surface area is 2πr², and the total surface area is 3πr².

Ratio-Unitary Method in Geometry

When two 3D shapes have equal volumes and identical base radii, equating their volume formulas directly provides the ratio of their heights, significantly speeding up complex fractional calculations.

Shortcut Trick

The volume of any pyramid is directly one-third of the product of its base area and height.

Volume = (1 ÷ 3) × Base Area × Height

Volume = (1 ÷ 3) × 60 × 9 = 180 cm³

∴ The correct answer is 180 cm³.

Alternate Method

Given:

Base Area of the pyramid = 60 cm²

Height of the pyramid = 9 cm

Formula Used:

Volume of a Pyramid = (1 ÷ 3) × Base Area × Height

h = 9 cm Base Area = 60 sq. cm

Calculations:

⇒ Volume = (1 ÷ 3) × 60 × 9

⇒ Volume = 20 × 9

⇒ Volume = 180 cm³

∴ The correct answer is 180 cm³.

Additional Information

Prism vs Pyramid Volume

The volume of a prism is directly calculated as Base Area × Height, whereas a pyramid's volume is exactly one-third of the prism's volume having the same base and height.

Total Surface Area of a Pyramid

Total Surface Area = Lateral Surface Area + Base Area. For a regular pyramid, Lateral Surface Area = (1 ÷ 2) × Base Perimeter × Slant Height.

Cone as a Pyramid

A cone is a specific type of pyramid with a circular base. Its volume formula is exactly (1 ÷ 3) × πr²h, which follows the same foundational logic.

Shortcut Trick

For an equilateral triangle circumscribing a circle of radius r, the remaining area can be directly calculated as:

⇒ Remaining Area = r² × (3√3 − π)

⇒ Remaining Area = 7² × (3 × 1.732 − 3.142)

⇒ Remaining Area ≈ 49 × (5.196 − 3.142) = 49 × 2.054

⇒ Remaining Area ≈ 100.64 cm²

∴ The correct answer is 100.6 cm².

Alternate Method

Given:

Radius of the inscribed circular disc (r) = 7 cm

Formula Used:

Inradius of equilateral triangle = a ÷ 2√3 (where a is the side length)

Area of equilateral triangle = (√3 ÷ 4) × a²

Area of a circle = π × r²

r = 7 cm

a = 14√3 cm Remaining Area = Area of Triangle − Area of Circle

Calculations:

⇒ a ÷ 2√3 = 7

⇒ a = 14√3 cm

⇒ Area of the equilateral triangle = (√3 ÷ 4) × (14√3)²

⇒ Area of the equilateral triangle = (√3 ÷ 4) × 196 × 3 = 147√3 cm²

⇒ Taking √3 ≈ 1.732

⇒ Area of the triangle ≈ 147 × 1.732 = 254.604 cm²

⇒ Area of the circle = (22 ÷ 7) × 7² = 154 cm²

⇒ Remaining Area = Area of the triangle − Area of the circle

⇒ Remaining Area = 254.604 − 154 = 100.604 cm²

∴ The correct answer is 100.6 cm².

Additional Information

Circumcircle of an Equilateral Triangle

The circumradius (R) of an equilateral triangle is exactly double its inradius. Formula: R = a ÷ √3 = 2r.

Area Ratio of Incircle and Circumcircle

For any equilateral triangle, the ratio of the area of the incircle to the area of the circumcircle is always 1:4.

Height of an Equilateral Triangle

The height or median (h) divides the triangle symmetrically. Formula: h = (√3 ÷ 2) × a.

Shortcut Trick

According to the geometric properties of a triangle, the centroid divides the triangle into three smaller triangles of equal area.

These three triangles are ΔGAB, ΔGBC, and ΔGCA.

Since their areas are equal, the area of ΔGAB is exactly one-third of the total area of ΔABC.

Ratio = Area(ΔGAB) : Area(ΔABC) = 1 : 3.

∴ The correct answer is 1:3.

Alternate Method

A B C D E F G Area(ΔGAB) = 1/3 × Area(ΔABC)

Given:

⇒ AD, BE, and CF are the medians of ΔABC.

⇒ G is the centroid where all three medians intersect.

Property Used:

⇒ The medians of a triangle divide it into six smaller triangles of equal area.

⇒ Area(ΔAFG) = Area(ΔBFG) = Area(ΔBDG) = Area(ΔCDG) = Area(ΔCEG) = Area(ΔAEG)

Calculation:

⇒ Let the area of one small triangle, Area(ΔAFG) = 1 unit

⇒ Total Area(ΔABC) = 6 units

⇒ Area(ΔGAB) = Area(ΔAFG) + Area(ΔBFG)

⇒ Area(ΔGAB) = 1 + 1 = 2 units

⇒ Ratio = Area(ΔGAB) ÷ Area(ΔABC)

⇒ Ratio = 2 ÷ 6

⇒ Ratio = 1 ÷ 3

∴ The correct answer is 1:3.

Additional Information

Centroid Division Property

The centroid divides each median of the triangle in the ratio 2:1, with the longer segment always originating from the vertex (e.g., AG:GD = 2:1).

Apollonius Theorem

It establishes a relationship between the lengths of the sides of a triangle and its median. For median AD on side BC: AB² + AC² = 2 × (AD² + BD²).

Medians and Area Rule

A new triangle formed by taking the lengths of the three medians of an original triangle as its sides will have an area exactly equal to 3/4 of the area of the original triangle.

Shortcut Trick

Area of △ADE : Area of trapezoid DECB = 4 : 21

Total Area of △ABC = Area(△ADE) + Area(DECB) = 4 + 21 = 25 parts

Ratio of corresponding sides AD : AB = √(Area of △ADE) : √(Area of △ABC) = √4 : √25 = 2 : 5

Since point D is on AB, DB = AB − AD = 5 − 2 = 3 parts

∴ The correct answer is 2:3.

Alternate Method

A B C D E Area = 4 units Area = 21 units

Given:

DE is parallel to BC (DE || BC).

Area of △ADE : Area of trapezoid DECB = 4 : 21.

Formula Used:

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Area(△ADE) ÷ Area(△ABC) = (AD ÷ AB)²

Calculations:

⇒ Area of △ABC = Area of △ADE + Area of trapezoid DECB

⇒ Area of △ABC = 4 + 21 = 25 units

⇒ In △ADE and △ABC, DE || BC

⇒ ∠ADE = ∠ABC (Corresponding angles)

⇒ ∠AED = ∠ACB (Corresponding angles)

⇒ ∴ △ADE ~ △ABC (by AA similarity criterion)

⇒ Area(△ADE) ÷ Area(△ABC) = (AD ÷ AB)²

⇒ 4 ÷ 25 = (AD ÷ AB)²

⇒ AD ÷ AB = √(4 ÷ 25) = 2 ÷ 5

⇒ Let AD = 2x and AB = 5x

⇒ Since point D lies on AB, AB = AD + DB

⇒ DB = AB − AD

⇒ DB = 5x − 2x = 3x

⇒ Ratio of AD to DB = 2x : 3x = 2 : 3

∴ The correct answer is 2:3.

Additional Information

Area of Similar Triangles

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, altitudes, medians, or angle bisectors.

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. ⇒ AD ÷ DB = AE ÷ EC.

AA Similarity Criterion

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are structurally similar.

Shortcut Trick

Perimeter of larger triangle = 3 + 4 + 5 = 12.

Area of larger triangle = 1/2 × 3 × 4 = 6.

For similar triangles, the ratio of their areas equals the square of the ratio of their perimeters.

Area of smaller triangle = 6 × (6 ÷ 12)²

⇒ Area = 6 × (1/2)² = 6 ÷ 4 = 1.5.

∴ The correct answer is 1.5 sq. units.

Alternate Method

Given:

⇒ Sides of the larger right-angled triangle = 3, 4, and 5

⇒ Perimeter of the inscribed smaller similar triangle = 6

Formula Used:

⇒ Area of a right triangle = 1/2 × Base × Perpendicular Height

⇒ For two similar triangles: Area₁ / Area₂ = (Perimeter₁ / Perimeter₂)²

3 4 5

2 1.5 2.5

Calculations:

⇒ The perimeter of the larger triangle = 3 + 4 + 5 = 12

⇒ Area of the larger triangle = 1/2 × 3 × 4 = 6 sq. units

⇒ The ratio of their perimeters (Smaller : Larger) = 6 / 12 = 1/2

⇒ According to the similarity theorem, the ratio of their areas is the square of the scale factor.

⇒ Area Ratio = (1 / 2)² = 1 / 4

⇒ Let the area of the smaller triangle be A.

⇒ A / 6 = 1 / 4

⇒ A = 6 × (1 / 4) = 1.5 sq. units

∴ The correct answer is 1.5 sq. units.

Additional Information

Similarity Area Theorem

For any two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides, altitudes, or perimeters: Area₁ / Area₂ = (P₁ / P₂)².

Medial Triangle Property

The triangle formed by connecting the midpoints of the sides of a given triangle is called the medial triangle. It perfectly represents a similar inscribed triangle whose area is exactly 1/4^th of the original area, and perimeter is half of the original.

Right Angled Triangle Basics

The area of a right-angled triangle is simply 1/2 × Base × Perpendicular Height. The three sides strictly follow the Pythagorean theorem relation: a² + b² = c².

Shortcut Trick For Direct Common Tangent (DCT) and distance between centers (d): d2 = DCT2 + (R − r)2 ⇒ d = √(152 + 82) ⇒ d = √(225 + 64) = √289 = 17 cm ∴ The correct answer is 17 cm. Alternate Method

C1 C2 R = 12 cm r = 4 cm DCT = 15 cm d = ? T1 T2

Given: ⇒ Radius of larger circle (R) = 12 cm ⇒ Radius of smaller circle (r) = 4 cm ⇒ Length of Direct Common Tangent (DCT) = 15 cm Formula Used: ⇒ DCT = √[d2 − (R − r)2] Where 'd' is the distance between the centers of the two circles. Calculations: ⇒ 15 = √[d2 − (12 − 4)2] ⇒ 15 = √[d2 − 82] Squaring both sides: ⇒ 225 = d2 − 64 ⇒ d2 = 225 + 64 ⇒ d2 = 289 ⇒ d = 17 cm ∴ The correct answer is 17 cm. Additional Information Transverse Common Tangent (TCT) A tangent that crosses the line joining the centers of the circles. Formula: TCT = √[d2 − (R + r)2] Touching Circles Externally When circles touch each other externally, the distance between centers is d = R + r, and the DCT length simplifies directly to 2√(R × r). Number of Common Tangents If circles are completely disjoint (d > R + r), they have exactly 4 common tangents (2 direct and 2 transverse).

Shortcut Trick

The standard identity relating the direct common tangent (L), transverse common tangent (M), distance between centers (d) and radii (r₁, r₂) is:

L² + M² = 2d² − 2(r₁² + r₂²)

Given: L² + M² = 200, and r₁² + r₂² = 100

⇒ 200 = 2d² − 2(100)

⇒ 2d² = 400 ⇒ d² = 200 ⇒ d = 10√2

∴ The correct answer is 10√2 units.

Alternate Method

Given:

Distance between centers = d

Direct common tangent = L

Transverse common tangent = M

Sum of squares of common tangents, L² + M² = 200

Sum of squares of radii, r₁² + r₂² = 100

L (Direct Tangent) M (Transverse Tangent) d C₁ C₂

Formula:

Length of direct common tangent (L) = √[d² − (r₁ − r₂)²]

Length of transverse common tangent (M) = √[d² − (r₁ + r₂)²]

Calculations:

Squaring both formulas to eliminate the square roots:

⇒ L² = d² − (r₁² + r₂² − 2r₁r₂)

⇒ M² = d² − (r₁² + r₂² + 2r₁r₂)

Adding both equations together:

⇒ L² + M² = d² − r₁² − r₂² + 2r₁r₂ + d² − r₁² − r₂² − 2r₁r₂

⇒ L² + M² = 2d² − 2(r₁² + r₂²)

Substitute the given values into the derived equation:

⇒ 200 = 2d² − 2(100)

⇒ 200 = 2d² − 200

⇒ 2d² = 400

⇒ d² = 200

⇒ d = √200 = 10√2

∴ The correct answer is 10√2 units.

Additional Information

Direct Common Tangent (DCT)

A common tangent is called a direct tangent if it does not intersect the line segment joining the centers of the circles. Its length is greater than the transverse common tangent.

Transverse Common Tangent (TCT)

A common tangent is called a transverse tangent if it intersects the line segment connecting the centers of the two circles.

Number of Common Tangents

Two disjoint circles have exactly 4 common tangents (2 DCT and 2 TCT). If the circles touch externally, they have 3 common tangents. Intersecting circles only have 2 direct common tangents.

Shortcut Trick

The half-lengths of the two chords are 16 ÷ 2 = 8 cm and 12 ÷ 2 = 6 cm.

In a circle, the radius, half-chord, and perpendicular distance from the center form a right-angled triangle.

Notice that the lengths 6, 8, and 10 form a standard Pythagorean triplet (6² + 8² = 10²).

The distances from the center to the chords are 6 cm and 8 cm, and their difference is 8 − 6 = 2 cm, matching the given spacing perfectly.

Therefore, the hypotenuse, which represents the circle's radius, must be 10 cm.

∴ The correct answer is 10 cm.

Alternate Method

Given:

Length of larger chord (AB) = 16 cm

Length of smaller chord (CD) = 12 cm

Space between the chords = 2 cm

O

C D

A B

P Q

8 6 2 x r r

Formula Used:

A perpendicular drawn from the center of a circle to a chord bisects the chord.

According to Pythagoras theorem: r² = (Perpendicular distance)² + (Half of chord)²

Calculations:

Let the perpendicular distance of the larger chord from the center be x cm.

Since both chords are on the same side of the center, the distance of the smaller chord from the center will be (x + 2) cm.

Half of larger chord = 16 ÷ 2 = 8 cm

Half of smaller chord = 12 ÷ 2 = 6 cm

Applying the Pythagoras theorem for the larger chord:

⇒ r² = x² + 8²

⇒ r² = x² + 64 --- (Equation 1)

Applying the Pythagoras theorem for the smaller chord:

⇒ r² = (x + 2)² + 6²

⇒ r² = x² + 4x + 4 + 36

⇒ r² = x² + 4x + 40 --- (Equation 2)

Equating (Equation 1) and (Equation 2):

⇒ x² + 64 = x² + 4x + 40

⇒ 4x = 64 − 40

⇒ 4x = 24

⇒ x = 6 cm

Substituting the value of x into Equation 1:

⇒ r² = 6² + 64

⇒ r² = 36 + 64 = 100

⇒ r = √100 = 10 cm

∴ The correct answer is 10 cm.

Additional Information

Perpendicular Bisector Theorem

A perpendicular dropped from the center of a circle to a chord bisects the chord into two equal halves.

Parallel Chords on Opposite Sides

If two parallel chords lie on opposite sides of the center, the perpendicular distance between them is the sum of their individual distances from the center (d₁ + d₂).

Pythagorean Triplets

Many geometry problems can be solved significantly faster by recognizing common triplets like (3, 4, 5), (6, 8, 10), or (5, 12, 13) which perfectly satisfy the equation a² + b² = c².

Shortcut Trick

Direct formula for the Area of a Segment:

⇒ Area = r² × [(π × θ ÷ 360°) − (sin θ ÷ 2)]

⇒ Area = 12² × [(π × 150° ÷ 360°) − (sin 150° ÷ 2)]

⇒ Area = 144 × [(5π ÷ 12) − (1 ÷ 4)]

⇒ Area = 60π − 36 cm²

∴ The correct answer is 60π−36 cm².

Alternate Method

150° 12 cm 12 cm O A B Segment Area of Segment = Area of Sector − Area of Triangle

Given:

⇒ Radius of the circle (r) = 12 cm

⇒ Angle subtended at the center (θ) = 150°

Formula Used:

⇒ Area of Sector = (θ ÷ 360°) × πr²

⇒ Area of Triangle = (1 ÷ 2) × r² × sin θ

⇒ Area of Segment = Area of Sector − Area of Triangle

Calculations:

⇒ Area of Sector = (150° ÷ 360°) × π × 12²

⇒ Area of Sector = (5 ÷ 12) × π × 144 = 60π cm²

⇒ To find the area of the triangle, we use sin(150°) = sin(180° − 30°) = sin(30°) = 1 ÷ 2

⇒ Area of Triangle = (1 ÷ 2) × 12² × (1 ÷ 2)

⇒ Area of Triangle = (1 ÷ 2) × 144 × (1 ÷ 2) = 36 cm²

⇒ Area of Segment = 60π − 36 cm²

∴ The correct answer is 60π−36 cm².

Additional Information

Area of Major Segment

It is calculated by subtracting the area of the minor segment from the total area of the circle: Area = πr² − Area of Minor Segment.

Length of an Arc

The arc length corresponding to a central angle θ is a fraction of the circle's circumference: L = (θ ÷ 360°) × 2πr.

Perimeter of a Sector

The boundary length of a sector comprises the arc length and two radii: Perimeter = L + 2r.