Geometry Questions

Geometry is an essential chapter in NCERT Math textbooks, regardless of which class you’re in. This article breaks down complex geometry concepts into understandable questions with solutions. Covering everything from basic geometry shapes and formulas to the principles of lines and angles, this list is designed to simplify geometry for students of all levels.

Geometry Questions with Solutions

Question 1: Calculate the area of a rectangle with a length of 8 cm and a width of 5 cm.

Solution:

Area of a rectangle = length × width.
= 8 cm × 5 cm = 40 cm².

Question 2: Find the perimeter of a square with a side length of 6 meters.

Solution:

Perimeter of a square = 4 × side length.
= 4 × 6 m = 24 meters.

Question 3: A circle has a radius of 4 inches. Calculate its circumference.

Solution:

Circumference of a circle = 2π × radius.
= 2 × π × 4 in ≈ 25.13 inches (using π ≈ 3.14159).

Question 4: A right triangle has one leg of 3 cm and another leg of 4 cm. Find the length of the hypotenuse.

Solution:

Use the Pythagorean theorem: hypotenuse² = leg₁² + leg₂².
= 3 cm² + 4 cm² = 9 cm² + 16 cm² = 25 cm².
Hypotenuse = √25 cm = 5 cm.

Question 5: What is the area of a circle with a diameter of 10 cm?

Solution:

Radius of the circle = diameter / 2 = 10 cm / 2 = 5 cm.
Area of a circle = π × radius².
= π × 5 cm² ≈ 78.54 cm² (using π ≈ 3.14159).

Question 6: Find the volume of a cube with a side length of 5 cm.

Solution:

Volume of a cube = side length³
= 5 cm × 5 cm × 5 cm = 125 cm³.

Question 7: A triangle has base 6 meters and height 4 meters. Calculate its area.

Solution:

Area of a triangle = 1/2 × base × height.
= 1/2 × 6 m × 4 m = 12 m².

Question 8: What is the sum of the interior angles of a pentagon?

Solution:

Sum of interior angles of a polygon = (n – 2) × 180°, where n is the number of sides.
For a pentagon, n = 5.
Sum = (5 – 2) × 180° = 3 × 180° = 540°.

Question 9: Calculate the length of the arc of a circle with a radius of 7 cm and a central angle of 45°.

Solution:

Length of an arc = (central angle / 360°) × 2π × radius.
= (45° / 360°) × 2 × π × 7 cm ≈ 5.50 cm (using π ≈ 3.14159).

Question 10: A cylinder has a radius of 3 cm and a height of 10 cm. Find its surface area.

Solution:

Surface area of a cylinder = 2πr² + 2πrh, where r is the radius and h is the height.
= 2 × π × 3 cm² + 2 × π × 3 cm × 10 cm ≈ 245.04 cm² (using π ≈ 3.14159).

Question 11: Find the area of a trapezoid with bases of 8 cm and 5 cm and a height of 4 cm.

Solution:

Area of a trapezoid = 1/2 × (sum of the lengths of the two parallel sides) × height.
= 1/2 × (8 cm + 5 cm) × 4 cm = 1/2 × 13 cm × 4 cm = 26 cm².

Question 12: Calculate the area of an equilateral triangle with a side length of 6 cm.

Solution:

Area of an equilateral triangle = (√3/ 4) × side length².
= (√3 / 4) × 6 cm² ≈ 15.59 cm² (using √3)≈ 1.732).

Question 13: A rhombus has diagonals of 10 cm and 8 cm. Find its area.

Solution:

Area of a rhombus = (product of the diagonals) / 2.
= (10 cm × 8 cm) / 2 = 80 cm² / 2 = 40 cm².

Question 14: What is the total surface area of a hemisphere with a radius of 7 cm?

Solution:

Total surface area of a hemisphere = 3πr², where r is the radius.
= 3 × π × 7 cm² ≈ 461.81 cm² (using π ≈ 3.14159).

Question 15: Find the length of a diagonal of a rectangle with a length of 8 cm and a width of 6 cm.

Solution:

Length of the diagonal = √(length² + width²).
= √(8 cm² + 6 cm²) = √(64 cm² + 36 cm²) = √(100 cm²) = 10 cm.

Question 16: Calculate the volume of a rectangular prism with length 10 cm, width 4 cm, and height 5 cm.

Solution:

Volume of a rectangular prism = length × width × height.
= 10 cm × 4 cm × 5 cm = 200 cm³.

Question 17: A cylindrical tank has a radius of 7 meters and a height of 10 meters. Find the volume of the tank.

Solution:

Volume of a cylinder = πr²h, where r is the radius and h is the height.
= π × 7 m² × 10 m ≈ 1540 m³ (using π ≈ 3.14159).

Question 18: Find the volume of a cone with a radius of 3 cm and a height of 7 cm.

Solution:

Volume of a cone = 1/3πr²h.
= 1/3 × π × 3 cm² × 7 cm ≈ 66 cm³ (using π ≈ 3.14159).

Question 19: A sphere has a radius of 6 cm. Calculate its volume.

Solution:

Volume of a sphere = 4/3πr³, where r is the radius.
= 4/3 × π × 6 cm³ ≈ 904.78 cm³ (using π ≈ 3.14159).

Question 20: Determine the volume of a frustum of a cone with a top radius of 3 cm, a bottom radius of 5 cm, and a height of 6 cm.

Solution:

Volume of a frustum of a cone = 1/3πh(R² + r² + Rr), where R and r are the radii of the circular ends and h is the height.
= 1/3 × π × 6 cm (5 cm² + 3 cm² + 5 cm × 3 cm) ≈ 150.8 cm³ (using π ≈ 3.14159).

Question 21: A sphere of radius 6 cm is inscribed in a cube. Find the volume of the space inside the cube but outside the sphere.

Solution:

The side length of the cube = diameter of the sphere = 2 × 6 cm = 12 cm.
Volume of the cube = 12 cm × 12 cm × 12 cm = 1728 cm³.
Volume of the sphere = 4/3π × radius³ = 4/3π × 6 cm³ ≈ 904.78 cm³.
Volume of the space = Volume of the cube – Volume of the sphere ≈ 1728 cm³ – 904.78 cm³ ≈ 823.22 cm³.

Question 22: Calculate the total surface area of a regular octahedron with an edge length of 5 cm.

Solution:

Area of one equilateral triangle = (√3/4) × side length² = (√3/4) × 5 cm² ≈ 10.83 cm².
Total surface area = Area of one triangle × 8 ≈ 10.83 cm² × 8 ≈ 86.6 cm².

Question 23: A cylinder of radius 4 cm and height 5 cm is completely filled with water. This water is poured into a conical flask with a radius of 2 cm. Find the height of the water in the flask.

Solution:

Volume of the cylinder = πr²h = π × 4 cm² × 5 cm = 80π cm³.
Volume of the cone = 1/3πr²h, where h is the height of the water in the flask.
Since the volume of water remains constant, set the cylinder’s volume equal to the cone’s volume:
80π cm³ = 1/3π × 2 cm² × h.
Solving for h gives: h = (80 × 3) / 2 = 120 cm.
Therefore, the height of the water in the flask is 120 cm.

Question 24: Find the surface area of a hollow sphere with an external radius of 10 cm and a thickness of 2 cm.

Solution:

External surface area = 4π × external radius² = 4π × 10 cm² ≈ 1256 cm².
Internal radius = external radius – thickness = 10 cm – 2 cm = 8 cm.
Internal surface area = 4π × internal radius² = 4π × 8 cm² ≈ 804.25 cm².
Total surface area = External surface area + Internal surface area ≈ 1256 cm² + 804.25 cm² ≈ 2060.25 cm².

Question 25: A right-angled triangle with sides 3 cm, 4 cm, and 5 cm is rotated about the longest side. Calculate the volume of the solid formed.

Solution:

The solid formed is a circular cone. Radius of the base = one of the shorter sides (3 cm), height = the other shorter side (4 cm).
Volume of the cone = 1/3πr²h = 1/3π × 3 cm² × 4 cm ≈ 37.70 cm³.

Question 26: A regular tetrahedron (a pyramid with a triangular base) has an edge length of 10 cm. Calculate the volume and total surface area of the tetrahedron.

Solution:

Volume = (√2 / 12) × edge length³ = (√2)/ 12) × 10 cm³.
Total surface area = √3 × edge length² = √3 × 10 cm².

Question 27: A cylindrical drill with a radius of 3 cm is used to bore a hole through the center of a sphere of radius 5 cm. Find the volume of the remaining portion of the sphere.

Solution:

Volume of the sphere = 4/3π × radius of sphere³ = 4/3π × (5 cm)³ = 523.6 cm³.
Volume of the cylindrical hole = π × radius of drill² × height of the cylinder = π × (3 cm)² × (2 × 5 cm) = 282.74 cm³.
Remaining volume = Volume of sphere – Volume of cylindrical hole = 523.6 cm³ – 282.74 cm³ = 240.86 cm³.

Question 28: A cone and a hemisphere share the same base with a radius of 6 cm. If the cone’s height equals the diameter of the hemisphere, find the ratio of their volumes.

Solution:

Volume of the cone = 1/3πr²h, where h = 2 × radius of the hemisphere.
Volume of the hemisphere = 2/3πr³.
Ratio = Volume of cone / Volume of hemisphere.

Question 29: A right circular cone with a radius of 4 cm and a height of 9 cm is cut by a plane parallel to its base, 3 cm from the base. Calculate the volume of the smaller cone formed at the top.

Solution:

Using similar triangles, find the ratio of the height of the smaller cone to the larger cone.
Then, use the ratio to find the radius of the smaller cone.
Volume of the smaller cone = 1/3π × radius² × height.

Question 30: An equilateral triangular prism has a side length of 10 cm and a height of 20 cm. Find the shortest distance from one of the vertices of the triangular base to the opposite face of the prism.

Solution:

The triangular base of the prism is an equilateral triangle with side length 10 cm.
The height of the prism is 20 cm.
To find the shortest distance from a vertex of the triangular base to the opposite face, we can consider the triangular base and the opposite face as two congruent right triangles.
The hypotenuse of each right triangle is the height of the prism, which is 20 cm.
The base of each right triangle is half the side length of the equilateral triangle, which is 10 cm / 2 = 5 cm.
Using the Pythagorean Theorem, we can find the height (shortest distance) of each right triangle:
Height (h) = √((20 cm)² – (5 cm)²) = √((400 cm² – 25 cm²)) = √(375 cm²) = 5√15 cm.
So, the shortest distance from a vertex of the triangular base to the opposite face is 5√15 cm.

Conclusion

To sum up, geometry shapes, principles, and formulas are important to learn for students of all classes. The questions provided with their solutions above will help you solve many geometry problems with ease.