# Mensuration Questions

Mensuration questions with answers are available for students. The problems have been solved in an elaborate manner to make each and every student understand the concept easily. The questions here are based on Class 8 and Class 10 Maths syllabus. The problems have been prepared, as per the latest NCERT guidelines.

**Also, read**: Mensuration

Area of Rectangle = Length x Breadth Area of Square = Side Area of Triangle = ½ Base x Height Area of Parallelogram = Base x Height Area of circle = πr Perimeter of polygons = Sum of their sides Circumference of circle = 2πr |

## Questions on Mensuration with Solutions

**1. A circle has a radius of 21 cm. Find its circumference and area. (Use π = 22/7)**

Solution: We know,

Circumference of circle = 2πr = 2 x (22/7) x 21 = 2 x 22 x 3 = 132 cm

Area of circle = πr^{2} = (22/7) x 21^{2} = 22/7 x 21 x 21 = 22 x 3 x 21

Area of circle with radius, 21cm = 1386 cm^{2}

**2. If one side of a square is 4 cm, then what will be its area and perimeter?**

Solution: Given,

Length of side of square = 4 cm

Area = side^{2} = 4^{2} = 4 x 4 = 16 cm^{2}

Perimeter of square = sum of all its sides

Since, all the sides of the square are equal, therefore;

Perimeter = 4+4+4+4 = 16 cm

Where d is the diagonal of quadrilateral dividing it into two triangles h
Where d
Where a and b are the two parallel sides of trapezium h is the distance between a and b. |

**Also, read:**

**3. Suppose a quadrilateral having a diagonal of length 10 cm, which divides the quadrilateral into two triangles and the heights of triangles with diagonals as the base, are 4 cm and 6 cm. Find the area of the quadrilateral.**

Solution: Given,

Diagonal, d = 10 cm

Height of one triangle, h_{1} = 4cm

Height of another triangle, h_{2} = 6cm

Area of quadrilateral = ½ d(h_{1}+h_{2}) = ½ x 10 x (4+6) = 5 x 10 = 50 sq.cm.

**4. A rhombus having diagonals of length 10 cm and 16 cm, respectively. Find its area.**

Solution: d_{1} = 10 cm

d_{2} = 16 cm

Area of rhombus = ½ d_{1} d_{2}

A = ½ x 10 x 16

A= 80 cm^{2}

**5. The area of a trapezium shaped field is 480 m ^{2}, the distance between two parallel sides is 15 m and one of the parallel sides is 20 m. Find the other parallel side.**

Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel side be b, height h = 15 m.

The given area of trapezium = 480 m^{2}

We know, by formula;

Area of a trapezium = ½ h (a+b)

480 = ½ (15) (20+b)

20 + b = (480×2)/15

b = 64 – 20 = 44 m

TSA of Cuboid = 2(lb + bh + hl) TSA of Cube = 6l TSA of Cylinder = 2πr (r + h) |

**6. The height, length and width of a cuboidal box are 20 cm, 15 cm and 10 cm, respectively. Find its area.**

Solution: Total surface area = 2 (20 × 15 + 20 × 10 + 10 × 15)

TSA = 2 ( 300 + 200 + 150) = 1300 cm^{2}

**7. If a cube has its side-length equal to 5cm, then its area is?**

Solution: Given,

l = 5 cm

Area = 6l^{2} = 6 x 5 x5 = 150 sq.cm

**8. Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm ^{2}.**

Solution: : Let height of the cylinder = h, radius = r = 7cm

Total surface area = 2πr (h + r)

TSA = 2 x (22/7) x 7 x (7+h) = 968

h = 15 cm

V of cube = l Volume of cuboid = l × b × h Volume of cylinder = πr |

**Check:** Mensuration Formulas Class 10

**9. Find the height of a cuboid whose volume is 275 cm ^{3} and base area is 25 cm^{2}.**

Solution: Volume of cuboid = l × b × h

Base area = l × b = 25 cm^{2}

Hence,

275 = 25 × h

h = 275/25 = 11 cm

**10. A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.**

Solution: Length of the paper will be the perimeter of the base of the cylinder and width will be its height.

Circumference of base of cylinder = 2πr = 11 cm

2 x 22/7 x r = 11 cm

r = 7/4 cm

Volume of cylinder = πr^{2}h = (22/7) x (7/4)^{2} x 4

= 38.5 cm^{3}