CG Board Class 8 Mathematics Chapter 14 Solutions: Mensuration-I (क्षेत्रमिति-1) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 8 Mathematics Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 8 Mathematics Chapter 14 Solutions: Mensuration-I (क्षेत्रमिति-1) in Bilingual Format (English & Hindi Medium).

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Area = 1/2 × d1 × d2 or Area = Base × Height

Area of Path = Outer Rectangle Area – Inner Rectangle Area

Area of Path = π (R2 – r2)

SECTION / EXERCISE

EXERCISE 14.1 / 14.1

Question 1

Find out the area of a parallelogram whose…


📝 Question:

Find out the area of a parallelogram whose base and height are as follows /

💡 Solution:

(i) Base = 15 cm, Height = 10 cm**

Area = Base × Height = 15 × 10 = 150 cm2

(ii) Base = 90 cm, Height = 8 cm

Area = 90 × 8 = 720 cm2

(iii) Base = 120 cm, Height = 15 cm

Area = 120 × 15 = 1800 cm2

Question 2

Find the area of the parallelogram where base…


📝 Question:

Find the area of the parallelogram where base is 26.5 cm and perpendicular is 7 cm. / 26.5 7

💡 Solution:

Area = Base × Height = 26.5 × 7 = 185.5 cm2

Question 3

Find out the base of the parallelogram whose…


📝 Question:

Find out the base of the parallelogram, whose area is 390 cm² and the perpendicular is 26 cm. / 390 26

💡 Solution:

Base = (Area)/(Height) = 390/26 = 15 cm

Question 4

Find out the perpendicular of parallelogram with area…


📝 Question:

Find out the perpendicular of parallelogram with area 1200 m² and base 60 m. / 1200 60

💡 Solution:

Height = (Area)/(Base) = (1200)/(60) = 20 m

SECTION / EXERCISE

EXERCISE 14.2 / 14.2

Question 1

Find out the area of a triangle with…


📝 Question:

Find out the area of a triangle with base 12 cm and corresponding height of 7 cm. / 12 7

💡 Solution:

Area = 1/2 × Base × Height = 1/2 × 12 × 7 = 42 cm2

Question 2

Find out the area of a triangle with…


📝 Question:

Find out the area of a triangle with base 25 cm and perpendicular length of 1.5 cm. / 25 1.5

💡 Solution:

Area = 1/2 × 25 × 1.5 = (37.5)/(2) = 18.75 cm2

Question 3

Find out the perpendicular length of a triangle…


📝 Question:

Find out the perpendicular length of a triangle where base is 6.5 cm and area is 26 cm². / 6.5 26

💡 Solution:

Height = (2 × Area)/(Base) = (2 × 26)/(6.5) = 52/6.5 = 8 cm

Question 4

Find out the area of a triangle where…


📝 Question:

Find out the area of a triangle where base is 120 dm and height is 75 dm. / 120 75

💡 Solution:

Area = 1/2 × 120 × 75 = 60 × 75 = 4500 dm2

SECTION / EXERCISE

EXERCISE 14.3 / 14.3

Question 1

Find the area of a rhombus where diagonals…


📝 Question:

Find the area of a rhombus where diagonals are 24 cm & 10 cm. / 24 10

💡 Solution:

Area = 1/2 × d1 × d2 = 1/2 × 24 × 10 = 120 cm2

Question 2

Find the area of a rhombus where one…


📝 Question:

Find the area of a rhombus where one side is 7.5 cm and perpendicular is 4 cm. / 7.5 4

💡 Solution:

Since a rhombus is also a parallelogram

Area = Base × Height = 7.5 × 4 = 30 cm2

Question 3

Find the area of a trapezium where parallel…


📝 Question:

Find the area of a trapezium where parallel sides are 20 m and 8 m and distance between them is 12 m. / 20 8 12

💡 Solution:

Area = 1/2 × (b1 + b2) × h = 1/2 × (20 + 8) × 12 = 14 × 12 = 168 m2

Question 4

What will be the area of the trapezium…


📝 Question:

What will be the area of the trapezium where bases are 30 cm and 24.4 cm, if its perpendicular height is 1.5 cm? / 30 24.4 1.5 ?

💡 Solution:

Area = 1/2 × (30 + 24.4) × 1.5 = 1/2 × 54.4 × 1.5 = 27.2 × 1.5 = 40.8 cm2

Question 5

The area of a trapezium is 105 cm²…


📝 Question:

The area of a trapezium is 105 cm² and its height is 7 cm. If one of the parallel sides is 6 cm more than the other, find the length of the two parallel sides. / 105 7 6 ,

💡 Solution:

Let the smaller parallel side = x cm.

  • Then the other parallel side = (x + 6) cm.
  • Formula

Area = 1/2 × (b1 + b2) × h

105 = 1/2 × (x + x + 6) × 7

105 × 2 = (2x + 6) × 7

210 = (2x + 6) × 7 ⇒ 2x + 6 = 210/7 = 30

2x = 30 – 6 = 24 ⇒ x = 12 cm

  • Parallel sides are 12 cm and 12 + 6 = 18 cm.

SECTION / EXERCISE

EXERCISE 14.4 / 14.4

Question 1

A 25 cm long and 10 cm wide…


📝 Question:

A 25 cm long and 10 cm wide picture is surrounded by a 2 cm wide frame on all sides; find the area of the frame. / 25 10 , 2

💡 Solution:

Dimensions of picture (Inner rectangle): Length = 25 cm, Width = 10 cm.

Inner Area = 25 × 10 = 250 cm2

  • Dimensions with frame (Outer rectangle)

Outer Length = 25 + 2 + 2 = 29 cm

Outer Width = 10 + 2 + 2 = 14 cm

Outer Area = 29 × 14 = 406 cm2

  • Area of frame = 406 – 250 = 156 cm2.

Question 2

A rectangular playground measures 35 m 25 m…


📝 Question:

A rectangular playground measures 35 m × 25 m. A road 3 m wide along the length and 2 m wide along the width goes through the middle of the ground. Find the area of the road. / 35 × 25 – 3 2

💡 Solution:

Area of road parallel to length = 35 × 3 = 105 m2.

  • Area of road parallel to width = 25 × 2 = 50 m2.
  • Area of common central intersection = 3 × 2 = 6 m2.
  • Total Area of road = 105 + 50 – 6 = 149 m2.

Question 3

A basketball ground is 28 m long and…


📝 Question:

A basketball ground is 28 m long and 15 m wide. A 5 m wide gallery for spectators has to be built on all sides of the ground. Find the area of the spectator gallery and the cost of making that gallery at Rs. 5.25 per square meter. / 28 15 5 5.25

💡 Solution:

Inner Area (Ground) = 28 × 15 = 420 m2.

  • Outer Length = 28 + 5 + 5 = 38 m.

Outer Width = 15 + 5 + 5 = 25 m.

Outer Area = 38 × 25 = 950 m2.

  • Area of Gallery = 950 – 420 = 530 m2.
  • Cost of construction = 530 × 5.25 = Rs. 2782.50.

SECTION / EXERCISE

EXERCISE 14.5 / 14.5

Question 1

The radius of two concentric circles are 9…


📝 Question:

The radius of two concentric circles are 9 cm and 12 cm respectively. Find the area of the circular path between the two circles. / 9 12

💡 Solution:

Given: R = 12 cm, r = 9 cm.

  • Area of circular path

Area = π (R2 – r2) = 22/7 (122 – 92) = 22/7 (144 – 81) = 22/7 × 63 = 22 × 9 = 198 cm2

Question 2

The area of a circle is 616 cm²…


📝 Question:

The area of a circle is 616 cm². There is a 2 m wide road on its edge. What will be the area of that road? / 616 2 ?

💡 Solution:

Area of inner circle = π r2 = 616 m2 (taking units as meters)

22/7 × r2 = 616 ⇒ r2 = (616 × 7)/(22) = 28 × 7 = 196 ⇒ r = 14 m

  • Outer radius R = 14 + 2 = 16 m.
  • Area of outer circle

Outer Area = 22/7 × 162 = (22 × 256)/(7) = (5632)/(7) ≈ 804.57 m2

  • Area of road = 804.57 – 616 = 188.57 m2.

Question 3

The radius of a circular cricket ground is…


📝 Question:

The radius of a circular cricket ground is 60 m. Around the ground a 7 m circular gallery is to be built for spectators. What will be its area? / 60 7 ?

💡 Solution:

Given: Inner radius r = 60 m, width of gallery = 7 m.

  • Outer radius R = 60 + 7 = 67 m.
  • Area of spectator gallery

Area = π (R2 – r2) = 22/7 (672 – 602) = 22/7 (67 + 60)(67 – 60)

Area = 22/7 × 127 × 7 = 22 × 127 = 2794 m2

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