Latest Maharashtra State Board (SSC & HSC) 2026-27 Syllabus Digest & Solutions Updated!
MHBoardSolutionsSSC & HSC Digest
Class 7Mathematics & Statistics2026-27 Syllabus

Chapter 12 Perimeter and Area Practice Set 47 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 12 Perimeter and Area Practice Set 47. Step-by-step solved exercises, numerical problems, and digest answers.

6 Solved Questions• 819 words

Perimeter and Area Class 7 Practice Set 47 Answers Solutions Chapter 12

Question 1 Maharashtra Board Solution
Find the total surface area of cubes having the following sides: i. 3 cm ii. 5 cm iii. 7.2 m iv. 6.8 m v. 5.5 m
Solution & Step-by-Step Answer:
i. Total surface area of cube = 6l² = 6 × (3)² = 6 × 9 = 54 sq. cm.

ii. Total surface area of cube = 6l²
= 6 × 5²
= 6 × 25
= 150 sq. cm.

iii. Total surface area of cube = 6l²
= 6 × (7.2)²
= 6 × 51.84
= 311.04 sq. m.

iv. Total surface area of cube = 6l²
= 6 × (6.8)²
= 6 × 46.24
= 277.44 sq. m.

v. Total surface area of cube = 6l²
= 6 × (5.5)²
= 6 × 30.25
= 181.5 sq. m.

Question 2 Maharashtra Board Solution
Find the total surface area of the cuboids of length, breadth and height as given below: i. 12 cm, 10 cm, 5 cm ii. 5 cm, 3.5 cm, 1.4 cm iii. 2.5 m, 2 m, 2.4 m iv. 8 m, 5 m, 3.5 m
Solution & Step-by-Step Answer:
i. Total surface area of cuboid = 2 (lb + bh + lh) = 2 (12 × 10 + 10 × 5 + 12 × 5) = 2 (120 + 50 + 60) = 2 × 230 = 460 sq. cm.

ii. Total surface area of cuboid
= 2 (lb + bh + lh)
= 2 (5 × 3.5 + 3.5 × 1.4 + 5 × 1.4)
= 2(17.5 + 4.9 + 7)
= 2 × 29.4
= 58.8 sq. cm.

iii. Total surface area of cuboid = 2 (lb + bh + lh)
= 2(2.5 × 2 + 2 × 2.4 + 2.5 × 2.4)
= 2 (5 + 4.8 + 6)
= 2 × 15.8
= 31.6 sq. m.

iv. Total surface area of cuboid = 2 (lb + bh + lh)
= 2 (8 × 5+ 5 × 3.5 + 8 × 3.5)
= 2(40 + 17.5 + 28)
= 2 × 85.5
= 171 sq. m.

Question 3 Maharashtra Board Solution
A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height. Its outer sides are to be covered exactly with craft paper. How much paper will be required to do so?
Solution & Step-by-Step Answer:
Length of the matchbox (l) = 4 cm, breadth (b) = 2.5 cm, height (h) = 1.5 cm ∴ Total surface area of the matchbox = 2 (lb + bh + lh) = 2 (4 × 2.5 + 2.5 × 1.5 + 4 × 1.5) = 2 (10 + 3.75 + 6) = 2 × 19.75 = 39.5 sq. cm. ∴ 39.5 sq. cm paper will be required.
Question 4 Maharashtra Board Solution
An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box? The inside and outside surface of the box is to be painted with rust proof paint. At a rate of Rs 150 per sq. m, how much will it cost to paint the box?
Solution & Step-by-Step Answer:
Length of the box (l) = 1.5 m, breadth (b) = 1 m, height (h) = 1 m Since, the box is open at top, ∴ Sheet required to make the box = total surface area of the box – area of the top = 2 (lb + bh + lh) – lb = 2lb + 2bh + 2lh – lb = lb + 2bh + 2lh = 1.5 × 1 + 2 × 1 × 1 + 2 × 1.5 × 1 = 1.5 + 2 + 3 = 6.5 sq. m. Since, the inside and outside surface of the box are to be painted. ∴ Area to be painted = 2 × Area of the box = 2 × 6.5 = 13 sq. m. Total cost of painting = area to be painted × rate per sq. m. = 13 × 150 = Rs 1950 ∴ 6.5 sq. m. sheet of metal will be required and the cost of painting the box will be Rs 1950.

Maharashtra Board Class 7 Maths Chapter 12 Perimeter and Area Practice Set 47 Intext Questions and Activities

Question 1 Maharashtra Board Solution
Measure the length and breadth of the courts laid out for games such as kho-kho, kabaddi, tennis, badminton, etc. Find out their perimeters and areas. (Textbook pg. no. 81)
Solution & Step-by-Step Answer:
(Students should attempt the above activities on their own)
Question 2 Maharashtra Board Solution
Take mobile handsets of different sizes and find the area of their screens. (Textbook pg. no. 82)
Solution & Step-by-Step Answer:
(Students should attempt the above activities on their own)
Previous Chapter
Chapter 12 Perimeter and Area Practice Set 46
Next Chapter
Chapter 13 Pythagoras’ Theorem Practice Set 48