Maharashtra Board Class 7 Maths Chapter 12 Set 47 Perimeter and Area Solutions

Question 1. 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
Answer:
(i) Total surface area of cube = \(6l^2\)
= \(6 \times (3)^2\)
= \(6 \times 9\)
= 54 sq. cm.
(ii) Total surface area of cube = \(6l^2\)
= \(6 \times 5^2\)
= \(6 \times 25\)
= 150 sq. cm.
(iii) Total surface area of cube = \(6l^2\)
= \(6 \times (7.2)^2\)
= \(6 \times 51.84\)
= 311.04 sq. m.
(iv) Total surface area of cube = \(6l^2\)
= \(6 \times (6.8)^2\)
= \(6 \times 46.24\)
= 277.44 sq. m.
(v) Total surface area of cube = \(6l^2\)
= \(6 \times (5.5)^2\)
= \(6 \times 30.25\)
= 181.5 sq. m.
In simple words: The total surface area of a cube is found by squaring the side length and multiplying by 6, as a cube has six identical square faces. Each part applies this formula with the given side length.

🎯 Exam Tip: Remember the formula for the total surface area of a cube (6l²) and be careful with decimal calculations, especially when squaring numbers.

Question 2. 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
Answer:
(i) Total surface area of cuboid
= \(2 (lb + bh + lh)\)
= \(2 (12 \times 10 + 10 \times 5 + 12 \times 5)\)
= \(2 (120 + 50 + 60)\)
= \(2 \times 230\)
= 460 sq. cm.
(ii) Total surface area of cuboid
= \(2 (lb + bh + lh)\)
= \(2 (5 \times 3.5 + 3.5 \times 1.4 + 5 \times 1.4)\)
= \(2(17.5 + 4.9 + 7)\)
= \(2 \times 29.4\)
= 58.8 sq. cm.
(iii) Total surface area of cuboid = \(2 (lb + bh + lh)\)
= \(2(2.5 \times 2 + 2 \times 2.4 + 2.5 \times 2.4)\)
= \(2 (5 + 4.8 + 6)\)
= \(2 \times 15.8\)
= 31.6 sq. m.
(iv) Total surface area of cuboid = \(2 (lb + bh + lh)\)
= \(2 (8 \times 5 + 5 \times 3.5 + 8 \times 3.5)\)
= \(2(40 + 17.5 + 28)\)
= \(2 \times 85.5\)
= 171 sq. m.
In simple words: The total surface area of a cuboid is calculated by summing the areas of all six faces. This is achieved using the formula \(2(lb + bh + lh)\), where l, b, and h are the length, breadth, and height, respectively.

🎯 Exam Tip: Accurately substitute the given length, breadth, and height values into the cuboid surface area formula. Pay close attention to unit consistency (cm vs. m) in your final answer.

Question 3. 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?
Answer:
Length of the matchbox (l) = 4 cm, breadth (b) = 2.5 cm, height (h) = 1.5 cm
Total surface area of the matchbox = \(2 (lb + bh + lh)\)
= \(2 (4 \times 2.5 + 2.5 \times 1.5 + 4 \times 1.5)\)
= \(2 (10 + 3.75 + 6)\)
= \(2 \times 19.75\)
= 39.5 sq. cm.
39.5 sq. cm paper will be required.
In simple words: To find the amount of craft paper needed, calculate the total surface area of the matchbox using its given length, breadth, and height, as all outer sides need covering.

🎯 Exam Tip: This problem is a direct application of the cuboid total surface area formula. Ensure correct multiplication and addition, especially with decimals, to get the precise amount of paper needed.

Question 4. 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?
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
= \(2 (lb + bh + lh) - lb\)
= \(2lb + 2bh + 2lh - lb\)
= \(lb + 2bh + 2lh\)
= \(1.5 \times 1 + 2 \times 1 \times 1 + 2 \times 1.5 \times 1\)
= \(1.5 + 2 + 3\)
= 6.5 sq. m.
Since, the inside and outside surface of the box are to be painted.
Area to be painted = \(2 \times\) Area of the box = \(2 \times 6.5\) = 13 sq. m.
Total cost of painting = area to be painted \(\times\) rate per sq. m.
= \(13 \times 150\)
= Rs. 1950
6.5 sq. m. sheet of metal will be required and the cost of painting the box will be Rs. 1950.
In simple words: First, calculate the metal needed for the open box by subtracting the top surface area from the total surface area of a closed cuboid. Then, double this area for painting (inside and outside) and multiply by the per-square-meter rate to find the total painting cost.

🎯 Exam Tip: Be careful to adjust the surface area formula for an "open box" by excluding the area of one face (the top). Also, remember to consider both inside and outside surfaces when calculating the total area to be painted.

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

Question 1. 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)
Answer:
(Students should attempt the above activities on their own)
In simple words: This activity requires you to practically measure the dimensions of various sports courts and then calculate their perimeter and area using the measured values.

🎯 Exam Tip: When performing practical measurements, ensure accuracy and use appropriate units. Clearly state the formulas used for perimeter and area in your calculations.

Question 2. Take mobile handsets of different sizes and find the area of their screens. (Textbook pg. no. 82)
Answer:
(Students should attempt the above activities on their own)
In simple words: For this activity, select several mobile phones, measure the length and width of their screens, and then compute the area of each screen.

🎯 Exam Tip: Remember that most phone screens are rectangular, so their area can be found by multiplying length by width. Be precise with measurements to get accurate area values.