Welcome dear students! Today we are going to learn about Perimeter and Area from Class 6 Maths. Do you remember what the perimeter of a closed plane figure is? Let us refresh our understanding! The perimeter of any closed plane figure is the distance covered along its boundary when you go around it once. For a polygon, that is, a closed plane figure made up of line segments, the perimeter is simply the sum of the lengths of all its sides, which means the total distance along its outer boundary. So, the perimeter of a polygon equals the sum of the lengths of all its sides. Let us revise the formulas for the perimeter of rectangles, squares, and triangles. [CHECKPOINT]
First, let us look at the perimeter of a rectangle. Consider a rectangle A B C D whose length and breadth are 12 cm and 8 cm, respectively. What is its perimeter? The perimeter of the rectangle equals the sum of the lengths of its four sides, which is A B plus B C plus C D plus D A. In this rectangle, side A B is 12 cm and side B C is 8 cm. Since opposite sides of a rectangle are always equal, A B equals C D and A D equals B C. So we can write the perimeter as A B plus B C plus A B plus B C. This simplifies to 2 times A B plus 2 times B C, which is 2 times the quantity A B plus B C. Substituting the values, we get 2 times 12 cm plus 8 cm, which is 2 times 20 cm, giving us 40 cm. From this example, we see that the perimeter of a rectangle equals 2 times the quantity length plus breadth. The perimeter of a rectangle is twice the sum of its length and breadth. [CHECKPOINT]
Next, let us find the perimeter of a square. Debojeet wants to put coloured tape all around a square photo frame of side 1 m as shown in the figure. What will be the length of the coloured tape he requires? Since he wants to put the tape all around the square photo frame, he needs to find its perimeter. Thus, the length of the tape required equals the sum of the lengths of all four sides of the square. That is 1 m plus 1 m plus 1 m plus 1 m, which equals 4 m. Now, we know that all four sides of a square are equal in length. Therefore, instead of adding each side, we can simply multiply the length of one side by 4. Thus, the length of the tape required equals 4 times 1 m, which is 4 m. From this example, we see that the perimeter of a square equals 4 times the length of a side. The perimeter of a square is quadruple the length of its side. [CHECKPOINT]
Now, let us look at the perimeter of a triangle. Consider a triangle having three given sides of lengths 4 cm, 5 cm and 7 cm. To find its perimeter, we simply add them up. The perimeter of the triangle equals 4 cm plus 5 cm plus 7 cm, which equals 16 cm. The perimeter of a triangle equals the sum of the lengths of its three sides. Let us move on to some worked examples. Example one: Akshi wants to put lace all around a rectangular tablecloth that is 3 m long and 2 m wide. Find the length of the lace required. Solution: The length of the rectangular table cover is 3 m. The breadth is 2 m. Since the lace goes all around, the length required equals the perimeter of the tablecloth. Now, the perimeter equals 2 times the quantity length plus breadth. Substituting the values, we get 2 times 3 m plus 2 m, which is 2 times 5 m, giving 10 m. Hence, the length of the lace required is 10 m. [CHECKPOINT]
Example two: Find the distance travelled by Usha if she takes three rounds of a square park of side 75 m. Solution: First, we find the perimeter of the square park. The perimeter equals 4 times the length of a side, which is 4 times 75 m, giving 300 m. The distance covered in one round is 300 m. Therefore, the total distance travelled in three rounds equals 3 times 300 m, which is 900 m. Now, let us solve the first set of exercises. Question one: Find the missing terms. Part a: Perimeter of a rectangle is 14 cm and breadth is 2 cm. We know perimeter equals 2 times length plus breadth. So 14 equals 2 times length plus 2. Dividing by 2, we get 7 equals length plus 2. Subtracting 2, the length is 5 cm. Part b: Perimeter of a square is 20 cm. Since perimeter equals 4 times side, 20 equals 4 times side. Dividing by 4, the side length is 5 cm. Part c: Perimeter of a rectangle is 12 m and length is 3 m. Using the formula, 12 equals 2 times 3 plus breadth. So 6 equals 3 plus breadth. The breadth is 3 m. [CHECKPOINT]
Question two: A rectangle having side lengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square? First, find the perimeter of the rectangle. It equals 2 times 5 cm plus 3 cm, which is 16 cm. The wire length is 16 cm. When bent into a square, the perimeter remains 16 cm. So 4 times side equals 16 cm. The side length is 4 cm. Question three: Find the length of the third side of a triangle having a perimeter of 55 cm and two sides of length 20 cm and 14 cm. The perimeter equals the sum of all three sides. So 55 equals 20 plus 14 plus third side. That is 55 equals 34 plus third side. The third side is 21 cm. Question four: What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs 40 rupees per metre? First, find the perimeter: 2 times 150 m plus 120 m equals 2 times 270 m, which is 540 m. The cost is 540 times 40 rupees, which equals 21600 rupees. [CHECKPOINT]
Question five: A piece of string is 36 cm long. What will be the length of each side, if it is used to form a square, a triangle with all sides of equal length, and a hexagon with sides of equal length? For the square, 4 times side equals 36 cm, so each side is 9 cm. For the equilateral triangle, 3 times side equals 36 cm, so each side is 12 cm. For the hexagon, 6 times side equals 36 cm, so each side is 6 cm. Question six: A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope. What is the total length of rope needed? First, find the perimeter: 2 times 230 m plus 160 m equals 2 times 390 m, which is 780 m. For 3 rounds, multiply by 3: 780 m times 3 equals 2340 m. Now, let us look at the Matha Pachchi activity. Akshi and Toshi start running along rectangular tracks. The outer track for Akshi has length 70 m and breadth 40 m. The inner track for Toshi has length 60 m and breadth 30 m. Akshi completes 5 rounds. Toshi completes 7 rounds. Let us find out who ran more. [CHECKPOINT]
First, find the total distance Akshi covered. The perimeter of Akshi track is 2 times 70 m plus 40 m, which equals 220 m. For 5 rounds, 5 times 220 m equals 1100 m. Second, find the total distance Toshi covered. The perimeter of Toshi track is 2 times 60 m plus 30 m, which equals 180 m. For 7 rounds, 7 times 180 m equals 1260 m. Comparing the two, Toshi ran a longer distance. Now, think and mark the positions as directed. Part a: Mark A where Akshi will be after 250 m. Since one round is 220 m, after 220 m she is back at the start. She runs 30 m more along the 70 m length. So A is 30 m from the starting point along the first long side. Part b: Mark B where Akshi will be after 500 m. 500 divided by 220 is 2 full rounds with a remainder of 60 m. So B is 60 m from the start along the first long side. Part c: Akshi ran 1000 m. 1000 divided by 220 is 4 full rounds with a remainder of 120 m. She has finished 4 full rounds. Mark her position C. 120 m from start means she covers the 70 m length, then 50 m along the 40 m breadth. So C is 50 m along the second side. [CHECKPOINT]
Part d: Mark X where Toshi will be after 250 m. Toshi round is 180 m. 250 minus 180 is 70 m. So X is 70 m from start. She runs 60 m along the length, then 10 m along the 30 m breadth. Part e: Mark Y where Toshi will be after 500 m. 500 divided by 180 is 2 full rounds with remainder 140 m. Y is 140 m from start. That covers 60 m length, 30 m breadth, and 50 m along the second length side. So Y is 50 m along the second length side. Part f: Toshi ran 1000 m. 1000 divided by 180 is 5 full rounds with remainder 100 m. She has finished 5 full rounds. Mark Z at 100 m from start, which is 60 m plus 30 m plus 10 m along the second length side. Now, let us dive deeper into races. Usually there is a common finish line. Here are two square running tracks with the inner track of 100 m each side and outer track of 150 m each side. The common finishing line is shown by flags in the center of one of the sides. If the total race is 350 m, we must find where the starting positions should be so both finish together. [CHECKPOINT]
The inner track perimeter is 4 times 100 m equals 400 m. To run 350 m, the runner on the inner track must start 50 m before the finish line. Mark this starting point as A. The outer track perimeter is 4 times 150 m equals 600 m. To run 350 m, the runner on the outer track must start 250 m before the finish line. Mark this starting point as B. Now, let us do an estimate and verify activity. Take a rough sheet of paper or a sheet of newspaper. Make a few random shapes by cutting the paper in different ways. Estimate the total length of the boundaries of each shape, then use a scale or measuring tape to measure and verify the perimeter for each shape. Next, consider a debate. Akshi says that the perimeter of a triangle shape is 9 units. Toshi says it cannot be 9 units and the perimeter will be more than 9 units. What do you think? The figure has lines of two different unit lengths. Measure the lengths of a red line and a blue line. Are they the same? We will call the red lines straight lines and the blue lines diagonal lines. So, the perimeter of this triangle is 6 straight units plus 3 diagonal units. We can write this in a short form as 6s plus 3d units. Since diagonal lines are longer than straight lines, the perimeter is indeed more than 9 units. [CHECKPOINT]
Now, look at the next figures in your textbook. Count the straight and diagonal lines for each, and write their perimeters in the short form, just like we did above. Let us move to the perimeter of a regular polygon. Like squares, closed figures that have all sides and all angles equal are called regular polygons. Examples of regular polygons are the equilateral triangle, where all three sides and all three angles are equal, and the regular pentagon, where all five sides and all five angles are equal. For the perimeter of an equilateral triangle, we know that for any triangle its perimeter is the sum of all three sides. Using this understanding, we can find the perimeter of an equilateral triangle. Perimeter of an equilateral triangle equals A B plus B C plus A C. Since all sides are equal, this equals A B plus A B plus A B, which is 3 times length of one side. So, perimeter of an equilateral triangle equals 3 times length of a side. What is a similarity between a square and an equilateral triangle? Both are regular polygons with all sides equal. [CHECKPOINT]
Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalise your understanding for the perimeter of other regular polygons. The perimeter of any regular polygon equals the number of sides multiplied by the length of one side. Now, let us do a split and rejoin activity. A rectangular paper chit of dimension 6 cm times 4 cm is cut into two equal pieces. These two pieces are joined in different ways. In arrangement a, the two 6 cm by 2 cm rectangles are joined end to end to form a longer rectangle of 12 cm by 2 cm. The perimeter is 28 cm. In arrangement b, they form an L-shape. By tracing the outer boundary, you will find the perimeter is 24 cm. In arrangement c, they form a T-shape, and the perimeter is 28 cm. In arrangement d, they are arranged with a vertical offset, giving a perimeter of 30 cm. Try arranging the two pieces to form a figure with a perimeter of 22 cm by overlapping them partially along the longer side to reduce the boundary. [CHECKPOINT]
Now, let us move to section 6.2 Area. We have studied the areas of closed figures in previous grades. Let us recall some key points. The amount of region enclosed by a closed figure is called its area. In previous grades, we arrived at the formula for the area of a rectangle and a square using square grid paper. Do you remember? Area of a square equals side times side. Area of a rectangle equals length times width. Let us see some real-life problems. Example: A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted. Solution: Length of the floor is 5 m. Width of the floor is 4 m. Area of the floor equals length times width, which is 5 m times 4 m, giving 20 sq m. Length of the square carpet is 3 m. Area of the carpet equals length times length, which is 3 m times 3 m, giving 9 sq m. Therefore, the area of the floor that is not carpeted is area of the floor minus the area of the carpet, which is 20 sq m minus 9 sq m, giving 11 sq m. [CHECKPOINT]
Example: Four square flower beds each of side 4 m are in four corners on a piece of land 12 m long and 10 m wide. Find the area of the remaining part of the land. Solution: Length of the land is 12 m. Width of land is 10 m. Area of the whole land equals 12 m times 10 m, giving 120 sq m. The side length of each flower bed is 4 m. Area of one flower bed equals 4 m times 4 m, giving 16 sq m. Hence, the area of the four flower beds equals 4 times 16 sq m, giving 64 sq m. Therefore, the area of the remaining part of the land is 120 sq m minus 64 sq m, giving 56 sq m. Now, solve the next exercises. Question one: The area of a rectangular garden 25 m long is 300 sq m. What is the width? Area equals length times width. So 300 equals 25 times width. Width equals 300 divided by 25, which is 12 m. Question two: What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of 8 rupees per hundred sq m? Area equals 500 m times 200 m, giving 100000 sq m. Number of hundred sq m units is 100000 divided by 100, which is 1000. Cost equals 1000 times 8 rupees, giving 8000 rupees. [CHECKPOINT]
Question three: A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted? Area of grove equals 100 m times 50 m, giving 5000 sq m. Number of trees equals 5000 divided by 25, giving 200 trees. Question four: By splitting the following figures into rectangles, find their areas. For figure a, we split it into two rectangles. The top rectangle is 3 m by 1 m, area 3 sq m. The bottom rectangle is 4 m by 3 m, area 12 sq m. Total area is 15 sq m. For figure b, we split it into two rectangles. The left rectangle is 5 m by 3 m, area 15 sq m. The right rectangle is 3 m by 2 m, area 6 sq m. Total area is 21 sq m. Now, let us explore the tangram activity. Cut out the tangram pieces given at the end of your textbook. The pieces are labeled A through G. A is a large purple triangle. B is a large blue triangle. C is a small teal triangle. D is a green square. E is a small pink triangle. F is a purple parallelogram. G is an orange parallelogram. Question one: Explore and figure out how many pieces have the same area. Shapes A and B have the same area. Shapes C and E have the same area. [CHECKPOINT]
Question two: How many times bigger is Shape D as compared to Shape C? Shape D can be exactly covered using Shapes C and E, so Shape D is twice as big as Shape C. Question three: Which shape has more area: Shape D or F? They have the same area. You can verify by cutting and rearranging. Question four: Which shape has more area: Shape F or G? They have the same area. Question five: What is the area of Shape A as compared to Shape G? Is it twice as big? Four times as big? Shape A is four times as big as Shape C. Shape G is twice as big as Shape C. So Shape A is twice as big as Shape G. Question six: Can you now figure out the area of the big square formed with all seven pieces in terms of the area of Shape C? The big square contains 16 small triangles of size C. So its area is 16 times the area of Shape C. Question seven: Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the area of Shape C now? The area remains the same, 16 times the area of Shape C, because we are just rearranging the same pieces. Question eight: Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? They are different. The rectangle will have a longer perimeter because the pieces are arranged in a more elongated shape, exposing more boundary length. [CHECKPOINT]
Now, look at the figures below and guess which one has a larger area. We can estimate the area of any simple closed shape by using a sheet of squared paper or graph paper where every square measures 1 unit by 1 unit or 1 square unit. To estimate the area, we trace the shape onto transparent paper, place it on graph paper, and follow conventions. One: The area of one full small square is taken as 1 sq unit. Two: Ignore portions less than half a square. Three: If more than half of a square is in a region, count it as 1 sq unit. Four: If exactly half the square is counted, take its area as half sq unit. Find the area of the following figures using these rules. Let us explore why area is generally measured using squares. Draw a circle on a graph sheet with diameter of length 3. Count the squares to estimate the area. Circles cannot be packed tightly without gaps, making accurate measurement difficult. If we pack a rectangle with circles, the first way has 42 circles and the second has 44 circles. We cannot use circles instead of squares because of the gaps. Try using different shapes like triangles and rectangles to fill space without overlaps and gaps. Squares are best because they tile perfectly, have equal sides, and make calculations simple. [CHECKPOINT]
Find the area in square metres of the floor outside of the corridor. If the total floor is 10 m by 8 m, area is 80 sq m. The corridor is 2 m by 8 m, area is 16 sq m. Outside area is 64 sq m. Find the area occupied by your school playground. Measure its length and width, then multiply. Let us explore rectangles on grid paper. Make as many rectangles as you can whose lengths and widths are whole numbers such that the area is 24 square units. The possible pairs are 1 by 24, 2 by 12, 3 by 8, and 4 by 6. Math talk: Which rectangle has the greatest perimeter? The 1 by 24 rectangle has perimeter 2 times 1 plus 24, which is 50 units. Which has the least perimeter? The 4 by 6 rectangle has perimeter 2 times 4 plus 6, which is 20 units. If you take a rectangle of area 32 sq cm, the greatest perimeter is for 1 by 32, giving 66 cm. The least is for 4 by 8, giving 24 cm. Given any area, the rectangle with the greatest perimeter is the most elongated one, where one side is 1 unit. The least perimeter is when the sides are closest to each other, approaching a square. [CHECKPOINT]
Now, section 6.3: Area of a Triangle. Draw a rectangle on paper and draw one of its diagonals. Cut along the diagonal to get two triangles. Check whether they overlap exactly. They do. They have the same area. Try with more rectangles and a square. The inference is that a diagonal divides a rectangle into two triangles of equal area. Now, see the figures. Is the area of the blue rectangle more or less than the area of the yellow triangle? They are the same. The yellow triangle is formed by combining two triangles that each make up half of two smaller rectangles, which together make the blue rectangle. Use grid paper to calculate. Find the area of blue triangle B A D. It covers 12 full squares and 4 half squares, so 14 sq units. Find the area of red triangle A B E. It also covers 14 sq units. Area of rectangle A B C D is 28 sq units. So, the area of triangle B A D is half of the area of rectangle A B C D. For triangle A B E, it is made of triangle A E F and triangle B E F. Area of A E F is half of rectangle A F E D. Area of B E F is half of rectangle B F E C. Thus, area of triangle A B E equals half of the area of rectangle A F E D plus half of the area of rectangle B F E C, which equals half of the sum of the areas of the rectangles A F E D and B F E C, which equals half of the area of rectangle A B C D. Conclusion: The area of any triangle is half the product of its base and height, or half the area of the rectangle that encloses it. [CHECKPOINT]
Now, find the areas of the figures below by dividing them into rectangles and triangles. For figure a, split into a 5 by 3 rectangle and a triangle of base 5 and height 2. Area is 15 plus 5 equals 20 sq units. For figure b, split into a 4 by 4 square and a triangle of base 4 and height 3. Area is 16 plus 6 equals 22 sq units. For figure c, split into two triangles. Each has base 6 and height 4. Area is 12 plus 12 equals 24 sq units. For figure d, split into a 3 by 5 rectangle and a triangle of base 3 and height 2. Area is 15 plus 3 equals 18 sq units. For figure e, split into a 6 by 4 rectangle and a triangle of base 6 and height 2. Area is 24 plus 6 equals 30 sq units. Let us make it more or less. Observe two figures made with 9 unit squares. The first has perimeter 12 units. The second has perimeter 20 units. Arrange 9 unit squares to get other perimeters. Question one: What is the smallest perimeter possible? Arrange them in a 3 by 3 square. Perimeter is 2 times 3 plus 3, which is 12 units. Question two: What is the largest perimeter possible? Arrange them in a 1 by 9 line. Perimeter is 2 times 1 plus 9, which is 20 units. Question three: Make a figure with a perimeter of 18 units. Arrange them in a 1 by 8 line with one square attached to the side. Question four: Can you make other shapes for each perimeter? Yes, by rearranging the squares while keeping the same boundary length. [CHECKPOINT]
Now, let us do something tricky. We have a figure with perimeter 24 units. Without calculating again, observe what happens to the perimeter if a new square is attached on the right. If attached to a flat side, two new sides are added and two old sides are covered, so the perimeter stays the same. Experiment by placing the new square at different places. Can you place the square so that the perimeter increases? Yes, attach it to a corner where it only covers one side. Can you place it so that it decreases? Yes, attach it in a notch where it covers three sides. Can you place it so that it stays the same? Yes, attach it to a straight edge where it covers one side and adds one. Now, look at Charan's house plan. It is a rectangular plot. The dimensions given are: Master Bedroom 15 ft by 15 ft, area 225 sq ft. Toilet 5 ft by 10 ft. Kitchen 15 ft by 12 ft, area 180 sq ft. Small Bedroom 15 ft by 12 ft, area 180 sq ft. The total vertical dimension is 30 ft. Notice that the Hall, Garden, Parking, and Utility have missing measurements marked with blanks in your textbook. To find these, use the total plot size and subtract the known room lengths and widths. For example, the total height is 30 ft. If you know the heights of the rooms in one column, you can find the missing height for the Hall. Similarly, use the total width to find the missing widths for the Garden, Parking, and Utility. Once you find the dimensions, calculate each area by multiplying length and width, then add all the room areas together to find the total area of his house. Try solving this on your own! [CHECKPOINT]
Next, let us look at Sharan's home plan. The total width of the house is 42 ft. The top row rooms are: Master Bedroom 12 ft by 15 ft, Kitchen 18 ft by 10 ft, and Utility with an area of 70 sq ft. The bottom row has: Small Bedroom 12 ft by 10 ft, Hall with a length of 23 ft, and an Entrance. Notice that several dimensions and areas are left blank for you to find. To solve this, remember that the total width of 42 ft is shared by the rooms in each row. For the top row, add the known widths and subtract from 42 ft to find the missing widths. For the bottom row, do the same. The heights of the rooms in each row must add up to the total height of the house. Use the given areas and lengths to find the missing heights. Once you have all dimensions, calculate the missing areas and sum them to find the total area of Sharan's house. Compare your results with Charan's house to see which one has a larger area and perimeter. Now, solve the Area Maze Puzzles. In puzzle a, four rectangles in a 2 by 2 grid. Top left is 13 sq cm, top right is 26 sq cm, bottom left is 15 sq cm. The ratio of top areas is 13 to 26, or 1 to 2. So the bottom right must be twice the bottom left. 2 times 15 sq cm equals 30 sq cm. [CHECKPOINT]
In puzzle b, we have an L-shaped arrangement of three rectangles. The bottom rectangle has an area of 10 sq cm. The middle vertical rectangle also has an area of 10 sq cm. To find the area of the top right shaded rectangle, we use the shared sides. First, determine the height of the middle rectangle from the bottom rectangle's dimensions. Once you find the shared height, apply it to the top right rectangle's given width of 3 cm to calculate its area. Following this logical deduction, the top right rectangle's area is 6 sq cm. In puzzle c, three stacked rectangles. Total height 15 cm. Bottom is 60 sq cm, extends 5 cm beyond middle. Middle is 42 sq cm, height 6 cm. Top is ? sq cm, width 3 cm. Middle width is 42 divided by 6, which is 7 cm. Bottom width is 7 plus 5 equals 12 cm. Bottom height is 60 divided by 12, which is 5 cm. Top height is 15 minus 6 minus 5, which is 4 cm. Top area is 3 times 4, which is 12 sq cm. In puzzle d, two rectangles forming an L-shape. Left is 38 sq cm, top width ? cm. Right is 18 sq cm, width 5 cm. Vertical gap is 4 cm. Right height is 18 divided by 5, which is 3.6 cm. Total height is 3.6 plus 4 equals 7.6 cm. Left width is 38 divided by 7.6, which is 5 cm. So the missing width is 5 cm. [CHECKPOINT]
Now, the final set of exercises. Question one: Give the dimensions of a rectangle whose area is the sum of the areas of two rectangles having measurements 5 m by 10 m and 2 m by 7 m. First area is 50 sq m. Second is 14 sq m. Sum is 64 sq m. Possible dimensions are 8 m by 8 m, or 4 m by 16 m. Question two: The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width. Width equals 1000 divided by 50, which is 20 m. Question three: The floor of a room is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area that is not carpeted. Floor area is 20 sq m. Carpet area is 9 sq m. Not carpeted area is 11 sq m. Question four: Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn? Garden area is 15 times 12 equals 180 sq m. Each flower bed area is 2 times 1 equals 2 sq m. Four beds area is 8 sq m. Available area is 180 minus 8 equals 172 sq m. Question five: Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes. Shape A can be a 1 by 18 rectangle, perimeter 38. Shape B can be a 4 by 5 rectangle, perimeter 18. [CHECKPOINT]
Question six: On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border? Since textbook page sizes vary, you will need to measure your own book page first. Let us say your page measures L cm in length and W cm in width. The border is 1 cm from the top and bottom, so the inner length is L minus 2 cm. It is 1.5 cm from the left and right, so the inner width is W minus 3 cm. The perimeter of the border is 2 times the quantity inner length plus inner width. Measure your page, substitute your values into this formula, and calculate the exact perimeter for your book. Question seven: Draw a rectangle of size 12 units by 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area. Outer area is 96 sq units. Inner area must be 48 sq units. Possible whole number dimensions for 48 are 6 by 8 or 4 by 12, but both would touch the outer edges if placed centrally. To strictly not touch, you can use non-integer dimensions or simply draw a 6 by 8 rectangle and leave a small visible margin from the border. The exercise highlights how area and boundary constraints interact. [CHECKPOINT]
Question eight: A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, which statement is always true? Let side be s. Square perimeter is 4s. Each rectangle has dimensions s by s/2. Perimeter of one rectangle is 2 times s plus s/2, which is 3s. Two rectangles perimeter sum is 6s. Option a is false. Option b is false. Option c says perimeters added is 1.5 times square perimeter. 6s divided by 4s is 1.5. So c is true. Option d is false. Therefore, statement c is always true. Let us review the summary. The perimeter of a polygon is the sum of the lengths of all its sides. The perimeter of a rectangle is twice the sum of its length and width. The perimeter of a square is four times the length of any one of its sides. The area of a closed figure is the measure of the region enclosed by the figure. Area is generally measured in square units. The area of a rectangle is its length times its width. The area of a square is the length of any one of its sides multiplied by itself. Two closed figures can have the same area with different perimeters, or the same perimeter with different areas. Areas of regions can be estimated or determined exactly by breaking up such regions into unit squares, or into more general-shaped rectangles and triangles whose areas can be calculated. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]