Welcome dear students! Today we are going to learn about Perimeter And Area from Class 7 Maths. We come across many shapes other than squares and rectangles. How will you find the area of a land which is a parallelogram in shape? Let us find a method to get the area of a parallelogram. Can a parallelogram be converted into a rectangle of equal area? Draw a parallelogram on graph paper as shown in Figure 9.1(i). Cut out the parallelogram. Draw a line from one vertex perpendicular to the opposite side, as shown in Figure 9.1(ii). Cut out the triangle and move it to the other side of the parallelogram. What shape do you get? You get a rectangle. Is the area of the parallelogram equal to the area of the rectangle formed? Yes. What are the length and breadth of the rectangle? We find that the length equals the base of the parallelogram and the breadth equals the height of the parallelogram, as shown in Figure 9.2. Thus, Area of parallelogram = Area of rectangle = length × breadth = l × b. But the length l and breadth b of the rectangle are exactly the base b and the height h of the parallelogram. Thus, Area of parallelogram = base × height = b × h.
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Any side of a parallelogram can be chosen as its base. The perpendicular dropped from the opposite vertex to that base is the height or altitude. In parallelogram ABCD, if DE is perpendicular to AB, then AB is the base and DE is the height. If BF is perpendicular to AD, then AD is the base and BF is the height. Now look at the parallelograms in Figure 9.3. Find their areas by counting enclosed squares and their perimeters by measuring the sides. Complete the table provided in your textbook. Row a has base 5 units, height 3 units, and area 15 square units. For rows b through g, measure the base and height from your figure, calculate the area using base × height, and measure the perimeter. You will discover that parallelograms can have equal areas but different perimeters. Now consider the parallelograms in Figure 9.4, each with sides 7 cm and 5 cm. Find their perimeter and area. You will find they have equal perimeters but different areas. To find the area, you only need the base and its corresponding height.
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Let us try these activities. Find the area of the parallelograms shown in your textbook. For each figure, identify the base and the perpendicular height, then multiply them. In parallelogram ABCD, AB = 7.2 cm and the perpendicular from C to AB is 4.5 cm. The area is 7.2 × 4.5 = 32.4 cm². Now we move to section 9.2, Area of a Triangle. A gardener wants to cover a triangular garden with grass. We need the area of the triangle. Draw a scalene triangle on paper, cut it out, and use it to trace and cut a second identical triangle. You now have two congruent triangles. Superpose them to verify they match. Join them along a pair of corresponding sides as shown in Figure 9.5. The figure formed is a parallelogram. The sum of the areas of both triangles equals the area of this parallelogram. The base and height of each triangle match those of the parallelogram. Therefore, Area of each triangle = 1/2 × (Area of parallelogram) = 1/2 × (base × height) = 1/2 bh.
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In Figure 9.6, all triangles share the same base AB of 6 cm. Their heights corresponding to base AB are equal, so their areas are equal, but the triangles are not congruent. We conclude that congruent triangles are equal in area, but triangles equal in area need not be congruent. Consider the obtuse-angled ∆ABC in Figure 9.7 with base 6 cm. Its height AD falls outside the triangle, but the area formula still applies as 1/2 × base × height. Let us solve Example 1. One side and corresponding height of a parallelogram are 4 cm and 3 cm. Area = 4 × 3 = 12 cm². Example 2. Find height x if area is 24 cm² and base is 4 cm. Using 24 = 4 × x, we divide by 4 to get x = 6 cm. Try these activities with different triangles and parallelograms. Divide parallelograms along their diagonals. The resulting triangles are congruent.
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Example 3. Two sides of parallelogram ABCD are 6 cm and 4 cm. Height to base CD is 3 cm. Find the area and the height to base AD. Area = 6 × 3 = 18 cm². For base AD of 4 cm, let height be x. 18 = 4 × x, so x = 18 ÷ 4 = 4.5 cm. Example 4. Find the area of the triangles in Figure 9.11. For the first, base QR is 4 cm, height PS is 2 cm. Area = 1/2 × 4 × 2 = 4 cm². For the second, base MN is 3 cm, height LO is 2 cm. Area = 1/2 × 3 × 2 = 3 cm². Example 5. Find BC if area of ∆ABC is 36 cm² and height AD is 3 cm. 36 = 1/2 × b × 3. Multiply by 2 to get 72 = 3b, so b = 24 cm. BC is 24 cm.
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Example 6. In ∆PQR, PR is 8 cm, QR is 4 cm, and PL is 5 cm. Find the area and QM. Using base QR and height PL, area = 1/2 × 4 × 5 = 10 cm². Using base PR and height QM, 10 = 1/2 × 8 × h. 10 = 4h, so h = 2.5 cm. QM is 2.5 cm. Now let us solve Exercise 9.1. Question 1 asks you to find the area of each parallelogram shown. Measure the base and corresponding height directly from the figures in your textbook, then apply Area = base × height. Question 2 asks for the area of each triangle shown. Measure the base and height from your figures, then apply Area = 1/2 × base × height. Question 3 asks you to find the missing values in the parallelogram table. For row a, base is 20 cm and area is 246 cm², so height = 246 ÷ 20 = 12.3 cm. For row b, height is 15 cm and area is 154.5 cm², so base = 154.5 ÷ 15 = 10.3 cm. For row c, base is 8.4 cm and area is 48.72 cm², so height = 48.72 ÷ 8.4 = 5.8 cm. For row d, height is 15.6 cm and area is 16.38 cm², so base = 16.38 ÷ 15.6 = 1.05 cm.
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Question 4 asks for missing values in the triangle table. For row a, base is 15 cm and area is 87 cm². Height = (87 × 2) ÷ 15 = 11.6 cm. For row b, height is 31.4 mm and area is 1256 mm². Base = (1256 × 2) ÷ 31.4 = 80 mm. For row c, base is 22 cm and area is 170.5 cm². Height = (170.5 × 2) ÷ 22 = 15.5 cm. Question 5. Parallelogram PQRS has SR = 12 cm and height QM = 7.6 cm. Area = 12 × 7.6 = 91.2 cm². If PS = 8 cm, then QN = 91.2 ÷ 8 = 11.4 cm. Question 6. Area of parallelogram ABCD is 1470 cm². AB = 35 cm, AD = 49 cm. For base AB, height DL = 1470 ÷ 35 = 42 cm. For base AD, height BM = 1470 ÷ 49 = 30 cm. Question 7. ∆ABC has ∠A = 90°. AB = 5 cm, BC = 13 cm, AC = 12 cm. Area = 1/2 × AB × AC = 1/2 × 5 × 12 = 30 cm². Using BC as base, 30 = 1/2 × 13 × AD. 60 = 13 × AD, so AD = 60/13 cm, which is approximately 4.62 cm.
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Question 8. Isosceles ∆ABC has AB = AC = 7.5 cm, BC = 9 cm, and height AD = 6 cm. Area = 1/2 × 9 × 6 = 27 cm². To find height CE to base AB, 27 = 1/2 × 7.5 × CE. 54 = 7.5 × CE, so CE = 7.2 cm. Now section 9.3, Circles. A racing track is semi-circular at both ends. We need a method for circular distances. Section 9.3.1 covers Circumference of a Circle. Tanya wants to put lace around curved cards. Mark a point on the edge, roll it on a table until the mark touches again. The distance rolled is the circumference. You can also wrap a string around it. The distance around a circular region is its circumference. Consider the table of circles in your textbook. Circle 1: radius 3.5 cm, diameter 7.0 cm, circumference 22.0 cm, ratio = 22/7 ≈ 3.14. Circle 2: radius 7.0 cm, diameter 14.0 cm, circumference 44.0 cm, ratio = 44/14 ≈ 3.14. Circle 3: radius 10.5 cm, diameter 21.0 cm, circumference 66.0 cm, ratio = 66/21 ≈ 3.14. Circle 4: radius 21.0 cm, diameter 42.0 cm, circumference 132.0 cm, ratio = 132/42 ≈ 3.14.
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Circle 5: radius 5.0 cm, diameter 10.0 cm, circumference 32.0 cm, ratio = 3.2. Circle 6: radius 15.0 cm, diameter 30.0 cm, circumference 94.0 cm, ratio ≈ 3.13. The ratio is approximately constant. Circumference is always more than three times the diameter. This constant is π, approximately 22/7 or 3.14. So C/d = π, or C = πd. Since d = 2r, C = 2πr. Let us do this activity. Take one quarter plate and one half plate. Roll each once on a table-top. The half plate covers more distance in one complete revolution and will take fewer revolutions to cover the table's length. Now, try these: complete the table in your textbook relating circumference and radius. Draw circles of different radii on graph paper, find the area by counting squares, then compare with the formula. In Figure 9.22, the outer square has the larger perimeter. The circle's circumference is larger than the inner square's perimeter.
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Example 7. Circumference of a circle with diameter 10 cm, taking π = 3.14. C = 3.14 × 10 = 31.4 cm. Example 8. Circumference of a disc with radius 14 cm, using π = 22/7. C = 2 × 22/7 × 14 = 88 cm. Example 9. Pipe radius 10 cm. Tape length for one wrap, π = 3.14. C = 2 × 3.14 × 10 = 62.8 cm. Example 10. Perimeter of the shape in Figure 9.23, using π = 22/7. The boundary is four semicircles. Diameter of each is 14 cm. Circumference of one semicircle = 1/2 × πd = 1/2 × 22/7 × 14 = 22 cm. Total perimeter = 4 × 22 = 88 cm. Example 11. Semicircular disc from radius 7 cm, using π = 22/7. Perimeter = curved part + diameter = πr + 2r = 22/7 × 7 + 14 = 22 + 14 = 36 cm. Section 9.3.2 covers Area of a Circle. A farmer needs fertiliser for a radius 7 m flower bed. We need the area. Divide a circle into 64 sectors and arrange them into a near-rectangle. Breadth is r. Length is half the circumference, which is πr. Area = length × breadth = πr × r = πr².
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Example 12. Area of a circle with radius 30 cm, using π = 3.14. Area = 3.14 × 30² = 3.14 × 900 = 2,826 cm². Example 13. Garden diameter 9.8 m. Radius is 4.9 m. Area = 22/7 × 4.9 × 4.9 = 75.46 m². Example 14. Concentric circles. Larger radius 10 cm, smaller radius 4 cm, using π = 3.14. Larger area = 3.14 × 10 × 10 = 314 cm². Smaller area = 3.14 × 4 × 4 = 50.24 cm². Shaded area = 314 - 50.24 = 263.76 cm². Now Exercise 9.2. Question 1. Circumference, using π = 22/7. (a) r = 14 cm. C = 2 × 22/7 × 14 = 88 cm. (b) r = 28 mm. C = 2 × 22/7 × 28 = 176 mm. (c) r = 21 cm. C = 2 × 22/7 × 21 = 132 cm. Question 2. Area. (a) r = 14 mm, using π = 22/7. Area = 22/7 × 14 × 14 = 616 mm². (b) d = 49 m, so r = 24.5 m. Area = 22/7 × 24.5 × 24.5 = 1886.5 m². (c) r = 5 cm. The question specifies to take π = 3.14. Area = 3.14 × 5 × 5 = 78.5 cm².
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Question 3. Circumference is 154 m, using π = 22/7. 2 × 22/7 × r = 154. r = 154 × 7 / 44 = 24.5 m. Area = 22/7 × 24.5 × 24.5 = 1886.5 m². Question 4. Garden diameter 21 m. Circumference = 2 × 22/7 × 10.5 = 66 m. Two rounds = 132 m. Cost at ₹4 per meter = 132 × 4 = ₹528. Question 5. Outer radius 4 cm, inner radius 3 cm, using π = 3.14. Remaining area = 3.14 × (4² - 3²) = 3.14 × (16 - 9) = 3.14 × 7 = 21.98 cm². Question 6. Table diameter 1.5 m. Circumference = 2 × 3.14 × 0.75 = 4.71 m. Cost at ₹15 per meter = 4.71 × 15 = ₹70.65. Question 7. Find the perimeter of the adjoining figure, a semicircle including its diameter. Refer to the figure in your textbook to identify the diameter. Once you have the diameter d, the perimeter is calculated as πd/2 + d. Question 8. Table diameter 1.6 m, radius 0.8 m. Area = 3.14 × 0.8 × 0.8 = 2.0096 m². Cost at ₹15 per square meter = 2.0096 × 15 = ₹30.144.
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Question 9. Wire length 44 cm bent to a circle. 2 × 22/7 × r = 44. r = 7 cm. Area = 22/7 × 7 × 7 = 154 cm². Bent to a square, 4a = 44, so a = 11 cm. Area = 121 cm². The circle encloses more area. Question 10. Large sheet radius 14 cm. Two circles of radius 3.5 cm and a rectangle 3 cm by 1 cm are removed. Large area = 22/7 × 14 × 14 = 616 cm². Two small circles = 2 × 22/7 × 3.5 × 3.5 = 77 cm². Rectangle area = 3 cm². Remaining area = 616 - 77 - 3 = 536 cm². Question 11. Square side 6 cm. Circle radius 2 cm cut out, using π = 3.14. Leftover area = 6² - 3.14 × 2² = 36 - 12.56 = 23.44 cm². Question 12. Circumference 31.4 cm, using π = 3.14. 2 × 3.14 × r = 31.4. r = 5 cm. Area = 3.14 × 5² = 78.5 cm². Question 13. Flower bed diameter 66 m, radius 33 m. Path width 4 m. Outer radius 37 m. Path area = 3.14 × (37² - 33²) = 3.14 × (1369 - 1089) = 3.14 × 280 = 879.2 m².
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Question 14. Garden area 314 m². 3.14r² = 314, so r² = 100, r = 10 m. Sprinkler radius is 12 m. Since 12 is greater than 10, the sprinkler waters the entire garden. Question 15. Find the circumference of the inner and outer circles in the adjoining figure, using π = 3.14. Refer to your textbook figure to read the exact inner and outer radii. Apply the formula C = 2πr for each. Question 16. Wheel radius 28 cm. Circumference = 2 × 22/7 × 28 = 176 cm = 1.76 m. Distance 352 m. Rotations = 352 / 1.76 = 200. Question 17. Minute hand length 15 cm. Distance in 1 hour = 2 × 3.14 × 15 = 94.2 cm. What have we discussed? Area of parallelogram = base × height. Area of triangle = 1/2 × base × height. The distance around a circular region is its circumference, given by C = πd or C = 2πr, where π is approximately 22/7 or 3.14. Area of a circle = πr². Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]