Page 1 Points to Remember : Region. A plane figure together with its interior is called the region enclosed by the plane figure. Area. The measurement of the region encloed by a plane figure is called the area of the plane figure. The standard unit of area is cm 2 . Source Formulae : 1. Area of rectangle = (length × breadth) Sq. units. 2. Length of a rectangle F H G I K J Area Breadth units 3. Breadth of a rectangle F H G I K J Area Length units 4. Perimeter of a rectangle = 2 × (length + breadth) units. 5. Area of a square = (Side) 2 sq. units. 6. Perimeter of a square = (4 × side) units. Units of Area. The standard unit of area is cm 2 . For measuring areas of lands and fields, we use larger units such as m 2 , areas and hectares etc. Conversion of Units : Units of Length Units of Area 1 cm = 10 mm 1 cm 2 = (10 × 10) mm 2 = 100 mm 2 1 m = 100 cm. 1 m 2 = (100 × 100) cm 2 = 10000 cm 2 1 dm = 10 cm 1 dm 2 = (10 × 10) cm 2 = 100 cm 2 1 dam = 10 m 1 dam 2 = (10 × 10) m 2 = 100 m 2 1 km = 10 hm 1 km 2 = (10 × 10) hm 2 = 100 hm 2 1 km = 1000 m 1 km 2 = (1000 × 1000) m 2 = 1000000 m 2 Page 2 Points to Remember : Region. A plane figure together with its interior is called the region enclosed by the plane figure. Area. The measurement of the region encloed by a plane figure is called the area of the plane figure. The standard unit of area is cm 2 . Source Formulae : 1. Area of rectangle = (length × breadth) Sq. units. 2. Length of a rectangle F H G I K J Area Breadth units 3. Breadth of a rectangle F H G I K J Area Length units 4. Perimeter of a rectangle = 2 × (length + breadth) units. 5. Area of a square = (Side) 2 sq. units. 6. Perimeter of a square = (4 × side) units. Units of Area. The standard unit of area is cm 2 . For measuring areas of lands and fields, we use larger units such as m 2 , areas and hectares etc. Conversion of Units : Units of Length Units of Area 1 cm = 10 mm 1 cm 2 = (10 × 10) mm 2 = 100 mm 2 1 m = 100 cm. 1 m 2 = (100 × 100) cm 2 = 10000 cm 2 1 dm = 10 cm 1 dm 2 = (10 × 10) cm 2 = 100 cm 2 1 dam = 10 m 1 dam 2 = (10 × 10) m 2 = 100 m 2 1 km = 10 hm 1 km 2 = (10 × 10) hm 2 = 100 hm 2 1 km = 1000 m 1 km 2 = (1000 × 1000) m 2 = 1000000 m 2 Q. 1. Find the perimeter of a rectangle in which : (i) length = 16·8 cm and breadth = 6·2 cm (ii) length = 2 m 25 cm and breadth = 1 m 50 cm (iii) length = 8 m 5 dm and breadth = 6 m 8 dm Sol. (i) Length (l) = 16·8 cm Breadth (b) = 6·2 cm Perimeter = 2 (l + b) = 2 (16·8 + 6·2) cm = 2 × 23 = 46 cm Ans. (ii) Length of rectanlge (l) = 2 m, 25 cm = 2.25 m and breadth (b) = 1 m 50 cm = 1.50 m D C A B 2m 25cm 1m 50cm Perimeter = 2 (l + b) = 2 (2.25 + 1.50) m = 2 × 3.75 m = 7.50 m = 7 m 50 cm (iii) Length of a rectangle (l) = 8 m 5 dm = 8.5 m and breadth (b) = 6 m 8 dm = 6.8 m D C A B 8m 5dm 6m 8dm Perimeter = 2 (l + b) = 2 (8.5 + 6.8) m = 2 × 15.3 m = 30.6 m = 30 m 6 dm Q.2. Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs. 16 per metre. Sol. Length of rectangular field (l) = 62 m and breadth (b) = 33 m C D B A 33 m 62 m Perimeter = 2 (l + b) = 2 (62 + 33) = 2 × 95 m = 190 m Rate of fencing = Rs. 16 per m Total cost of fencing the field = Rs. 16 × 190 = Rs. 3040 Q.3. The length and the breadth of a rectangular field are in the ratio 5 : 3. If its perimeter is 128 m, find the dimensions of the field. Sol. Perimeter of field = 128 m Length + Breadth = 2 128 = 64 m Ratio in length and breadth = 5 : 3 Let length (l) = 5x Then breadth = 3x 5x + 3x = 64 8x = 64 x = 8 64 = 8 Length of the field = 5x = 5 × 8 = 40 m and breadth = 3x = 3 × 8 = 24 m Q.4. The cost of fencing a rectangular field at Rs. 18 per metre is Rs. 1980. If the width of the field is 23 m, find its length. Page 3 Points to Remember : Region. A plane figure together with its interior is called the region enclosed by the plane figure. Area. The measurement of the region encloed by a plane figure is called the area of the plane figure. The standard unit of area is cm 2 . Source Formulae : 1. Area of rectangle = (length × breadth) Sq. units. 2. Length of a rectangle F H G I K J Area Breadth units 3. Breadth of a rectangle F H G I K J Area Length units 4. Perimeter of a rectangle = 2 × (length + breadth) units. 5. Area of a square = (Side) 2 sq. units. 6. Perimeter of a square = (4 × side) units. Units of Area. The standard unit of area is cm 2 . For measuring areas of lands and fields, we use larger units such as m 2 , areas and hectares etc. Conversion of Units : Units of Length Units of Area 1 cm = 10 mm 1 cm 2 = (10 × 10) mm 2 = 100 mm 2 1 m = 100 cm. 1 m 2 = (100 × 100) cm 2 = 10000 cm 2 1 dm = 10 cm 1 dm 2 = (10 × 10) cm 2 = 100 cm 2 1 dam = 10 m 1 dam 2 = (10 × 10) m 2 = 100 m 2 1 km = 10 hm 1 km 2 = (10 × 10) hm 2 = 100 hm 2 1 km = 1000 m 1 km 2 = (1000 × 1000) m 2 = 1000000 m 2 Q. 1. Find the perimeter of a rectangle in which : (i) length = 16·8 cm and breadth = 6·2 cm (ii) length = 2 m 25 cm and breadth = 1 m 50 cm (iii) length = 8 m 5 dm and breadth = 6 m 8 dm Sol. (i) Length (l) = 16·8 cm Breadth (b) = 6·2 cm Perimeter = 2 (l + b) = 2 (16·8 + 6·2) cm = 2 × 23 = 46 cm Ans. (ii) Length of rectanlge (l) = 2 m, 25 cm = 2.25 m and breadth (b) = 1 m 50 cm = 1.50 m D C A B 2m 25cm 1m 50cm Perimeter = 2 (l + b) = 2 (2.25 + 1.50) m = 2 × 3.75 m = 7.50 m = 7 m 50 cm (iii) Length of a rectangle (l) = 8 m 5 dm = 8.5 m and breadth (b) = 6 m 8 dm = 6.8 m D C A B 8m 5dm 6m 8dm Perimeter = 2 (l + b) = 2 (8.5 + 6.8) m = 2 × 15.3 m = 30.6 m = 30 m 6 dm Q.2. Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs. 16 per metre. Sol. Length of rectangular field (l) = 62 m and breadth (b) = 33 m C D B A 33 m 62 m Perimeter = 2 (l + b) = 2 (62 + 33) = 2 × 95 m = 190 m Rate of fencing = Rs. 16 per m Total cost of fencing the field = Rs. 16 × 190 = Rs. 3040 Q.3. The length and the breadth of a rectangular field are in the ratio 5 : 3. If its perimeter is 128 m, find the dimensions of the field. Sol. Perimeter of field = 128 m Length + Breadth = 2 128 = 64 m Ratio in length and breadth = 5 : 3 Let length (l) = 5x Then breadth = 3x 5x + 3x = 64 8x = 64 x = 8 64 = 8 Length of the field = 5x = 5 × 8 = 40 m and breadth = 3x = 3 × 8 = 24 m Q.4. The cost of fencing a rectangular field at Rs. 18 per metre is Rs. 1980. If the width of the field is 23 m, find its length. Sol. Cost of fencing a rectangular field = Rs. 18 per m Total cost = Rs. 1980 Perimeter of the field = Rate cost Total = 18 1980 = 110 m and length + breadth = 2 110 = 55 m Width of the field = 23 m Length = 55 – 23 = 32 m Q.5. The length and the breadth of a rectangular field are in the ratio 7 : 4. The cost of fencing the field at Rs. 25 per metre is Rs. 3300. Find the dimensions of the field. Sol. Total cost of fencing a rectangular field = Rs. 3300 Rate of fencing = Rs. 25 per m Perimeter = Rate cost Total = 25 3300 = 132 m and length + breadth = 2 Perimeter = 2 132 = 66 m Now ratio in length and breadth = 7 : 4 Let length = 7x Then breadth = 4x 7x + 4x = 66 11x = 66 x = 11 66 = 6 Length of the field = 7x = 7 × 6 = 42 m and breadth = 4x = 4 × 6 = 24 m Q.6. Find the perimeter of a square, each of whose sides measures : (i) 3.8 cm (ii) 4.6 m (iii) 2 m 5 dm Sol. (i) Side of square = 3.8 cm Perimeter = 4 × side = 4 × 3.8 cm = 15.2 cm (ii) Side of a square = 4.6 m Perimeter = 4 × side = 4 × 4.6 m = 18.4 m (iii) Side of a square = 2 m 5 dm = 2.5 m Perimeter = 4 × side = 4 × 2.5 m = 10 m Q.7. The cost of putting a fence around a square field at Rs. 35 per metre is Rs. 4480. Find the length of each side of the field. Sol. Total cost of fencing a square field = Rs. 4480 Rate of fencing = Rs. 35 per m Perimeter of square field = Rate cost Total = 35 4480 = 128 m Side of the square = 4 Perimeter = 4 128 = 32 m Q. 8. Each side of a square field measures 21 m, Adjacent to this field, there is a rectangular field having its sides in the ratio 4 : 3. If the perimeters of both of the fields are equal, find the dimensions of the rectangular field. Sol. Side of a square field (a) = 21 m Perimeter = 4 a = 4 × 21 = 84 m Perimeter of rectangular field = 84 m Ratio in length and breadth = 4 : 3 Let length (l) = 4 x and breadth (b) = 3 x Perimeter = 2 (l + b) Page 4 Points to Remember : Region. A plane figure together with its interior is called the region enclosed by the plane figure. Area. The measurement of the region encloed by a plane figure is called the area of the plane figure. The standard unit of area is cm 2 . Source Formulae : 1. Area of rectangle = (length × breadth) Sq. units. 2. Length of a rectangle F H G I K J Area Breadth units 3. Breadth of a rectangle F H G I K J Area Length units 4. Perimeter of a rectangle = 2 × (length + breadth) units. 5. Area of a square = (Side) 2 sq. units. 6. Perimeter of a square = (4 × side) units. Units of Area. The standard unit of area is cm 2 . For measuring areas of lands and fields, we use larger units such as m 2 , areas and hectares etc. Conversion of Units : Units of Length Units of Area 1 cm = 10 mm 1 cm 2 = (10 × 10) mm 2 = 100 mm 2 1 m = 100 cm. 1 m 2 = (100 × 100) cm 2 = 10000 cm 2 1 dm = 10 cm 1 dm 2 = (10 × 10) cm 2 = 100 cm 2 1 dam = 10 m 1 dam 2 = (10 × 10) m 2 = 100 m 2 1 km = 10 hm 1 km 2 = (10 × 10) hm 2 = 100 hm 2 1 km = 1000 m 1 km 2 = (1000 × 1000) m 2 = 1000000 m 2 Q. 1. Find the perimeter of a rectangle in which : (i) length = 16·8 cm and breadth = 6·2 cm (ii) length = 2 m 25 cm and breadth = 1 m 50 cm (iii) length = 8 m 5 dm and breadth = 6 m 8 dm Sol. (i) Length (l) = 16·8 cm Breadth (b) = 6·2 cm Perimeter = 2 (l + b) = 2 (16·8 + 6·2) cm = 2 × 23 = 46 cm Ans. (ii) Length of rectanlge (l) = 2 m, 25 cm = 2.25 m and breadth (b) = 1 m 50 cm = 1.50 m D C A B 2m 25cm 1m 50cm Perimeter = 2 (l + b) = 2 (2.25 + 1.50) m = 2 × 3.75 m = 7.50 m = 7 m 50 cm (iii) Length of a rectangle (l) = 8 m 5 dm = 8.5 m and breadth (b) = 6 m 8 dm = 6.8 m D C A B 8m 5dm 6m 8dm Perimeter = 2 (l + b) = 2 (8.5 + 6.8) m = 2 × 15.3 m = 30.6 m = 30 m 6 dm Q.2. Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs. 16 per metre. Sol. Length of rectangular field (l) = 62 m and breadth (b) = 33 m C D B A 33 m 62 m Perimeter = 2 (l + b) = 2 (62 + 33) = 2 × 95 m = 190 m Rate of fencing = Rs. 16 per m Total cost of fencing the field = Rs. 16 × 190 = Rs. 3040 Q.3. The length and the breadth of a rectangular field are in the ratio 5 : 3. If its perimeter is 128 m, find the dimensions of the field. Sol. Perimeter of field = 128 m Length + Breadth = 2 128 = 64 m Ratio in length and breadth = 5 : 3 Let length (l) = 5x Then breadth = 3x 5x + 3x = 64 8x = 64 x = 8 64 = 8 Length of the field = 5x = 5 × 8 = 40 m and breadth = 3x = 3 × 8 = 24 m Q.4. The cost of fencing a rectangular field at Rs. 18 per metre is Rs. 1980. If the width of the field is 23 m, find its length. Sol. Cost of fencing a rectangular field = Rs. 18 per m Total cost = Rs. 1980 Perimeter of the field = Rate cost Total = 18 1980 = 110 m and length + breadth = 2 110 = 55 m Width of the field = 23 m Length = 55 – 23 = 32 m Q.5. The length and the breadth of a rectangular field are in the ratio 7 : 4. The cost of fencing the field at Rs. 25 per metre is Rs. 3300. Find the dimensions of the field. Sol. Total cost of fencing a rectangular field = Rs. 3300 Rate of fencing = Rs. 25 per m Perimeter = Rate cost Total = 25 3300 = 132 m and length + breadth = 2 Perimeter = 2 132 = 66 m Now ratio in length and breadth = 7 : 4 Let length = 7x Then breadth = 4x 7x + 4x = 66 11x = 66 x = 11 66 = 6 Length of the field = 7x = 7 × 6 = 42 m and breadth = 4x = 4 × 6 = 24 m Q.6. Find the perimeter of a square, each of whose sides measures : (i) 3.8 cm (ii) 4.6 m (iii) 2 m 5 dm Sol. (i) Side of square = 3.8 cm Perimeter = 4 × side = 4 × 3.8 cm = 15.2 cm (ii) Side of a square = 4.6 m Perimeter = 4 × side = 4 × 4.6 m = 18.4 m (iii) Side of a square = 2 m 5 dm = 2.5 m Perimeter = 4 × side = 4 × 2.5 m = 10 m Q.7. The cost of putting a fence around a square field at Rs. 35 per metre is Rs. 4480. Find the length of each side of the field. Sol. Total cost of fencing a square field = Rs. 4480 Rate of fencing = Rs. 35 per m Perimeter of square field = Rate cost Total = 35 4480 = 128 m Side of the square = 4 Perimeter = 4 128 = 32 m Q. 8. Each side of a square field measures 21 m, Adjacent to this field, there is a rectangular field having its sides in the ratio 4 : 3. If the perimeters of both of the fields are equal, find the dimensions of the rectangular field. Sol. Side of a square field (a) = 21 m Perimeter = 4 a = 4 × 21 = 84 m Perimeter of rectangular field = 84 m Ratio in length and breadth = 4 : 3 Let length (l) = 4 x and breadth (b) = 3 x Perimeter = 2 (l + b) 84 = 2 (4 x + 3 x) = 2 × 7 x = 14 x x 84 14 6 Length = 4 x = 4 × 6 = 24 metre and breadth = 3 x = 3 × 6 = 18 metre Ans. Q.9. Find the perimeter of (i) a triangle of sides 7.8 cm, 6.5 cm and 5.9 cm, (ii) an equilateral triangle of side 9.4 cm, (iii) an isosocels triangle with equal sides 8.5 cm each and third side 7 cm. Sol. (i) Sides of a triangle are 7.8 cm, 6.5 cm and 5.9 cm A B C 7.8 cm 6.5 cm 5.9 cm Perimeter = Sum of sides = (7.8 + 6.5 + 5.9) cm = 20.2 cm (ii) Each side of an equilateral triangles = 9.4 cm A B C 9.4 cm 9.4 cm 9.4 cm Perimeter = 3 × side = 3 × 9.4 = 28.2 cm (iii) Each of equal sides = 8.5 cm and third side = 7 cm A C B 8.5 cm 7 cm 8.5 cm Perimeter = Sum of sides = (8.5 + 8.5 + 7) cm = 24 cm Q.10. Find the perimeter of (i) a regular pentagon of side 8 cm, (ii) a regular octagon of side 4.5 cm, (iii) a regular decagon of side 3.6 cm. Sol. (i) Each side of a regular pentagon = 8 cm Perimter = 5 × Side = 5 × 8 = 40 cm (ii) Each side of an octagon = 4.5 cm Perimeter = 8 × Side = 8 × 4.5 = 36 cm (iii) Each side of a regular decagon = 3.6 cm Perimeter = 10 × Side = 10 × 3.6 = 36 cm Q.11. Find the perimeter of each of the following figures : 35cm 45 cm 27 cm 35cm 18 cm 18 cm 18 cm 18 cm 4 cm 16 cm 16 cm 8 cm 4 cm 12 cm 12 cm Page 5 Points to Remember : Region. A plane figure together with its interior is called the region enclosed by the plane figure. Area. The measurement of the region encloed by a plane figure is called the area of the plane figure. The standard unit of area is cm 2 . Source Formulae : 1. Area of rectangle = (length × breadth) Sq. units. 2. Length of a rectangle F H G I K J Area Breadth units 3. Breadth of a rectangle F H G I K J Area Length units 4. Perimeter of a rectangle = 2 × (length + breadth) units. 5. Area of a square = (Side) 2 sq. units. 6. Perimeter of a square = (4 × side) units. Units of Area. The standard unit of area is cm 2 . For measuring areas of lands and fields, we use larger units such as m 2 , areas and hectares etc. Conversion of Units : Units of Length Units of Area 1 cm = 10 mm 1 cm 2 = (10 × 10) mm 2 = 100 mm 2 1 m = 100 cm. 1 m 2 = (100 × 100) cm 2 = 10000 cm 2 1 dm = 10 cm 1 dm 2 = (10 × 10) cm 2 = 100 cm 2 1 dam = 10 m 1 dam 2 = (10 × 10) m 2 = 100 m 2 1 km = 10 hm 1 km 2 = (10 × 10) hm 2 = 100 hm 2 1 km = 1000 m 1 km 2 = (1000 × 1000) m 2 = 1000000 m 2 Q. 1. Find the perimeter of a rectangle in which : (i) length = 16·8 cm and breadth = 6·2 cm (ii) length = 2 m 25 cm and breadth = 1 m 50 cm (iii) length = 8 m 5 dm and breadth = 6 m 8 dm Sol. (i) Length (l) = 16·8 cm Breadth (b) = 6·2 cm Perimeter = 2 (l + b) = 2 (16·8 + 6·2) cm = 2 × 23 = 46 cm Ans. (ii) Length of rectanlge (l) = 2 m, 25 cm = 2.25 m and breadth (b) = 1 m 50 cm = 1.50 m D C A B 2m 25cm 1m 50cm Perimeter = 2 (l + b) = 2 (2.25 + 1.50) m = 2 × 3.75 m = 7.50 m = 7 m 50 cm (iii) Length of a rectangle (l) = 8 m 5 dm = 8.5 m and breadth (b) = 6 m 8 dm = 6.8 m D C A B 8m 5dm 6m 8dm Perimeter = 2 (l + b) = 2 (8.5 + 6.8) m = 2 × 15.3 m = 30.6 m = 30 m 6 dm Q.2. Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs. 16 per metre. Sol. Length of rectangular field (l) = 62 m and breadth (b) = 33 m C D B A 33 m 62 m Perimeter = 2 (l + b) = 2 (62 + 33) = 2 × 95 m = 190 m Rate of fencing = Rs. 16 per m Total cost of fencing the field = Rs. 16 × 190 = Rs. 3040 Q.3. The length and the breadth of a rectangular field are in the ratio 5 : 3. If its perimeter is 128 m, find the dimensions of the field. Sol. Perimeter of field = 128 m Length + Breadth = 2 128 = 64 m Ratio in length and breadth = 5 : 3 Let length (l) = 5x Then breadth = 3x 5x + 3x = 64 8x = 64 x = 8 64 = 8 Length of the field = 5x = 5 × 8 = 40 m and breadth = 3x = 3 × 8 = 24 m Q.4. The cost of fencing a rectangular field at Rs. 18 per metre is Rs. 1980. If the width of the field is 23 m, find its length. Sol. Cost of fencing a rectangular field = Rs. 18 per m Total cost = Rs. 1980 Perimeter of the field = Rate cost Total = 18 1980 = 110 m and length + breadth = 2 110 = 55 m Width of the field = 23 m Length = 55 – 23 = 32 m Q.5. The length and the breadth of a rectangular field are in the ratio 7 : 4. The cost of fencing the field at Rs. 25 per metre is Rs. 3300. Find the dimensions of the field. Sol. Total cost of fencing a rectangular field = Rs. 3300 Rate of fencing = Rs. 25 per m Perimeter = Rate cost Total = 25 3300 = 132 m and length + breadth = 2 Perimeter = 2 132 = 66 m Now ratio in length and breadth = 7 : 4 Let length = 7x Then breadth = 4x 7x + 4x = 66 11x = 66 x = 11 66 = 6 Length of the field = 7x = 7 × 6 = 42 m and breadth = 4x = 4 × 6 = 24 m Q.6. Find the perimeter of a square, each of whose sides measures : (i) 3.8 cm (ii) 4.6 m (iii) 2 m 5 dm Sol. (i) Side of square = 3.8 cm Perimeter = 4 × side = 4 × 3.8 cm = 15.2 cm (ii) Side of a square = 4.6 m Perimeter = 4 × side = 4 × 4.6 m = 18.4 m (iii) Side of a square = 2 m 5 dm = 2.5 m Perimeter = 4 × side = 4 × 2.5 m = 10 m Q.7. The cost of putting a fence around a square field at Rs. 35 per metre is Rs. 4480. Find the length of each side of the field. Sol. Total cost of fencing a square field = Rs. 4480 Rate of fencing = Rs. 35 per m Perimeter of square field = Rate cost Total = 35 4480 = 128 m Side of the square = 4 Perimeter = 4 128 = 32 m Q. 8. Each side of a square field measures 21 m, Adjacent to this field, there is a rectangular field having its sides in the ratio 4 : 3. If the perimeters of both of the fields are equal, find the dimensions of the rectangular field. Sol. Side of a square field (a) = 21 m Perimeter = 4 a = 4 × 21 = 84 m Perimeter of rectangular field = 84 m Ratio in length and breadth = 4 : 3 Let length (l) = 4 x and breadth (b) = 3 x Perimeter = 2 (l + b) 84 = 2 (4 x + 3 x) = 2 × 7 x = 14 x x 84 14 6 Length = 4 x = 4 × 6 = 24 metre and breadth = 3 x = 3 × 6 = 18 metre Ans. Q.9. Find the perimeter of (i) a triangle of sides 7.8 cm, 6.5 cm and 5.9 cm, (ii) an equilateral triangle of side 9.4 cm, (iii) an isosocels triangle with equal sides 8.5 cm each and third side 7 cm. Sol. (i) Sides of a triangle are 7.8 cm, 6.5 cm and 5.9 cm A B C 7.8 cm 6.5 cm 5.9 cm Perimeter = Sum of sides = (7.8 + 6.5 + 5.9) cm = 20.2 cm (ii) Each side of an equilateral triangles = 9.4 cm A B C 9.4 cm 9.4 cm 9.4 cm Perimeter = 3 × side = 3 × 9.4 = 28.2 cm (iii) Each of equal sides = 8.5 cm and third side = 7 cm A C B 8.5 cm 7 cm 8.5 cm Perimeter = Sum of sides = (8.5 + 8.5 + 7) cm = 24 cm Q.10. Find the perimeter of (i) a regular pentagon of side 8 cm, (ii) a regular octagon of side 4.5 cm, (iii) a regular decagon of side 3.6 cm. Sol. (i) Each side of a regular pentagon = 8 cm Perimter = 5 × Side = 5 × 8 = 40 cm (ii) Each side of an octagon = 4.5 cm Perimeter = 8 × Side = 8 × 4.5 = 36 cm (iii) Each side of a regular decagon = 3.6 cm Perimeter = 10 × Side = 10 × 3.6 = 36 cm Q.11. Find the perimeter of each of the following figures : 35cm 45 cm 27 cm 35cm 18 cm 18 cm 18 cm 18 cm 4 cm 16 cm 16 cm 8 cm 4 cm 12 cm 12 cm Sol. We know that perimeter of a closed figure or a polygon = Sum of its sides (i) In the figure, sides of a quadrilateral are 45 cm, 35 cm, 27 cm, 35 cm Its perimeter = Sum of its sides = (45 + 35 + 27 + 35) cm = 142 cm (ii) Sides of a quadrilateral (rhombus) are 18 cm, 18 cm, 18 cm, 18 cm i.e., each side = 18 cm Perimeter = 4 × Side = 4 × 18 = 72 cm (iii) Sides of the polgyon given are 16 cm, 4 cm, 12 cm, 12 cm, 4 cm, 16 cm and 8 cm Its perimeter = Sum of its sides = (16 + 4 + 12 + 12 + 4 + 16 + 8) cm = 72 cm Q. 1. Find the circumference of a circle whose radius is (i) 28 cm (ii) 10·5 cm (iii) 3.5 m Sol. (i) Radius of the circle (r) = 28 cm Circumference = 2 r 2 22 7 28 cm = 176 cm Ans. (ii) Radius of the circle (r) = 10·5 cm Circumference = 2 r 2 22 7 10 5 . cm = 44 × 1·5 = 66 cm Ans. (iii) Radius of a circle (r) = 3.5 m Circumference = 2 r = 2 × 7 22 × 3.5 m = 10 7 35 22 2 = 22 m Ans. Q. 2. Find the circumference of a circle whose diameter is (i) 14 cm (ii) 35 cm (iii) 10.5 m Sol. (i) Diameter of the circle (d) = 14 cm Circumference = d 22 7 14 44 cm Ans. (ii) Diameter of the circle (d) = 35 cm Circumference = d 22 7 35 110cm Ans. (iii) Diameter of a circle (d) = 10.5 m Circumference = d = 7 22 × 10.5 m = 33 m Ans. Q. 3. Find the radius of a circle whose circumference is 176 cm. Sol. Circumference of the circle = 176 cm Let r be the radius, then r Circumference 2 176 7 2 22 4 7 28 cm Radius = 28 cm Ans. Q. 4. Find the diameter of a wheel whose circumference is 75 3 7 cm. Sol. Find the diameter of a wheel whose circumference is 264 cm. Sol. Circumference of a wheel = 264 cm Let d be its diameter, then d = 264 7 22 d = 264Read More

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