NCERT Solutions for class 7 Maths Chapter 11: Perimeter and AreaExercise 11.11. The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find: (i) Its area Answer: 150000 m2 Explanation: Area of the rectangular land = Length × Breadth Length of the rectangular land = 500m Breadth of the rectangular land = 300m Area of the rectangular land = 500m × 300m = 150000 m2 (ii) The cost of the land, if 1 m2 of the land costs ₹ 10,000. Answer: ₹ 1,500,000,000 Explanation: Area of the rectangular land = 500m × 300m = 150000 m2 Cost of the land = Area × Cost of the land per 1m2 Cost of the land = 150000 m2 × ₹ 10,000 Cost of the land = ₹ 1,500,000,000 2. Find the area of a square park whose perimeter is 320 m. Answer: 6400 m2 Explanation: Perimeter is the sum of all the sides of the given closed polygon. Perimeter of a square park = 4 × Side of the square park 320 m = 4 × Side of the square park Side of the square park = 320/4 = 80 m Area of the square park = Side × Side or (Side)2 Area of the square park = 80 m × 80 m Area of the square park = 6400 m2 3. Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter. Answer: 20 m, 84 m Explanation: Length of the rectangular land = 22 m Area of the rectangular land = Length × Breadth 440 m2= 22 m × Breadth Breadth = 440 m2/22m Breadth = 20 m Perimeter of the rectangular land = 2 × (Length + Breadth) Perimeter of the rectangular land = 2 × (22 + 20) m Perimeter of the rectangular land = 2 × 42 m Perimeter of the rectangular land = 84 m 4. The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area. Answer: 15 cm, 525 cm2 Explanation: Perimeter of the rectangular sheet = 2 × (Length + Breadth) 100 cm = 2 × (35 + Breadth) cm (35 + Breadth) = 100/2 (35 + Breadth) = 50 Breadth = 50 - 35 Breadth = 15 cm Area of the rectangular sheet = Length × Breadth Area of the rectangular sheet = 35 cm × 15 cm Area of the rectangular sheet = 525 cm2 5. The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park. Answer: 40 m Explanation: Length of the rectangular park = 90 m Area of the rectangular park = Length × Breadth Area of the rectangular park = 90 m × Breadth Side of the square park = 60 m Area of the square park = Side × Side Area of the square park = 60 m × 60 m = 3600 m2 Area of the square park = Area of the rectangular park 3600 m2 = 90 m × Breadth Breadth of the rectangular park = 3600 m2/ 90 m Breadth of the rectangular park = 40 m 6. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? Also find which shape encloses more area? Answer: 31 cm; Square Explanation: Length = 40 cm Breadth = 22 cm Perimeter of the rectangle = 2 × (Length + Breadth) Perimeter of the rectangle = 2 × (40 + 22) cm Perimeter of the rectangle = 2 × 62 cm Perimeter of the rectangle = 124 cm Note: When a wire is bent in the form of any shape, its perimeter is always considered. Perimeter is the sum of all sides of a given polygon.Perimeter of a square = 4 × Side The same wire when bent in the form of a square will have the same perimeter. 124 cm = 4 × Side Side = 124 cm/4 Side = 31 cm The side of the square is 31 cm. Area of the rectangle = Length × Breadth Area of the rectangle = 40 cm × 22 cm Area of the rectangle = 880 cm2 Area of the square = Side × Side Area of the square = 31 cm × 31 cm = 961 cm2 Area of square > Area of the rectangle Hence, square encloses more area. 7. The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle. Answer: 43 cm; 1050 cm2 Explanation: Perimeter of the rectangle = 130 cm Length =? Breadth = 30 cm Perimeter of the rectangle = 2 × (Length + Breadth) 130 cm = 2 × (Length + 30) cm (Length + 30) cm = 130/2 (Length + 30) cm = 65 cm Length = 65 cm - 30 cm Length = 35 cm Area of the rectangle = Length × Breadth Area of the rectangle = 30 cm × 35 cm Area of the rectangle = 1050 cm2 8. A door of length 2 m and breadth 1m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (Fig11.6). Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m2? Answer: ₹ 284 Explanation: Length of the door = 2 m Breadth of the door = 1 m Area of the door = Length × Breadth Area of the door = 2 m × 1 m Area of the door = 2 m2 Length of the wall = 4.5 m Breadth of the wall = 3.6 m Area of the wall = Length × Breadth Area of the wall = 4.5 m × 3.6 m Area of the wall = 16.2 m2 Area of whitewashing the wall = Area of the wall - Area of the door Area of whitewashing the wall = 16.2 m2 - 2 m2 = 14. 2 m2 Cost of whitewashing the wall = Area of whitewashing the wall × Rate per m2 Cost of whitewashing the wall = 14. 2 m2 × ₹ 20 Cost of whitewashing the wall = ₹ 284 Exercise 11.21. Find the area of each of the following parallelograms: (a) Answer: 28 cm2 Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = 7 cm Height of the parallelogram = 4 cm Area of the parallelogram = 7 cm × 4 cm Area of the parallelogram = 28 cm2 (b) Answer: 15 cm2 Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = 5 cm Height of the parallelogram = 3 cm Area of the parallelogram = 5 cm × 3 cm Area of the parallelogram = 15 cm2 (c) Answer: 8.75 cm2 Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = 2.5 cm Height of the parallelogram = 3.5 cm Area of the parallelogram = 2.5 cm × 3.5 cm Area of the parallelogram = 8.75 cm2 (d) Answer: 24 cm2 Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = 5 cm Height of the parallelogram = 4.8 cm Area of the parallelogram = 5 cm × 4.8 cm Area of the parallelogram = 24 cm2 (e) Answer: 8.8 cm2 Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = 2 cm Height of the parallelogram = 4.4 cm Area of the parallelogram = 2 cm × 4.4 cm Area of the parallelogram = 8.8 cm2 2. Find the area of each of the following triangles: (a) Answer: 6 cm2 Explanation: Area of triangle = ½ × Base × Height Base of the triangle = 4 cm Height of the triangle = 3 cm Area of triangle = ½ × 4 cm × 3 cm Area of triangle = ½ × 12 cm2 Area of triangle = 6 cm2 (b) Answer: 8 cm2 Explanation: Area of triangle = ½ × Base × Height Base of the triangle = 5 cm Height of the triangle = 3.2 cm Area of triangle = ½ × 5 cm × 3.2 cm Area of triangle = ½ × 16 cm2 Area of triangle = 8 cm2 (c) Answer: 6 cm2 Explanation: Area of triangle = ½ × Base × Height Base of the triangle = 3 cm Height of the triangle = 4 cm Area of triangle = ½ × 3 cm × 4 cm Area of triangle = ½ × 12 cm2 Area of triangle = 6 cm2 (d) Answer: 3 cm2 Explanation: Area of triangle = ½ × Base × Height Base of the triangle = 3 cm Height of the triangle = 2 cm Area of triangle = ½ × 3 cm × 2 cm Area of triangle = ½ × 6 cm2 Area of triangle = 3 cm2 3. Find the missing values: (a) Answer:
Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = 20 cm Height of the parallelogram =? Area of the parallelogram = 246 cm2 Area of the parallelogram = Base × Height 246 cm2 = 20 cm × Height Height = 246 cm2/20 cm Height = 12.3 cm (b) Answer:
Explanation: Area of the parallelogram = Base × Height Base of the parallelogram =? Height of the parallelogram = 15 cm Area of the parallelogram = 154.5 cm2 Area of the parallelogram = Base × Height 154.5 cm2 = Base × 15 cm Base = 154.5 cm2/15 cm Height = 10.3 cm (c) Answer:
Explanation: Area of the parallelogram = Base × Height Base of the parallelogram =? Height of the parallelogram = 8.4 cm Area of the parallelogram = 48.72 cm2 Area of the parallelogram = Base × Height 48.72 cm2 = Base × 8.4 cm Base = 48.72 cm2/8.4 cm Height = 5.8 cm (d) Answer:
Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = 15.6 cm Height of the parallelogram =? Area of the parallelogram = 16.38 cm2 Area of the parallelogram = Base × Height 16.38 cm2 = 15.6 cm × Height Height = 16.38 cm2/15.6 cm Height = 1.05 cm 4. Find the missing values: (a) Answer:
Explanation: Area of triangle = ½ × Base × Height Base of the triangle = 15 cm Height of the triangle =? Area of triangle = 87 cm2 87 cm2 = ½ × 15 cm × Height Height = (87 cm2 × 2)/ 15 cm Height = 174 cm2/ 15 cm Height = 11.6 cm (b) Answer:
Explanation: Area of triangle = ½ × Base × Height Base of the triangle =? Height of the triangle = 31.4 mm Area of triangle = 1256 mm2 1256 mm2 = ½ × Base × 31.4 mm Base = (1256 mm2× 2)/ 31.4 mm Base = 2512 mm2/ 31.4 mm Base = 80 mm (c) Answer:
Explanation: Area of triangle = ½ × Base × Height Base of the triangle = 22 cm Height of the triangle =? Area of triangle = 170.5 cm2 170.5 cm2 = ½ × 22 cm × Height Height = (170.5 cm2 × 2)/ 22 cm Height = 341 cm2/ 22 cm Height = 15.5 cm 5. PQRS is a parallelogram (Fig 11.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find: (a) The area of the parallelogram PQRS Answer: 91.2 cm2 Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = SR = 12 cm Height of the parallelogram = QM = 7.6 cm Area of the parallelogram = 12 cm × 7.6 cm Area of the parallelogram = 91.2 cm2 (b) QN, if PS = 8 cm Answer: 11.4 cm Explanation: QM and QN are the perpendiculars or the height of the parallelogram. Area of the parallelogram = Base × Height Base of the parallelogram = PS = 8 cm Height of the parallelogram = QN =? Area of the parallelogram = 8 cm × Height 91.2 cm2 = 8 cm × QN QN = 91.2 cm2 /8 cm QN = 11.4 cm 6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL. Answer: BM = 30 cm, DL = 42 cm Explanation: Area of the parallelogram = Base × Height Base of the parallelogram = AB and AD Let's first find the length of the height DL. Base of the parallelogram = AB = 35 cm Height of the parallelogram = DL =? Area of the parallelogram = 35 cm × Height 1470 cm2 = 35 cm × DL DL = 1470 cm2 /35 cm DL = 42 cm Base of the parallelogram = AD = 49 cm Height of the parallelogram = BM =? Area of the parallelogram = 49 cm × Height 1470 cm2 = 49 cm × BM BM = 1470 cm2 /49 cm BM = 30 cm 7. ∆ABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ∆ABC. Also find the length of AD. Answer: Area of triangle = 30 cm2, Height AD = 60/13 cm Explanation:∆ABC is right angled at A and D. It means that AD and AB are the height of the triangle. Area of triangle = ½ × Base × Height Base of the triangle = AB = 12 cm Height of the triangle = AB = 5 cm Area of triangle = ½ × 12 cm × 5 cm Area of triangle = ½ × 60 cm2 Area of triangle = 30 cm2 Let's find the height AD. Height of the triangle = AD Base of the triangle = BC = 13 cm Area of triangle = 30 cm2 30 cm2 = ½ × 13 cm × Height AD = (30 cm2 × 2)/ 13 cm AD = 60 cm2/ 13 cm AD = 60/13 cm 8. ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE? Answer: Area of triangle = 27 cm2, CE = 7.2 cm Explanation: AD and CE are both the heights of the triangle ABC. Area of triangle = ½ × Base × Height Base of the triangle = BC = 9 cm Height of the triangle = AD = 6 cm Area of triangle = ½ × 9 cm × 6 cm Area of triangle = ½ × 54 cm2 Area of triangle = 27 cm2 Let's find the height CE. Height of the triangle = CE Base of the triangle = AB = 7.5 cm Area of triangle = 27 cm2 27 cm2 = ½ × 7.5 cm × Height CE = (27 cm2 × 2)/ 7.5 cm CE = 54 cm2/ 7.5 cm CE = 7.2 cm Exercise 11.31. Find the circumference of the circles with the following radius: (Take π = 22/7) (a) 14 cm Answer: 88 cm Explanation: Circumference of the circle = 2πr Where, π = 22/7 r = Radius of the circle r = 14 cm Circumference of the circle = 2 × π × r Circumference of the circle = 2 × 22/7 × 14 Circumference of the circle = 88 cm (b) 28 mm Answer: 176 mm Explanation: Circumference of the circle = 2πr Where, π = 22/7 r = Radius of the circle r = 28 mm Circumference of the circle = 2 × π × r Circumference of the circle = 2 × 22/7 × 28 Circumference of the circle = 176 mm (c) 21 cm Answer: 132 cm Explanation: Circumference of the circle = 2πr Where, π = 22/7 r = Radius of the circle r = 21 cm Circumference of the circle = 2 × π × r Circumference of the circle = 2 × 22/7 × 21 Circumference of the circle = 132 cm 2. Find the area of the following circles, given that: (a) Radius = 14 mm (Take π = 22/7) Answer: 616 mm2 Explanation: Area of the circle = πr2 Where, π = 22/7 r = Radius of the circle r = 14 mm Area of the circle = π × r × r Area of the circle = 22/7 × 14 mm × 14 mm Area of the circle = 616 mm2 (b) Diameter = 49 m Answer: 1886.5 m2 Explanation: Area of the circle = πr2 Where, π = 22/7 r = Radius of the circle D = Diameter of the circle D = 2r 49 m = 2r r = 49/2 m Area of the circle = π × r × r Area of the circle = 22/7 × 49/2 m × 49/2 m Area of the circle = (22 × 49 × 49)/ (7 × 2 × 2) Area of the circle = 1886.5 m2 (c) Radius = 5 cm Answer: 550/7 cm2 Explanation: Area of the circle = πr2 Where, π = 22/7 r = Radius of the circle r = 5 cm Area of the circle = π × r × r Area of the circle = 22/7 × 5 cm × 5 cm Area of the circle = 550/7 cm2 3. If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7) Answer: r = 24.5 m, Area = 1886.5 m2 Explanation: Circumference of the circle = 2πr Where, π = 22/7 r = Radius of the circle r =? Circumference of the circle = 2 × π × r 154 m = 2 × 22/7 × r r = (154 m × 7)/ (2 × 22) r = 49/2 m r = 24.5 m Area of the circle = πr2 Area of the circle = π × r × r Area of the circle = 22/7 × 49/2 m × 49/2 m Area of the circle = (22 × 49 × 49)/ (7 × 2 × 2) Area of the circle = 1886.5 m2 4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take π = 22/7) Answer: 132 m, ₹ 528 Explanation: Diameter = 21 m D = 2r 21 = 2r r = 21/2 m Length of the rope to fence the circular garden = Circumference of the garden Circumference of the circle = 2πr Circumference of the circle = 2 × π × r Circumference of the circle = 2 × 22/7 × 21/2 Circumference of the circle = 66 m Length of the rope to fence 1 round = 66 m Length of the rope to fence 2 rounds = 2 × 66 = 132 m Cost of the rope = Cost per meter × Total length of the rope Cost of the rope = ₹ 4 × 132 m Cost of the rope = ₹ 528 5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14) Answer: 21.98 cm2 Explanation: Area of the circle = πr2 Area of the circle = π × r × r Area of the circle with radius 4 cm = 3.14 × 4 cm × 4 cm Area of the circle with radius 4 cm = 50.26 cm2 Area of the circle with radius 3 cm = 3.14 × 3 cm × 3 cm Area of the circle with radius 3 cm = 28.27 cm2 Remaining area of the sheet = Area of the sheet with radius 4 cm - Area of the sheet with radius 3 cm Remaining area of the sheet = 50.26 cm2 - 28.27 cm2 Remaining area of the sheet = 21.98 cm2 6. Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14) Answer: 4.71 m; ₹ 70.65 Explanation: Diameter = 1.5 m d = 2r 1.5 = 2r r = 1.5/2 m Length of the lace to fence the circular table = Circumference of the table Circumference of the circle = π d Circumference of the circle = π × d Circumference of the circle = 3.14 × 1.5 Circumference of the circle = 4.71 m Length of the lace required to cover the edge of the table = 4.71 m Cost of the lace = Cost per meter × Total length of the lace Cost of the lace = ₹ 15 × 4.71 m Cost of the lace = ₹ 70.65 7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter. Answer: 25.70 cm Explanation: Perimeter of the semi-circle = Circumference of the full circle/2 Circumference of the circle = 2πr Circumference of the semi-circle = 2πr/2 = πr Diameter = 10 cm D = 2r 2r = 10 r = 10/2 = 5 cm Radius of the circle = 5 cm Circumference of the semi-circle = π × 5 Circumference of the semi-circle = 3.141 × 5 Circumference of the semi-circle = 15.70 cm Perimeter of the semi-circle including its diameter = Circumference of the semi-circle + Diameter = 15.70 cm + 10 cm = 25.70 cm 8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m2. (Take π = 3.14) Answer: ₹ 30.14 Explanation: Diameter = 1.6 m 2r = 1.6 m r = 1.6/2 r = 0.8 m Area of the circular table-top = πr2 = π × r × r Area of the circular table-top = 3.14 × 0.8 m × 0.8 m Area of the circular table-top = 2.0096 m2 Cost of polishing the circular table-top = Area of the table-top × Cost per meter square Cost of polishing the circular table-top = 2.0096 m2 × ₹ 15/m2 Cost of polishing the circular table-top = = ₹ 30.14 9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π = 22/7) Answer: 7 cm; 154 cm2; 11 cm; circle Explanation: Length of the wire bent into the shape of a circle = Perimeter of the circle Perimeter of the circle = 44 cm 44 cm = 2πr Where, r = radius of the circle π = 22/7 44 cm = 2 × 22/7 × r r = (44 × 7)/ (2 × 22) r = 7 cm Thus, the radius of the circle is 7 cm. Area of the circle = πr2 = π × r × r = 22/7 × 7 × 7 Area of the circle = 154 cm2 Length of the wire bent into the shape of a square = Perimeter of the square Perimeter of the square = 44 cm 44 cm = 4 × (Side of a square) Side of a square = 44/4 cm Side of a square = 11 cm Area of the square = Side × Side Area of the square = 11 cm × 11cm Area of the square = 121 cm2 Area of the circle > Area of the square 154 cm2 > 121 cm2 Thus, circle encloses more area. 10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = 22/7) Answer: 536 cm2 Explanation: Radius of card sheet = 14 cm Area of the card sheet = πr2 = π × r × r = 22/7 × 14 × 14 Area of the circle = 616 cm2 Radius of smaller circle = 3.5 cm Area of the smaller circle = πr2 = π × r × r = 22/7 × 3.5 cm × 3.5 cm Area of the smaller circle = 38.5 cm2 Area of rectangle = Length × breadth Length of the rectangle = 3 cm Breadth of the rectangle = 1 cm Area of the rectangle = 3 cm × 1 cm Area of the rectangle = 3 cm2 Area of the remaining sheet = Area of the card sheet - Area of the first circle - Area of the second circle - Area of the rectangle Area of the remaining sheet = 616 cm2 - 38.5 cm2 - 38.5 cm2 - 3 cm2 Area of the remaining sheet = 536 cm2 11. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14) Answer: 23.44 cm2 Explanation: Side of the square = 6 cm Area of the square = Side × Side Area of the square = 6 cm × 6 cm Area of the square = 36 cm2 Radius of the circle = 2 cm Area of the circle = πr2 = π × r × r Area of the circle = 3.14 × 2 cm × 2 cm Area of the circle = 12.56 cm2 Area of the leftover aluminium sheet = Area of the square - Area of the cut out circle Area of the leftover aluminium sheet = 36 cm2 - 12.56 cm2 Area of the leftover aluminium sheet = 23.44 cm2 12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14) Answer: 5 cm, 78.5 cm2 Explanation: Circumference of the circle = 31.4 cm Circumference of the circle = 2πr 2πr = 31.4 cm 2 × 3.14 × r = 31.4 cm r = (31.4)/ (2 × 3.14) r = 10/2 r = 5 cm Area of the circle = πr2 = π × r × r Area of the circle = 3.14 × 5 cm × 5 cm Area of the circle = 78.5 cm2 13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14) Answer:879.20 m2 Explanation: Diameter of the flower bed = 66 m Radius of the flower bed = 66/2 = 33 m The width of the surrounded path = 4 m Area of the path = Area of the flower bed with path width - Area of the flower bed without the path width Diameter of the flower bed with path width = 4 + 66 + 4 = 74 m Radius of the flower bed with path width = 74/2 = 37 m Area of the circle = πr2 = π × r × r Area of the path = Area of the flower bed with path width - Area of the flower bed without the path width Area of the path = 3.14 × 37 × 37 - 3.14 × 33 × 33 Area of the path = 3.14 × (37 × 37 - 33 × 33) Area of the path = 3.14 × (1369 - 1089) Area of the path = 3.14 × 280 Area of the path = 879.20 m2 14. A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14) Answer: Yes Explanation: Area of the circular flower garden = 314 m2 Radius of the circle that a sprinkler can cover = 12 m Area of the circle = πr2 = π × r × r Area of the circle = 3.14 × 12 × 12 Area of the circle = 452.16 m2 Area of the circle that a sprinkler can cover is greater than the area of the circular flower garden. Hence, the sprinkler can easily water the entire garden. The answer is 'Yes'. 15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14) Answer: 119.32 m, 56.52 m Explanation: Circumference of the circle = 2πr Radius of the outer circle = 19 m Width of the outer circle = 10 m Radius of the inner circle = Radius of the outer circle - Width of the outer circle Radius of the inner circle = 19 m - 10 m = 9 m Circumference of the inner circle = 2× 3.14 × 9 Circumference of the inner circle = 56.52 m Circumference of the outer circle = 2× 3.14 × 19 Circumference of the outer circle = 119.32 m 16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7) Answer: 200 times Explanation: 1 m = 100 cm 28 cm = 28/100 = 0.28 m Radius of the wheel = 0.28 m Circumference of the wheel = 2πr Circumference of the wheel = 2× 22/7 × 0.28 Circumference of the wheel = 1.76 m The number of times a wheel can rotate = 352 m/ Circumference of the wheel The number of times a wheel can rotate = 352 m/1.76 m The number of times a wheel can rotate = 200 times 17. The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14) Answer: 94.2 cm Explanation: Radius of the minute hand of a clock = 15 cm In 1 hour, the minute hand covers the full 12 clock hours, i.e., the full circle. The length of the minute hand moved in hour = Circumference of the clock Circumference of the clock = 2πr Circumference of the clock = 2× 3.14 × 15 Circumference of the clock = 94.2 cm Thus, the length of the minute hand moved in hour is 94.2 cm. Exercise 11.41. A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare. Answer: 1750 m2; 0.675 hectare Explanation: Length of the garden = 90 m Breadth of the garden = 75 m Area of the rectangular garden = Length × Breadth Area of the rectangular garden = 90 m × 75 m Area of the rectangular garden = 6750 m2 Path to be built outside the garden = 5 m Length of the garden with the width = 90 + 5 + 5 = 100 m Breadth of the garden with the width = 75 + 5 + 5 = 85 m Area of the rectangular garden = Length × Breadth Area of the rectangular garden = 100 m × 85 m Area of the rectangular garden = 8500 m2 Area of the path = Area of the rectangular garden with outside width - Area of the inside rectangular garden Area of the path = 8500 m2 - 6750 m2 Area of the path = 1750 m2 Area of the garden = 6750 m2 1 square meter = 0.0001 hectare = 6750 m2 = 0.0001 × 6750 = 0.675 hectare 2. A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path? Answer: 1176 m2 Explanation: Length of the park =125 m Breadth of the park = 65 m Area of the rectangular park = Length × Breadth Area of the rectangular park = 125 m × 65 m Area of the rectangular park = 8125 m2 Path to be built outside the park = 3 m Length of the park with the width = 125 + 3 + 3 = 131 m Breadth of the park with the width = 65 + 3 + 3 = 71 m Area of the rectangular park = Length × Breadth Area of the rectangular park = 131 m × 71 m Area of the rectangular park = 9301 m2 Area of the path = Area of the rectangular park with outside width - Area of the inside rectangular park Area of the path = 9301 m2 - 8125 m2 Area of the path = 1176 m2 3. A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin. Answer: 30 cm2 Explanation: Length of the cardboard = 8 cm Width of the cardboard = 5 cm Area of the cardboard = Length × Width Area of the cardboard = 8 cm × 5 cm Area of the cardboard = 40 cm2 Margin along the sides of the cardboard = 1.5 cm Length of the painting = 8 cm - 1.5 cm - 1.5 cm = 5 cm Width of the painting = 5 cm - 1.5 cm - 1.5 cm = 2 cm Area of the painting = Length × Width Area = 5 cm × 2 cm Area = 10 cm2 Area of the margin = Area of the cardboard- Area of the painting Area of the margin = 40 cm2 - 10 cm2 Area of the margin = 30 cm2 4. A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find: (i) The area of the verandah. Answer: 63 m2 Explanation: Length of the room = 5.5 m Width of the room = 4m Area of the room = Length × Width Area of the room = 5.5 m × 4 m Area of the room = 22 m2 Width of the verandah = 2.25 m Length of the room with verandah = 5.5 m + 2.25 m + 2.25 m = 10 m Width of the room with verandah = 4 m + 2.25 m + 2.25 m = 8.5 m Area of the room with verandah = Length × Width = 10 m × 8.5 m = 85 m2 Area of the verandah = Area of the room with verandah - Area of the room Area of the verandah = 85 m2 - 22 m2 Area of the verandah = 63 m2 (ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2 . Answer: ₹ 12600 Explanation: Cost of cementing the floor of the verandah = Area of the verandah × Rate per meter square Cost of cementing the floor of the verandah = 63 m2 × ₹ 200 Cost of cementing the floor of the verandah = ₹ 12600 Note: When the width is added outside any area, it is added twice to the length and the breadth of the given area. Similarly, if the width lies inside any area, it is subtracted twice from the length and the breadth.5. A path 1 m wide is built along the border and inside a square garden of side 30 m. Find: (i) The area of the path Answer: 116 m2 Explanation: Side of the garden = 30 m Area of the garden = Side × Side Area of the garden = 30 m × 30 m Area of the garden = 900 m2 Width of the path built inside the square garden = 1 m Side of the square built inside the garden = 30 m - 1 m - 1 m = 28 m Area of the inside garden = Side × Side Area of the inside garden = 28 m × 28 m Area of the inside garden = 784 m2 Area of the path = Area of the garden - Area of the inside garden Area of the path = 900 m2 - 784 m2 Area of the path = 116 m2 (ii) The cost of planting grass in the remaining portion of the garden at the rate of ₹ 40 per m2 Answer: ₹ 31360 Explanation: Cost of planting trees in the remaining portion of the garden = Area of the inside × Rate per meter square Cost of planting trees in the remaining portion of the garden = 784 m2 × ₹ 40 Cost of planting trees in the remaining portion of the garden = ₹ 31360 6. Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares. Answer: 0.99 hectare, 20.01 hectares Explanation: Length of the rectangular park = 700 m Breadth of the rectangular park = 300 m The width of the cross road = 10 m Area of the rectangular park = Length × Width Area of the rectangular park = 700 m × 300 m Area of the rectangular park = 210000 m2 Length of one side of the cross road = 300 m Width = 10 m Area = Length × Width Area = 300 m × 10 m Area = 3000 m2 Length of the other side of the cross road = 10 m Width = 700 m Area = Length × Width Area = 10 m × 700 m Area = 7000 m2 Area of the centre region (common area of the intersection of two crossroads) = 10 m × 10 m Area = 100 m2 Total area of the two cross roads = 3000 m2 + 7000 m2 - 100 m2 Total area of the two cross roads = 9900 m2 1 square meter = 0.0001 hectare = 9900 m2 = 0.0001 × 9900 Total area of the two cross roads = 0.99 hectare Area of the park excluding the cross road = Total area of the rectangular park - Area of the cross roads Area of the park excluding the cross road = 210000 m2 - 9900 m2 Area of the park excluding the cross road = 200100 m2 1 square meter = 0.0001 hectare = 200100 m2 = 0.0001 × 200100 Area of the park excluding the cross road = 20.01 hectares 7. Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find (i) The area covered by the roads. Answer: 441 m2 Explanation: Length of the rectangular field = 90 m Breadth of the rectangular field = 60 m The width of the cross road = 3 m Area of the rectangular park = Length × Width Area of the rectangular park = 90 m × 60 m Area of the rectangular park = 5400 m2 Length of one side of the cross road = 60 m Width = 3 m Area = Length × Width Area = 60 m × 3 m Area = 180 m2 Length of the other side of the cross road = 3 m Width = 90 m Area = Length × Width Area = 3 m × 90 m Area = 270 m2 Area of the centre region (common area of the intersection of two crossroads) = 3 m × 3 m Area = 9 m2 Total area of the two cross roads = 180 m2 + 270 m2 - 9 m2 Total area of the two cross roads = 441 m2 (ii) The cost of constructing the roads at the rate of ₹ 110 per m2. Answer: ₹ 48,510 Explanation: Cost of constructing the roads = Area of the cross roads × Rate per meter square Cost of constructing the roads = 441 m2 × ₹ 110 Cost of constructing the roads = ₹ 48,510 8. Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π = 3.14) Answer: Yes, 9. 12 cm cord length is left Explanation: Radius of the circle = 4 cm Side of the square = 4 cm The length of the cord required to cover a circular pipe = Perimeter of the circle The length of the cord required to cover a square = Perimeter of the square Perimeter of the circular pipe = 2 π r = Circumference of the circle Perimeter of the circular pipe = 2 × 3.14 × 4 Perimeter of the circular pipe = 25.12 cm So, length of the cord required to cover a circular pipe is 25.12 cm. Perimeter of the square = 4 × Side Perimeter of the square = 4 × 4 Perimeter of the square = 16 cm Perimeter of the square < Perimeter of the circle Chord left = Length of the cord required to cover the circle - Length of the cord required to cover the square Chord left = 25.12 cm - 16 cm Chord left = 9. 12 cm 9. The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: (i) The area of the whole land Answer: 62.56 m2 Explanation: Area of the whole land = Area of the rectangular lawn Length of the rectangle = 10 m Width of the rectangle = 5 m Area of the rectangular lawn = Length × Width Area of the rectangular lawn = 10 m × 5 m Area of the rectangular lawn = 50 m2 (ii) The area of the flower bed Answer: 12.56 m2 Explanation: Radius of the circular flower bed = 2 m Area of the circular flower bed = π r2 Area of the circular flower bed = 3.14 × 2 × 2 Area of the circular flower bed = 12.56 m2 (iii) The area of the lawn excluding the area of the flower bed Answer: 37.44 m2 Explanation: Area of the lawn excluding the area of the flower bed = Area of the rectangular lawn - Area of the circular flower bed Area of the rectangular lawn = 50 m2 Area of the circular flower bed = 12.56 m2 Area of the lawn excluding the area of the flower bed = 50 m2- 12.56 m2 Area of the whole land = 37.44 m2 (iv) The circumference of the flower bed. Answer: 12.56 m Explanation: Radius of the flower bed = 2 m Circumference of the flower bed = 2 π r Circumference of the flower bed = 2 × 3.14 × 2 Circumference of the flower bed =12.56 m 10. In the following figures, find the area of the shaded portions: (i) Answer: 110 cm2 Explanation: Area of the shaded portion = Area of the rectangle ABCD - Area of the triangle AEF - Area of the triangle BEC Length of the rectangle = 18 cm Width of the rectangle = 10 cm Area of the rectangle = Length × Width Area of the rectangle = 18 cm × 10 cm Area of the rectangle ABCD = 180 cm2 Area of the triangle AEF = ½ × Base × Height Base of the triangle AEF = 10 cm Height of the triangle AEF = 6 cm Area of the triangle AEF = ½ × 10 cm × 6 cm Area of the triangle AEF = 30 cm2 Area of the triangle BEC = ½ × Base × Height Base of the triangle BEC = 8 cm Height of the triangle BEC =10 cm Area of the triangle BEC = ½ × 8 cm × 10 cm Area of the triangle BEC = 40 cm2 Area of the shaded portion = Area of the rectangle ABCD - Area of the triangle AEF - Area of the triangle BEC Area of the shaded portion = 180 cm2 - 30 cm2 - 40 cm2 Area of the shaded portion = 110 cm2 (ii) Answer: 150 cm2 Explanation: Area of the shaded portion = Area of the rectangle PQRS - Area of the triangle PQT - Area of the triangle TUS - Area of the triangle QUR Length of the rectangle = 20 cm Width of the rectangle = 20 cm Area of the rectangle = Length × Width Area of the rectangle = 20 cm × 20 cm Area of the rectangle PQRS = 400 cm2 Area of the triangle PQT = ½ × Base × Height Base of the triangle PQT = 20 cm Height of the triangle PQT =10 cm Area of the triangle PQT = ½ × 20 cm × 10 cm Area of the triangle PQT = 100 cm2 Base of the triangle TUS = 10 cm Height of the triangle TUS =10 cm Area of the triangle TUS = ½ × 10 cm × 10 cm Area of the triangle TUS = 50 cm2 Base of the triangle QUR = 10 cm Height of the triangle QUR =20 cm Area of the triangle QUR = ½ × 10 cm × 20 cm Area of the triangle QUR = 100 cm2 Area of the shaded portion = Area of the rectangle PQRS - Area of the triangle PQT - Area of the triangle TUS - Area of the triangle QUR Area of the shaded portion = 400 cm2 - 100 cm2 - 50 cm2 - 100 cm2 Area of the shaded portion = 150 cm2 11. Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC Answer: 66 cm2 Explanation: Area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ACD Area of the triangle ABC = ½ × Base × Height of the perpendicular BM Base of the triangle ABC = 22 cm Height of the triangle ABC =3 cm Area of the triangle ABC = ½ × 22 cm × 3 cm Area of the triangle ABC = 33 cm2 Area of the triangle ACD = ½ × Base × Height of the perpendicular DN Base of the triangle ACD = 22 cm Height of the triangle ACD =3 cm Area of the triangle ACD = ½ × 22 cm × 3 cm Area of the triangle ACD = 33 cm2 Area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ACD Area of the quadrilateral ABCD = 33 cm2 + 33 cm2 Area of the quadrilateral ABCD = 66 cm2 |