If SR = 12 cm, PS = 8 cm and QM = 7.6 cm, find: (ii) the area of the unshaded region. Find the area of the shaded region. These solutions cover all topics included in the NCERT syllabus, prescribed by the CBSE board. There are a bigger semicircle and two small semicircles (b) The area of the finished quilt is about 3.4m2. Speed = 7.5 km/hr. Radius of each circle in it =  cm right-angled triangles of equal areas. Width of road made out side of it = 3.5 m Now are of ∆ =  × Base × Height AB = 3 cm , AC = 4 cm (v) If the perimeter of a parallelogram is 40 cm and the length of one side is 12 cm, then the length of the adjacent side is …… Find the area of the shaded region. (Take π = 3.14), 13. Both cross road have the same width = 10 m, Area of the ABCD cross road = length × breadth, Area of the EFGH cross road = length × breadth, Area of the IJKL at center = length × breadth, Area of the roads = Area of ABCD + Area of EFGH – Area of IJKL, Hence, area of roads in hectare = 9900/10000, Finally, Area of the park excluding roads = Area of park – Area of the roads, 7. (a) 2 : r All measurements are in centimetres. The diameter of a circular park is 84 m. On its outside, there a 3.5 m wide road. (True) Question 5. 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. (a) 1 unit (ii) Now Area of AEFD = l × b = 15 × 24 = 360 sq. The new length and breadth of the room when verandah is included is 10 m and 8.5 m respectively. Perimeter of rectangle = 2 (Length + Breadth). Take π =, In the given figure, ABCD is a square whose side is 14 cm. Perimeter = 2(l + b) = 2(18.7 + 14.3) cm = 2 × 33 = 66 cm If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ΔABC. If the area of a parallelogram is 54 cm2 and the length of one side is 7.5 cm, then the corresponding height is 6. Learn from the best math teachers and top your exams. Area of square = (Side)2 = (14)2 = 196 cm2 A circle of the maximum area is cut out from is (i) 7 cm Question 1. Area =  × b × h =  × (17.52 – 142) = 400 m2. Circumference = 2πr = 2 × 3.14 × 20 cm = 125.60 = 125.6 cm (Take π = 3.14), A circle of radius to be removed r = 3 cm, The area of the remaining sheet = πR2 – πr2. ... Ex.11.1 Q8 Perimeter and Area Solutions. and area of square = (Side)2 = (31)2 cm2 = 961 cm2 (ii) Area of ∆ABC = 27 cm2 = 131 × 71 – 125 × 65 = 9301 – 8125 = 1176 m2. Very popular with students, these worksheets (with solutions) are great for topic-based revision. = 130 – 50 From square cardboard, the biggest circle is cut out (π = 3.14), Perimeter of the square = 4 × side of the square, So, the length of cord left with Pragya = Perimeter of circular pipe – Perimeter of square. (c) 25 : 9 (b) 3 : 2 Perimeter (Length) of wire = 2(l + b) = 2(40 + 22) cm = 2 × 62 cm = 124 cm (b) 14.4 cm Your email address will not be published. The inner part of the carpet is blue in colour. Length of altitude =  =  = 8 cm Find: (ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m2. 8. 28 cm is = 28 × π = 28 ×  = 88 cm (b), Question 8. Width of a verandah = 2.25 m 4. =  Circumference = 2πr (d) 36 m2, Perimeter of a square = 24 cm 9. 2πr : πr2 = 2 : r (a), Question 9. Circumference of a circle = 44 cm Circumference = 2πr = 2 ×  × 28 mm = 176 mm If you have any doubts, please comment below. and height (h) = 90 cm =  =  m EC = 18 – 10 = 8 cm Distance travelled in 5 rounds = 600 × 5 = 3000 m = 3 km CBSE Class VII Maths Solutions, Mathematics Class 7 Perimeter And Area Chapter 11 Ex 11.2 NCERT Solutions How many times will the wheel of a car rotate in a journey of 88 km, given that the diameter of the wheel is 56 cm? Download pdf of Class 7 Chapter 11 in their respective links. Calculate the length of the boundary and the area of the shaded region in the following diagrams. (d) 4 units, Area of a circle is numerically equal to its = 196 – 154 = 42 cm2 Find the area of the shaded region. Length of blue carpet (l) = 5 m Base AB = 7.5 cm. Now the area of shaded portion (ii) The unit of measurement of the area is a square unit. (i) base = 2 m, height = 1.5 m Also find the cost of the rope, if it costs ₹ 4 per metre. Find the area of the circles, given that: Find the area of the remaining sheet. Also, find which shape encloses more area and by how much? Perimeter and Area - Solution for Class 7th mathematics, NCERT solutions for Class 7th Maths. (Take π = 3.14). Area and Perimeter of Circle: Area = Πr²; Circumference of the circle = 2Πr; Where r is the radius of the circle Π = 3.14 or 22/7. Question 11. of the table-top. (d) 44 cm, The perimeter of a semicircle If the breadth of the rectangle is. Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. 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? and radius of each of smaller semicircles =  = 7 cm BC = 5 cm (ii) Base of the triangle (b) = 3.4 m Area enclosed by two concentric circles = 770 cm2 (b) 29 cm Radius (R) =  = 14 cm and radius =  = 7 cm ΔABC is right angled at A (Fig 11.25). Question 12. and breadth (b) = 17 m Diameter of a circular garden = 21 m cm. The area of margin = Area of the cardboard when margin is including – Area of the. (c) 4 : 9 Area of 4 circles = 4 × πr2 A pattern is made from two similar trapeziums. (d) 9 : 4, Let radii be 2x and 3x Find the perimeter and the area of the shaded region in the following figures. Circle a statement. by inserting a dent in your shape the area gets reduced by the dent. (Take π = 3.14), Area of the circular flower garden = 314 m2, Sprinkler at the centre of the garden can cover an area that has a radius = 12 m. ∴Radius of the circular flower garden is 10 m. Since, the sprinkler can cover an area of radius 12 m. Hence, the sprinkler will water the whole garden. Perimeter of the a rectangular sheet = 100 cm, Perimeter of the rectangle = 2 (Length + Breadth), ∴Area of the rectangular sheet is 525 cm2. (i) the length of the boundary Side of square shaped wire = 27.5 cm (False) So the area of these triangles will be half of the area of a rectangle. (v) The area of a circle of diameter d is πd2. Side of equilateral triangular shaped wire = 44 cm Then diameter of each circle =  = 14 cm Total cost = ₹ 60 × 346.5 = ₹ 20790, Question 11. = 24 cm2 Area of right-angled triangle = 54 cm2 Length = – Breadth So, the radius of circle is 24.5 and area of circle is 1886.5. KSEEB Solutions for Class 5 Maths Chapter 9 Perimeter and Area October 20, 2020 January 8, 2020 by Prasanna Students can Download Maths Chapter 9 Perimeter and Area Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Perimeter of the rectangle = 2(Length + Breadth). In an isosceles ∆ABC = 180 – (40+ 30) If the ratio of the radii of two circles is 3 : 5, then the ratio of their areas is The perimeter of an isosceles triangle is 25 cm The length of each side, in cm, is a prime number. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Cost of 1 m carpet = ₹ 250 ∴Perimeter of the rectangular plot is 84 m. 4. (c) 1 : r Breadth = 7 cm π(3x)2 : π(5x)2 = 97πx2 : 25πx2 Rate of white washing = ₹ 20 per m2 and breadth (b) = 30 m (a) 3 : 5 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. Due to the fact that the area of a shape is calculated by multiplying a shape’s length by its width, it is measured in ‘square units‘ .For example, the area of a square which is 1 metre on each side is 1 metre x 1 metre = 1 square metre or m2. Side of the square ABCD = 24 cm Question 2. (d), Question 6. 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