ID : ae-7-mensuration-perimeter-area-volume [1] Grade 7 Mensuration - Perimeter, Area, Volume For more such worksheets visit www.edugain.com Answer the questions (1) A teacher gave a rectangular colouring sheet to each of his students. The sheet is 19 cm long and 14 cm wide. What is the area and perimeter of the sheet? (2) If a rectangle of area 1802 cm 2 is 34 cm broad, find its length. (3) A farmer wants to tie his buffalo with a rope so that it can graze the grass in an area of 108 m 2. What should be the length of the rope? (assume π = 3 for this question) (4) The length of a rectangle is four times its width. If the length of the diagonal is 18 17 cm, then find the perimeter of the rectangle. (5) Faisal has Dhs 3434080 with which he wants to buy some land. If he wants to buy a piece of land that is at least 32 meters wide and the cost of land is Dhs 1651 per square meter, then what will be the length of the longest piece of land that he can buy? (6) A rectangular playground is 130 metres long and 100 metres wide. Inside the playground, there is a 8 meters wide path that runs along the boundary of the playground. What is the total area of the path? (7) If the circle given below has an area 5544, then what will be perimeter of a semi-circle formed, if we cut the circle at AB (assume π = 22/7) (8) Find the area of the given figure. (All measurements are in meters).
(9) The following shape is made of several small cubes. Find the surface area of the shape, if each small cube is 1 cm 1 cm 1 cm. ID : ae-7-mensuration-perimeter-area-volume [2] (10) Find the area of the given figure (Area of each square on the graph paper is 1 cm 2 ). (11) Habib has a small plot of land that he wants to pave. If the size of the plot is 104.7 m by 61.5 m and he wants to pave it with square tiles, what will be the measure of the largest tile he can use? (12) Find the circumference of a circle of radius 2.1 cm (Assume π = 22/7). (13) Find the area of the given figure (all measurements are in meters). (14) A wire of length 16 m is to be folded in the form of a square. If each side of a square has to be an integer (measured in meters), what is the maximum number of squares that can be formed by folding the wire. Choose correct answer(s) from the given choices (15) The perimeter of a triangle is 75 inches. If side A is twice the length of side B, which is half the length of side C, what is the length of side B? a. 15 b. 18 c. 12 d. None of these
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Answers ID : ae-7-mensuration-perimeter-area-volume [4] (1) Area: 266 cm 2 Perimeter: 66 cm Perimeter means the sum of all sides. A rectangle has four sides, where two sides are of 19cm length, and the other two are 14cm long. So, here the perimeter will be the sum of all these lengths. When we add all the sides we get: 19 cm + 19 cm + 14 cm + 14 cm = 66 cm, which is the perimeter of the rectangle. Area of a rectangle is the product of length and the breadth, which is 19 cm 14 cm = 266 cm 2. Hence, Area of the rectangle = 266 cm 2 Perimeter of the rectangle = 66 cm (2) 53 cm We know that the area of a rectangle is equal to the product of its length and breadth. Here, we know the following: Area of the rectangle = 1802 cm 2 Breadth of the rectangle = 34 cm The length of the rectangle = Area Breadth = 1802 34 = 53 cm cm
(3) 6 meters ID : ae-7-mensuration-perimeter-area-volume [5] Let us assume the length of the rope as r meters by which the farmer wants to tie his buffalo. Since, buffalo is tied with the rope, the buffalo can graze the grass in a circular area, whose radius is equal to the length of the rope. According to the question, the buffalo can graze the grass in an area of 108 m 2. Or, πr 2 = 108 [Since, the area of the circle is πr 2 ] r 2 = 108 π r 2 = 108 3 r 2 = 36 r = 6 meters Therefore, the length of the rope should be 6 meters.
(4) 180 cm ID : ae-7-mensuration-perimeter-area-volume [6] Let us assume that ABCD is a rectangle. According to question, the length of the diagonal of the rectangle is 18 17 cm and the length of the rectangle is four times its width. Let us assume the width of rectangle to be w cm. Therefore, the length of the rectangle = 4w cm. On looking at the rectangle ABCD carefully, we notice that ABC is a right angled triangle where AB, BC, and AC are the width, length, and diagonal of the rectangle, respectively. In the right angled triangle ABC, AC 2 = AB 2 + BC 2 (18 17) 2 = w 2 + (4w) 2 (18) 2 ( 17) 2 = w 2 + 16w 2 324 17 = (1 + 16)w 2 5508 = 17w 2 17w 2 = 5508 w 2 = 5508 17 w 2 = 324 w 2 = (18) 2 w = 18 Therefore, the width of the rectangle is 18 cm and the length of the rectangle = 4w cm = 4 18 = 72 cm Step 5 Now, the perimeter of the rectangle = 2(length of the rectangle + width of the rectangle) = 2(72 + 18) = 2(90) = 180 cm.
(5) 65 m ID : ae-7-mensuration-perimeter-area-volume [7] If we read the question carefully, we notice that Faisal has Dhs 3434080 and in Dhs 1651 he can buy 1 square meter land. Therefore, the piece of land he can buy in Dhs 3434080 = 1 3434080 = 2080 square 1651 meter Hence, the area of the piece of land = 2080 square meter The width of the piece of land is 32 meters. Area of the piece of land = length width 2080 = length 32 2080 = length 32 65 = length length = 65 meters Hence, the width of the land is 32 meters.
ID : ae-7-mensuration-perimeter-area-volume [8] (6) 3424 m 2 Rectangular Playground Let us assume that ABCD is the rectangular playground and PQRS is the area inside the path. According to the question, the length of the rectangular playground ABCD = 130 metres. Width of the rectangular playground ABCD = 100 metres. Area of the rectangular playground ABCD = 130 100 = 13000 m 2 Since, the width of the path is 8 metres, we need to subtract 8 metres from both the length and width of the rectangular playground ABCD. Length of the rectangle PQRS = 130-16 = 114 metres. Width of the rectangle PQRS = 100-16 = 84 metres. Area of the rectangle PQRS = 114 84 = 9576 m 2 Total area of the path = Area of the rectangular playground ABCD - Area of the rectangle PQRS = 13000-9576 = 3424 m 2
(7) 216 ID : ae-7-mensuration-perimeter-area-volume [9] According to the question, the area of the circle is 5544. Let us assume that the radius of the circle is r. Now, the area of the circle = πr 2 πr 2 = 5544 22r2 7 = 5544 r 2 = 5544 7 22 r 2 = 1764 r = 42 Now, the perimeter of the given semi-circle = πr + 2r = 22 7 42 + 2 42 = 216 Therefore, the perimeter of the given semi-circle is 216. (8) 153 m 2 We need to find the area of the figure given in the question :
ID : ae-7-mensuration-perimeter-area-volume [10] Let us first divide this figure into two rectangles. Hence, the area of the whole figure will be the sum of the areas of these two rectangles. Rectangle 1 Rectangle 2 Let us now find the area of rectangles 1 and 2: Area of Rectangle 1 = Length Width = 4 22 = 88 m 2 Area of Rectangle 2 = Length Width = 13 5 = 65 m 2 Let us now add the areas of rectangle 1 and 2 to get the area of the original figure: Area of the given figure = Area of Rectangle 1 + Area of Rectangle 2 = 88 m 2 + 65 m 2 = 153 m 2
ID : ae-7-mensuration-perimeter-area-volume [11] (9) 42 cm 2 The given shape is a cuboid which is made of several small cubes. Since all edges of small cubes are of 1cm, we can easily find the width, height and length of bigger cuboid. Height of cuboid = 1 3 = 3 cm. Similarly, width and length will be 3 cm and 2 cm. Step 5 Surface area of cuboid, A = 2 (width length + length height + width height) A = 2 (3 2 + 2 3 + 3 3) A = 2 21 A = 42 cm 2 (10) 10 cm 2 If we count the squares and half-squares in the given figure, we will notice that the total number of squares are 10(count two half-squares as one small square). Since the area of each small square is 1 cm 2, the area of the given figure is 10 cm 2.
(11) 0.3 m ID : ae-7-mensuration-perimeter-area-volume [12] According to the question, the length and width of the plot are 104.7 m and 61.5 m, respectively. Since, 1 m = 100 cm Therefore, Length of the plot = 104.7 100 = 10470 cm Width of the plot = 61.5 100 = 6150 cm The measure of the largest tile he can use is equal to the H.C.F(Highest Common Factor) of the length and width of the plot. H.C.F of the length and width of the plot, i.e., 10470 and 6150 is 30. Hence, the measure of the largest tile = 30 cm = 0.3 m (12) 13.2 cm If we look at the question carefully, we notice that we have to find the circumference of the circle. Given: radius of the circle = 13.2 cm The circumference of a circle is the distance around the outside or the boundary. We know: C = 2πr (where, r is the radius of the circle and π = 22 7 ) = 2 22 7 2.1 = 13.2 cm Therefore, the circumference of the circle is 13.2 cm. (13) 252 m 2 We need to find the area of the given figure:
ID : ae-7-mensuration-perimeter-area-volume [13] First, let us divide the figure into three rectangles. The area of the figure in question will be the sum of the are of these three rectangles. Rectangle 1 Rectangle 2 Rectangle 3 Let us now find the area of each rectangle: Area of Rectangle 1 = Length Width = 5 m 19 m = 95 m 2 Area of Rectangle 2 = Length Width = 18 m 4 m = 72 m 2 Area of Rectangle 3 = Length Width = 5 m 17 m = 85 m 2
ID : ae-7-mensuration-perimeter-area-volume [14] Area of the given figure = Area of Rectangle 1 + Area of Rectangle 2 + Area of Rectangle 3 = 95 m 2 + 72 m 2 + 85 m 2 = 252 m 2 (14) 4 We have to maximize the numbers of squares that can be created with a given length of wire. Therefore, we need to keep the size of squares as small as possible. It is given that each side of the square has to be an integer (i.e. it cannot be 0.5m, 0.8m etc.), therefore the least possible length of the side can be 1 m. Perimeter of a square with each side of 1 m = 4 1 = 4 m. Since, 4 m wire is needed for 1 square. Therefore, number of squares that can be created with 16 m = 16 4 = 4 Step 5 We can create maximum 4 squares by folding the given wire.
(15) a. 15 ID : ae-7-mensuration-perimeter-area-volume [15] Let us assume that the length of side C is x. Since side B is half of side C, the length of side B = x 2 and side A is twice the length of side B, i.e., the length of side A = 2 x 2 = x Now, the perimeter of a triangle = length of side A + length of side B + length of side C = x + x 2 + x Since the perimeter of the triangle is 75 inches. Therefore, x + x 2 + x = 75 2x + x 2 = 75 2 2x + x 2 = 75 4x + x = 75 2 By cross multiplying both the sides, 5x = 2 75 5x = 150 x = 150 5 x = 30 Therefore, the length of side B = x 2 = 30 2 = 15.