Grade 7 Mensuration - Perimeter, Area, Volume

Transcription

1 ID : gb-7-mensuration-perimeter-area-volume [1] Grade 7 Mensuration - Perimeter, Area, Volume For more such worksheets visit Answer t he quest ions (1) A square and an equilateral triangle have the same perimeter. The diagonal of the square is 15 cm. What is the area of the triangle? (2) If the sides of a square are increased by 35%, then by what percent does the area of the square increase by? (3) A f armer wants to tie his goat with rope so that it can graze the grass in an area of 27 m 2, what should be the length of the rope? (assume π = 3 f or this question). (4) Find the radius of a circle if its area is m 2 (assume π = 22/7). Choose correct answer(s) f rom given choice (5) The cost of f encing a rectangular ground at 8.5 per meter is 918. If width of f ield is 15, what is the length of the ground? a. 34 m b. 39 m c. 49 m d. 36 m (6) Willow has a small plot of land that she wants to pave. If the size of the plot is 70.6 m by 29.6 metres, and she wants to pave it with square tiles, what will be the measure of the largest tile she can use? a m. b. 0.2 m. c m. d m. (7) if area of a circle is 9πr 2, f ind circumf erence of the circle. a. 6πr b. π r c. 9 π r d. None of these (8) Heidi changes a picture on her computer such that width of the image is 50% of original image and height of the image is 50% of original image. If area of new image is cm 2, f ind the area of original image. a. 120 cm 2 b. 124 cm 2 c. 118 cm 2 d. 121 cm 2

2 ID : gb-7-mensuration-perimeter-area-volume [2] Fill in the blanks (9) Area of shaded region = cm 2 (All measurements are in cm) (10) If radius of smaller circles are 3, 1, 4, and 3 meters, the area of shaded region = meters 2. (For this question assume π = 3). (11) Radius of two circle wheel shown are 15 cm and 30 cm. Wheels are connected as gear so they rotate with each other. If f irst wheel makes 900 revolutions, second wheel will make revolutions. (12) A f armer wants to tie his cow with rope so that it can graze the grass in an area of 75 m 2. The length of the rope should be meters. (assume π = 3 f or this question).

3 ID : gb-7-mensuration-perimeter-area-volume [3] (13) Teacher asks students to f ix f lags around the periphery of a circular ground of radius 63 meters. If f lags are f ixed at a distance of 3 meters apart, f lags would be required to cover periphery of the ground. (assume π = 22/7) (14) The lawn is 50 m long and 26 m wide. If a gardner charges to maintain a sq. m. of the lawn, then the cost to maintain the whole lawn is. (15) A wire f rame is in rectangular shape which is 106 cm long and 74 cm broad. If wire is reshaped in the semi-circular shape as shown below. T he radius of the semi-circle will be cm. (assume π = 22/7) Edugain ( All Rights Reserved Many more such worksheets can be generated at

4 Answers ID : gb-7-mensuration-perimeter-area-volume [4] (1) 50 3 Lets assume s and t are the sides of square and triangle respectively as shown in the f ollowing f igures. Square Equilateral triangle According to question the diagonal AC of the square is 15 cm. Now in right angle triangle ABC AB 2 + BC 2 = AC 2 s 2 + s 2 = (15) 2 2(s) 2 = (15) 2 s 2 = (15)2 2 s 2 = cm Perimeter of the given square = 4s Perimeter of the given equilateral triangle = 3t According to question the perimeter of a square is equal to the perimeter of an equilateral triangle. Theref ore 3t = 4s Squaring both sides (3t) 2 = (4s) 2 9t 2 = 16s 2 t = [Since s 2 = 112.5] 9 t 2 = 200 cm Now the area of an equilateral traingle = t2 3 4 = [Since t 2 = 200 ] 4 = 50 3 cm 2

5 Step 4 ID : gb-7-mensuration-perimeter-area-volume [5] Theref ore the area of the triangle is 50 3 cm 2. (2) 82.25% Lets assume s and A are the side of the square and area of the square respectively. According to question the sides of the square are increased by 35% 35 theref ore the side of the square = s + s s + 35s = 100 = 1.35s Now the area of the square = (1.35s) 2 = s 2 = A [Since A = s 2 ] The area of the square increase = A - A = A Step 4 Theref ore you can say that the area of the square increase = = 82.25% 100 %

6 (3) 3 meters ID : gb-7-mensuration-perimeter-area-volume [6] Lets assume the length of the rope is r meters by which the f armer wants to tie his goat. Since goat is tie with rope, the goat can graze the grass in a circular area, which radius is equals to the length of the rope. According to question goat can graze the grass in an area of 27 m 2. or πr 2 = 27 [Since area of a circle is = πr 2 ] r 2 = 27 π r 2 = 27 3 r 2 = 9 r = 3 meters Theref ore the length of the rope should be 3 meters.

7 (4) 32.9 m ID : gb-7-mensuration-perimeter-area-volume [7] Lets assume r and d are the radius and diameter of a circle, as shown in the f ollowing f igure. Circle The area of a circle = πr 2 According to question the area of a circle is m 2. Theref ore πr 2 = r 2 = π r 2 = 22 7 r 2 = r 2 = r 2 = (32.9) 2 r = 32.9 Theref ore the radius of a circle = 32.9 m.

8 (5) b. 39 m ID : gb-7-mensuration-perimeter-area-volume [8] If you look at the question caref ully, you will notice that the cost of f encing a rectangular ground at 8.5 per meter is 918 and the perimeter of a rectangle is the total distance around the outside of a rectangle, theref ore the perimeter of a rectangular ground = 918 = 108 meter 8.5 The width of f ield = 15 meter The perimeter of a rectangular ground = 2(length of the ground + breadth of the ground) The perimeter of a rectangular groung or length of the ground = - width of the ground 2 Theref ore the length of the ground = = = 39 meter (6) b. 0.2 m. According to question the length and width of the plot are 70.6 m and 29.6 m respectively. Since 1 m = 100 cm. Theref ore the length of the plot = = 7060 cm Width of the plot = = 2960 cm The measure of the largest tile she can use is equal to the HCF(Highest Common Factor) of length and width of the plot. HCF of length and width of the plot i.e 7060 and 2960 is 20. Step 4 Now you can say that the measure of the largest tile is = 20 cm = 0.2 m.

9 (7) a. 6πr ID : gb-7-mensuration-perimeter-area-volume [9] Lets assume R is the radius of a circle. The Area of a circle = πr (1) According to question the area of the circle = 9πr 2 = π(3r) (2) From equation (1) and (2) you will notice that radius of the circle(r) = 3r Now circumf erence of the circle = 2πR = 2π(3r) = 6πr Theref ore circumf erence of the circle is 6πr. (8) d. 121 cm 2 Let us assume that x and y be the width and height of the original image respectively. Area of image = width height = xy Since width of the image is 50% of original image and height of the image is 50% of original image. Theref ore the width of new image = x = 5x 10 height of new image = y = 5y 10 Area of new image = width height = 5x Step y 10 = 25xy 100 Since area of new image is cm 2 Theref ore this area is equal to 25xy xy = xy = 3025 xy = xy = 121 = Step 5 Theref ore, the area of the original image = xy = 121 cm 2

10 (9) 1440 ID : gb-7-mensuration-perimeter-area-volume [10] The area of the shaded region = The area of the rectangle - The area of the unshaded right angled triangles Area of the rectangle = = 2080 cm 2 Area of a right angled triangle = 1 2 (Base Height) Theref ore, the area of the unshaded right angled triangles = 1 2 (24 32) (16 32) = ( ) cm 2 Step 4 Hence, the area of shaded region is: ( ) = 1440 cm 2

11 (10) 258 ID : gb-7-mensuration-perimeter-area-volume [11] If you look at the f igure caref ully, you will notice that the radius of smaller circles are 3, 1, 4, and 3 meters. Area of a circle = πr 2 Now area of the smaller circles = π(3) 2 + π(1) 2 + π(4) 2 + π(3) 2 = π( ) = 3( ) = 3(35) = 105 meters 2 Radius of the larger circle is equal to the sum of radius of the smaller circles. Area of the larger circel = πr 2 = π(11 2 ) = 3(121) = 363 meters 2 Area of the shaded region = area of the larger circle - area of the smaller circles = = 258 meters 2 Step 4 Area of the shaded region is 258 meters 2.

12 (11) 450 ID : gb-7-mensuration-perimeter-area-volume [12] The distance covered by a wheel in a revolution is equal to the perimeter of the wheel. According to the question, the radius of the f irst and the second wheel is 15 cm and 30 cm respectively. The perimeter of the f irst wheel = 2π 15 = 30π cm. The distance covered by the f irst wheel in one revolution = 30π cm. The distance covered by the f irst wheel in 900 revolutions = 900π 30 = 27000π cm. The perimeter of the second wheel = 2π 30 = 60π cm. or we can say that, the number of revolutions covered by the second wheel in 60π cm = 1 revolution. 1 The number of revolutions covered by the second wheel in 1 cm = revolutions. 60π The number of revolutions covered by second wheel in 27000π cm = 450 revolutions. Step 4 Thus, the number of revolutions f or second wheel = 450 revolutions. 1 60π 27000π = (12) 5 Lets assume the length of the rope is r meters by which the f armer wants to tie his cow. Since cow is tie with rope, the cow can graze the grass in a circular area, which radius is equals to the length of the rope. According to question cow can graze the grass in an area of 75 m 2. or πr 2 = 75 [Since area of a circle is = πr 2 ] r 2 = 75 π r 2 = 75 3 r 2 = 25 r = 5 meters Theref ore the length of the rope should be 5 meters.

13 (13) 132 ID : gb-7-mensuration-perimeter-area-volume [13] According to question the students have to f ix the f lags around the periphery of a circular ground of radius 63 meters. Periphery of the circular ground = 2πr = 2π63 = = 396 meters Since the f lags are f ixed at a distance of 3 meters apart. The number of f lags required to cover periphery of the ground = Periphery of the ground Distance between the f lags = = 132 f lags Step 4 Theref ore the number of f lags required to cover periphery of the ground are 132 flags. (14) 97240

14 (15) 70 ID : gb-7-mensuration-perimeter-area-volume [14] According to question the length and width of the rectangular wire f rame are 106 cm and 74 cm respectively. The perimeter of the rectangular wire f rame = 2(Length + Width) = 2( ) = 360 cm The perimeter of the semi-circle = πd 2 + d [Where d is the diameter of the semi-circle.] = 22 7 d + d 2 = 22d 14 + d = 22d + 14d 14 = 36d 14 = 18d 7 Since the wire is reshaped in the semi-circular shape, the perimeter of the rectangle is equal to the perimeter of the semi-circle. Theref ore 18d = or d = = 140 cm Now the radius of the circle is = = 70 cm.